Quadratics

To sum up: Recycled quadratic II: (a-b)2 = a2-2ab+b2 Whichever of the above forms you encounter in a GMAT problem, changing the recycled quadratic into the other form will usually set you on the right path to solving the problem quickly. 1) Factored form:(a-b)2 Identification: look for (a-b)2 Solution: change the form: turn into quadratic form: square a and b, and subtract twice their product. Check using FOIL.

2) Expanded form: a2 - 2ab + b2 Identification - recognize the pattern: If the first and third terms are squares, figure out what they're squares of. Multiply those things, multiply that product by 2 and compare your result with the quadratic's middle term. If you've got a match, then you've got a recycled quadratic. If the middle term appears in a negative "-" sign, then it is specifically recycled quadratic II: a2-2ab+b2 = (a-b)2 Solution: change the form: turn it into square form - subtract the roots of the third term from that of the first term, and square the difference: (a-b)²

If −1<x<0, then which of the following has the greatest value? x²+x / x+1 x³ 1+2x+x² / x+1 1-x² / 1+x 1+ (x-1)(x+1)

Correct. This question is best solved by Plugging In. To make life easier and Plugging In quicker, simplify the answer choices using the quadratics formulas. Then plug in an easy number between zero and −1, such as −1/2, or −1/10. Plug the same number into all (simplified) answer choices to see which is of greatest value. For example, if you choose to plug in x=−1/10, then x²+x / x+1 = x(x+1) / (x+1) = x = -0.1 x³ = - 1/1000 1+2x+x² / x+1 = (x+1)² / (x+1) = x+1 = 0.9 1-x² / 1+x = (1-x)(1+x) / 1+x = 1-x = 1.1 1+ (x-1)(x+1) = 1 + x² -1 = x² = 1/100

Back in the happy days of high school, you may have encountered three special quadratics. We call them recycled quadratics, because the GMAC recycles - these quadratics appear in many algebra questions. Identifying them is often the key to a quick solution of a GMAT problem. Memorize these recycled quadratics: I (a+b)2 = a2+2ab+b2 II (a-b)2 = a2-2ab+b2 III (a+b)(a-b) = a2 - b2 We'll discuss how to identify them and what to do when you do in the near future.

The recycled quadratics are : The two perfect square trinominals: (a+b)2 = a2+2ab+b2 (a-b)2 = a2-2ab+b2 These two are quite similar in structure, and only differ by the + or - sign inside the squared term (a+b)2 or (a-b)2, which is mirrored by the same sign for the middle component on the right side of the equation. The third recycled quadratic is known as the difference of squares: (a+b)(a-b) = a2 - b2

If x=3 is a root of the equation x2=15−k·x, what is the value of k? 5 2 −2 −3 −5

Correct. Remember the meaning of factoring solutions: The solution x1 is the value that satisfies the equation ax2+bx+c=0. Plugging it into the equation will give a result of zero. Change the quadratic equation into the form of ax2+bx+c=0: x2=15−k·x --> x2+k·x-15=0 Plug x=3 into the quadratic equation x2+k·x-15=0 to form an easy equation with k: 32+3k-15=0 Solve for k: --> 3k=15-9=6 --> k=2

(m−n)(n−m)=? m2+n2−2m·n m2−n2−2m·n n2−m2−2m·n m2−n2 2m·n−m2−n2

Correct. Use FOIL - First, Out, In, Last - in order to move from factored to quadratic form. --> (m−n)(n−m) = --> mn-m2-n2+mn = --> 2mn-m2-n2

A used cars salesman receives an annual bonus if he meets a certain quota. This year, the salesman has so far sold 1/5 of last year's quota. If this year's quota is reduced by 25 percent, the quantity that he still needs to sell this year in order to receive the annual bonus is what fraction of the preceding year's car quota? 4/15 4/20 3/4 11/20 9/20

You slightly overestimated the time this question took you. You actually solved it in 1 minutes and 43 seconds. Correct. This question would be much easier to calculate if you knew the value of the quota. However, since the question is only interested in the fraction of the quota needed to receive the annual bonus, the actual quota is irrelevant. This is indeed an Invisible Plug-In question. Plug in an easy number for the quota. The question presents a fraction (1/5) and a percentage (25%) which could be translated into a fraction - 1/4. Thus, a good number will be divisible by both 5 and 4, such as 20. Assume that the quota is 20 cars and work the problem accordingly. So far this year, the salesman has sold 1/5×20 = 4 cars. Now, he gets the good news that this year's quota is reduced by 25% = 1/4. 1/4 × 20 = 5 cars, so he only needs to sell a total of 20-5=15 cars this year to qualify for the bonus. Thus, he still needs 15-4=11 cars. 11 cars out of the preceding year's car quota of 20 is 11/20. Thus, D is the right answer.

To sum up: Recycled quadratic I: (a+b)2 = a2+2ab+b2 Whichever of the above forms you encounter in a GMAT problem, changing the recycled quadratic into the other form will usually set you on the right path to solving the problem quickly. 1) Factored form:(a+b)2 Identification: look for(a+b)2 Solution: change the form: turn into quadratic form: square a and b, and add twice their product. Check using FOIL.

2) Expanded form: a2 + 2ab + b2 Identification - recognize the pattern: If the first and third terms are squares, figure out what they're squares of. Multiply those things, multiply that product by 2 and compare your result with the quadratic's middle term. If you've got a match, then you've got a recycled quadratic. If the middle term appears in a positive "+" sign, then it is specifically recycled quadratic I: a2+2ab+b2 = (a+b)2 Solution: change the form: turn it into square form - add the roots of the first and third term, and square the sum: (a+b)2

I (a+b)2 = a2+2ab+b2 This is one of the recycled quadratics - the square of (a+b), or (a+b)2 Whichever of the above forms you encounter in a GMAT problem, changing the quadratic into the other form will usually set you on the right path to solving the problem quickly. The trick lies in identifying the recycled quadratic in both its forms: 1) Factored form: (a+b)2 a. Identification: This is easy to identify: look for the square of the sum of two terms. For example: (x+5)2 b. First step to solution: change the form: turn it into expanded form - square a and b, and add twice their product: (x+5)2 = x2 + 2·x·5 + 52 = x2 + 10x + 25 You can always check that the new form is identical to the original by writing (x+5)2 as (x+5)(x+5) and expanding the parentheses using FOIL.

2) Expanded form: a2 + 2ab + b2 a. Identification - recognize the pattern: How do we recognize that a quadratic expression of the form of ax2 + bx + c is actually a perfect square recycled quadratic? The trick is really quite simple: If the first and third terms are squares, figure out what they're squares of. Multiply those things, multiply that product by 2 and compare your result with the quadratic's middle term. If you've got a match, then you've got a recycled quadratic. If the middle term appears in a addition "+" sign, then it is specifically recycled quadratic I: a2+2ab+b2 = (a+b)2 Example: is x2 + 6x + 9 a recycled quadratic? Well, the first term x2 is the square of x, and the third term 9 is the square of 3. multiply to get 3x. Multiply the product by 2 to get 6x. This matches the middle term, so we have a recycled quadratic on our hands. b. First step to solution: Change the form: turn it into factored form - add the roots of the first and third term, and square the sum. In the above example, the root of x2 is x, and the root of 9 is 3. So: x2 + 6x + 9 = (x+3)2

II (a-b)2 = a2-2ab+b2 This is one of the recycled quadratics - the square of (a-b), or (a-b)2. Whichever of the above forms you encounter in a GMAT problem, changing the quadratic into the other form will usually set you on the right path to solving the problem quickly. The trick lies in identifying the recycled quadratic in both its forms: 1) Factored form: (a-b)2 a. Identification: This is easy to identify: look for the square of the difference of two terms. For example: (x-5)2 b. First step to solution: change the form: Turn it into expanded form - square a and b, and subtract twice their product: (x-5)2 = x2 - 2·x·5 + 52 = x2 - 10x + 25 You can always check that the new form is identical to the original by writing (x-5)2 as (x-5)(x-5) and expanding the parentheses using FOIL.

2) Expanded form: a2 - 2ab + b2 a. Identification - recognize the pattern: How do we recognize that a quadratic expression of the form of ax2 + bx + c is actually a perfect square recycled quadratic? The trick is really quite simple: If the first and third terms are squares, figure out what they're squares of. Multiply those things, multiply that product by 2 and compare your result with the quadratic's middle term. If you've got a match, then you've got a recycled quadratic. If the middle term appears in a negative "-" sign, then it is specifically recycled quadratic II: a²-2ab+b² = (a-b)² Example: is x² - 6x + 9 a recycled quadratic? Well, the first term x² is the square of x, and the third term 9 is the square of 3. multiply to get 3x. Multiply the product by 2 to get 6x. This matches the middle term, so we have a recycled quadratic on our hands. Since the middle term appears with a negative sign (-6x), this is recycled quad II (x-3)². b. First step to solution: change the form: turn it into factored form - subtract the root of third term from the root of the first term, and square the difference. In the above example, the root of x² is x, and the root of 9 is 3. So: x² - 6x + 9 = (x-3)²

III (a+b)(a-b) = a2-b2 This is one of the recycled quadratics - also known as the difference of squares. It is GMAC's favorite, and appears in many questions in either of its forms. Whichever of the above forms you encounter in a GMAT problem, changing the quadratic into the other form will usually set you on the right path to solving the problem quickly. The trick lies in identifying the recycled quadratic in both its forms: 1) Factored form: (a+b)(a-b) a. Identification: Two sets of parentheses, one with the sum of two terms, the other with the difference of the same two terms. Example: (x+5)(x-5) b. First step to solution: change the form: turn it into expanded form - square a and b, and subtract the square of b from the square of a. Check that it works using FOIL: (x+5)(x-5) = x·x + x·-(5) + 5x + 5·(-5) = x2-52 = x2-25

2) Expanded form: a2 - b2 a. Identification - recognize the pattern: Look for the difference of even powers (square in particular) Example: is x2 - 1 a recycled quadratic? Well, the first term x2 is the square of x, and the second term 1 is the square of 1. Therefore, we have a recycled quadratic III on our hands. b. First step to solution: Change the form: turn it into factored form - find the roots of both terms, then put them in two parentheses - one with the sum of the roots, the other with their difference. In the above example, the root of x2 is x, and the root of 1 is 1. So: x2 - 1 = (x+1)(x-1)

The solutions formula is an all-purpose method - it will always tell you what the solutions of an equation are (provided it has solutions). However, it is also a long and cumbersome method to use, and many GMAT quadratics problems can be solved without it. Furthermore, it is a "brute force" technique, and does not help when dealing with questions that test for the subtle understanding of the reasoning behind quadratics. which is why we'll take a look at the second method of finding the solutions of a quadratic equation: Factoring the equation. Example: Find the solutions for x2+3x-10=0. 1) Write as (x+e)(x+f)=0. 2) Find e and f: we're looking for two integers whose product is c=-10. Possible pairs: (-2,5) (2,-5), (1,-10), (-1, 10). From these possible pairs, select the pair whose sum reaches b=3: (-2,5) is the only one that fits, so that pair is e and f. 3) Write e=-2 and f=5 in their appropriate places in the factorized form: (x+(-2))(x+5)=0 --> (x-2)(x+5)=0 4) Set each individual factor to zero and solve: If x-2=0, then x1=2 If x+5=0, then x2=-5 And there you have it. x1=2 and x2=-5 are the two solutions of the above equation.

2) Factoring quadratics: Factoring a quadratic equation is a matter of transforming the quadratic from the form of ax2+bx+c=0 into the form of (x+e)(x+f)=0. This is the factorized form of the quadratic, and each of the pair of parentheses is called a factor of the equation. The Factoring process goes as follows: 1) Your first order of business is finding e and f. If a=1 (which it almost always will in GMAT problems) think of two integers e and f which satisfy the following conditions: 1) Their product is equal to c. 2) Their sum is equal to b. 2) Once you've found e and f, write them in the appropriate places in the factorized form. 3) The equation is now expressed as the product of two factors. Since the product is zero, at least one of the factors must equal zero. Set each factor individually to zero and solve for x. The resulting values of x are the solutions of the original quadratic. To sum up: Factoring a quadratic of the form ax2+bx+c=0: 1) Write in factorized form (x+e)(x+f)=0. Each of the pairs of parentheses is called a factor of the equation. 2) Find e and f: a pair of integers whose product is c and whose sum is b. 3) Write e and f in the appropriate places in the factors. e.g. (x+5)(x+3)=0 4) Set each factor to zero and solve.

Which of the following equations has a root in common with x2−6x+8=0? x2+x−2=0 x2+4=0 x2−12x+16=0 32−2x2=0 x2−4x+16=0

32−2x2=0 Correct. Factor the equation x2−6x+8=0: The factored form is (x+e)(x+f)=x2+(e+f)x+e·f Therefore e+f=−6, and e·f=8. From e+f=−6 it is clear that either e or f are negative, and for e·f=8to be true, both e and f should have the same sign. Hence x2−6x+8=(x−2)(x−4). Next, set each factor to zero and solve: From (x−2)=0 you find one root, x=2, and from (x−4)=0 you find the other root, x=4. Now, check the answer choices. You DON'T have to solve each one of them! In order to check if either x=2 or x=4 is a root of an equation, plug each root in the equation to check if it fits. For example, plug in x=2 in A, x2+x−2=0. It reads 22+2−2, which does not equal zero. We then try the other root, x = 4. This gives us 42+4-2 = 18, which is also not equal to zero. Since neither root works, we can eliminate A. Go on plugging in either x=2 or x=4 until the equation is satisfied. This happens only for 32−2x2=0 with x=4.

(√5+3)(√5-3) = √5-9 √15-9 -4 4 9

Correct. First, identify the expression as recycled quadratic III - Two sets of parentheses, one with the sum of two terms, the other with the difference of the same two terms. Next, change the form: (√5+3)(√5-3) = (√5)2 - 32 = (5½)2 - 9 = 5½·2 - 9 = 5-9 = -4

If (x+2) is a factor of the equation x2−2x+p=0, what is the value of p? 8 4 −2 −4 −8

Correct. If (x+2) is a factor of the quadratic, then setting it to zero (x+2=0) and solving for x will yield the solution for the quadratic. Therefore, x=-2 is a solution or root of x2−2x+p=0. Remember the meaning of factoring solutions: The solution is the value that satisfies the equation ax2+bx+c=0. Plugging it into the equation will give a result of zero. Plug x=-2 into the quadratic equation x2−2x+p=0 to form an easy equation with p: (-2)2-2(-2)+p=0 Solve for p: --> 4+4+p=0 --> p=-8

44²-88*34+34² / 51²-49² = 0.5 1 1.5 2 2.5

Correct. Looks like a tough calculation. Before you start working out the value of 442, see if you can simplify the fraction by looking for common patterns. Start by breaking down 88 into 2×44. The numerator of the fraction then looks as follows: 44²-2*44*34+34² = The first and third terms of the numerator are squares of 44 and 34, respectively. The middle term is the product of 44 and 34 times 2. What you have here is a recycled quadratic. Since the the middle term appears in a negative "-" sign, then it is specifically recycled quadratic II: a2-2ab+b2 = (a-b)2 Change the form into square form: 44²-2*44*34+34² = (44-34)² = 10² = 100 Now, the denominator of the fraction is of the form of Recycled quadratic III: (a+b)(a-b) = a2-b2 Change the form into factored form: 51²-49² = (51+49)(51-49) = 100*2=200 At the end of this process, the fraction assumes the much more manageable form of: 100/200 = 0.5

If n=(6−4p)2, then n is lowest when p= −1.5 −2 0 1.5 2

Correct. Numbers in the answer choices? Specific question? Use reverse plugging in - plug in the answers! --> n=(6−4p)2=(6-4·1.5)2=0 Note: n cannot be negative, since it is a square of another number. If you remember this fact, no need to plug in all the other answer choices - p=1.5 is the only answer choice that results in n=0.

If a+b=3, and a2+b2=6+a·b, what is the value of a·b? −3 −2 −1 1 6

Correct. Okay. You COULD take the long road and use the two equations to find the values of a and b, then multiply the two together to get a·b. However, finding a and b is an unnecessary and time consuming step for this question. Instead, note that a2, b2 and a·b are components of the expanded form for recycled quad I. First square the first equation: --> a+b=3 --> (a+b)2=9 --> a2+2ab+b2=9 The second equation is: --> a2+b2=6+a·b Now plug in the second equation in the first - substitute 6+a·b for a2+b2 --> 6+ab+2ab=9 --> 3ab=3 --> ab=1

In the equation x2+n·x+12=0, x is a variable, and n is a constant. If x=2 is a solution for the equation x2+n·x+12=0, then which of the following is a factor of x2+n·x+12=0? (x+10) (x+8) (x+6) (x−6) (x−10) Alternative method: Remember: the solution for a quadratic equation of the form ax2+bx+c=0 is the value of x for which the quadratic equals zero. If x=2 is a solution for the quadratic, then plugging in x=2 into the equation will enable you to find n, then factor the equation to find the other factor.

Correct. Recall the method for factoring a quadratic of the form ax2+bx+c=0: 1) Write in factorized form (x+e)(x+f)=0. Each of the pairs of parentheses is called a factor of the equation. 2) Find e and f: a pair of integers whose product is c and whose sum is b. 3) Write e and f in the appropriate places in the factors. e.g. (x+5)(x+3)=0 4) Set each factor to zero and solve. Since x=2 is a solution for the quadratic, then (x-2) is a factor of the quadratic and e=-2. Find the corresponding f which fits the terms set in step 2 and proceed accordingly. For the quadratic x2+n·x+12=0, a=1; b=n; c=12; ?Ask a tutor The product of e and f must be equal to c=12. If e=-2 is one of the values of the factors, then (-2)·f=12, or f=-6. Plug in e and f into the expanded form of the quadratic: (x-2)(x-6)=0. Therefore, x-6 is the other factor of the equation.

If m=√(m/6), which of the following could be the value of m? 1/6 6 36 63 64

Correct. Root in the question? Square both sides to get rid of the root sign, then solve. Square both sides of m=√(m/6) i.e. m2= (√(m/6))2 Focus on the right side of the equation: Powers and roots of the same order cancel each other. Delete the power of 2 and the square root √, and (√(m/6))2 simply becomes m/6. --> m2= (m/6) Now divide both sides of the equation by m to get --> m= (1/6) The other solution for the above quadratic equation (m=0) is not one of the answer choices. Hence, this is the correct answer.

(2+√7)=? 3/5 √3/(√7+2) 3/(√7+2) √3/(√7−2) 3/(√7 − 2)

Correct. Start with the answers. A, B, and C can be POEd rather easily - they are all smaller than 1, while the expression in the question is greater than 1. Between D and E, Try to get rid of √7 by expanding them using Recycled quadratic III: (a+b)(a-b) = a2-b2. Expand the fraction: multiply the numerator and denominator by (√7 + 2) to get recycled quad III in the denominator and be rid of the √7: --> 3/(√7 − 2) = [3(√7 + 2)] / [(√7 -2)(√7 + 2)] = --> [3(√7 + 2)]/(√72 -22) = --> [3(√7 + 2)]/(7-4) = --> [3(√7 + 2)]/3 = --> 2+√7

If y=(x2−4x+4) / (x−2), and x≠2, then, in terms of x, y2 is which of the following? Another way to work this question is to plug in a good number for x. Look at all these factors of 2 in the stem and answer choices. This is a sign not to plug in 1, 2 (which is forbidden anyway) or 4. Go for something completely different, yet simple. How about x=5? y = (52−4·5+4)/(5−2) = (25−20+4)/(3) = 9 / 3 = 3 y2=9 Now, quickly eliminate any answer choice that will not equal 9 when x is replaced by 5. Only C fits.

Correct. The issue here is recycled quadratics. The immediate suspect should be (x2−4x+4). There are no fractions in the answer choices, so the fraction in the definition of y ought to be reduced. Use the recycled quadratics formulas to simplify y and find y2. Alternatively, if all these variables make you dizzy, consider replacing every x with a specific number. We call that plugging in... y = (x2−4x+4) / (x−2) = --> (x−2)2 / (x−2) = --> (x−2) Now don't forget you are after y2, which would be (x−2)².

Which of the following equations does NOT have a root in common with x2+5x+6=0? x2−x−6=0 18−2x2=0 x2−x−2=0 x2+1.5x−1=0 x2+2x−3=0

Correct. The issue of this question is factoring a quadratic equation. Factor x2+5x+6=0. Move to expanded form (x+e)(x+f). You're looking for two numbers e and f, whose product is 6 and whose sum is 5. It is pretty obvious that (2,3) fit these requirements. Plug these two numbers into the factored form: (x+e)(x+f) = (x+2)(x+3) = 0. Set each factor to zero to find the roots: (x+2)=0 --> x = −2 (x+3)=0 --> x = −3 There is no need to repeat this process for each of the answer choices - once you've factored the quadratic in the question stem, plug the solutions into the quadratics in the answer choices. Eliminate answer choices for which one of the roots satisfies the equation. This equation is not true for either of the solutions x=−2,−3: (−2)2−(−2)−2 = 4+2−2 ≠ 0 (−3)2−(−3)−2 = 9+3−2 ≠ 0 Thus, x2−x−2=0 has no common root with x2+5x+6=0.

Is x=√8? (1) x2−4√2·x+8=0 (2) x2+2√2·x−16=0

Correct. This is a DS Yes\No question. Answering a definite "Yes" (x equals only √8) or a definite "No" (x never equals √8) means Sufficient. If the answer is sometimes "Yes" and sometimes "No" (i.e there are a few possible solutions, one of which is x=√8), it means Maybe, which means Insufficient. This means that the real issue in the question isn't factoring the quadratics, but rather finding the number of solutions for the quadratics. If they have a single solution, then that fact alone is sufficient - regardless of whether that single solution is indeed x=√8, you will be able to give a definite answer to the question "Is x=√8?". Since the quadratics in the statements have all three coefficients - a, b and c, plug in those values into b2-4ac to see how many solutions the equations have. Remember the following about b2-4ac (the discriminant): If the discriminant is positive, (b2-4ac > 0), then x1 and x2 will have different values - 2 different solutions. If the discriminant is equal to zero (b2-4ac = 0), then x1=x2 =-b / 2a, then the quadratic has only one solution. Stat.(1): Plug in a=1, b=-4√2, and c=8. b2-4ac=32-32=0. The equation has only one solution, so the statement is Sufficient->AD. If you want to be sure, proceed: x=√8, so Stat.(1)->Yes->S->AD. Stat.(2): Plug in a=1, b=2√2, and c=-16 . b2-4ac=8+64=72. So the quadratic has two solutions. Plug a, b and c into the solutions formula to find that one of the solutions is x=√8 and that the other is not, so the answer is Maybe, i.e., according to Stat.(2) one cannot determine whether x=√8. So, Stat.(2)->Maybe->IS->A.

If a is an integer and |a|≠1, which of the following must be an integer? (a4−2a2+1)/(a2+2a+1) (a2+2a+1)/(a−1) (a+1)/(a2+2a+1) (a2+2a+1)/(a2−2a+1) (a4−1)/(a2−2a+1)

Correct. Use recycled quad I in the numerator and recycled quad II in the denominator. The numerator is a bit tougher to recognize here. Remember - If the first and third terms are squares, figure out what they're squares of. Multiply those things, multiply that product by 2 and compare your result with the quadratic's middle term. If you've got a match, then you've got a recycled quadratic. If the middle term appears in a negative "-" sign, then it is specifically recycled quadratic II: a2-2ab+b2 = (a-b)2 --> (a4−2a2+1)/(a2+2a+1) = Note that in the numerator a4 is the square of a2 and 1 is the square of 1: --> = (a2-1)2/(a+1)2 = Now use recycled quad III in the numerator: --> ((a-1)(a+1))2/(a+1)2 = In the numerator you now have a product of two powers with the same exponent. Split the product and reduce: --> (a-1)2×(a+1)2/(a+1)2 = (a-1)2 This expression has to be an integer, since a is an integer.

64−44=? 20 52 104 520 1040

Correct. Use recycled quadratic III: (a+b)(a-b) = a2-b2 --> 64−44= (62)2 - (42)2 --> (62−42)×(62+42) = --> (36−16)×(36+16) = --> 20×52 = 1040

In the equation x2+b·x+c=0, b and c are constants, and x is a variable. If m is a root of x2+b·x+c=0, and (x+n) is a factor of x2+b·x+c=0, then b= m·n m−n m+n n−m m

Correct. Use what the question gives you: if x=m is a root of x2+b·x+c=0, then (x-m) is a factor of the quadratic equation. Now write the equation in Factored Form. x2+b·x+c=(x+n)(x-m) Expand the parentheses using FOIL: (x+n)(x-m) = x2 - mx + nx - nm Collect like terms: x2 + (n-m)x - nm = x2+b·x+c Therefore, b = (n-m)

What is the value of a2+b2? (1) ab=77 (2) (a-b)2=16

Correct. The question stem itself should make you think of the issue of recycled quadratics - a2 + b2 are the main pieces of the puzzle of recycled quad I. The third piece of the puzzle, ab, is given in stat. (1); If you missed all this, stat (2) is a dead giveaway, as it contains the recycled quad in factor form. Stat. (1): at this point, all you have is a single equation with two unknowns - ab. No way to find the individual values of a and b, or to reach a recycled quad. Therefore, stat (1)->IS->BCE. Stat. (2): change the form to expanded form: (a-b)2 = a2 - 2ab + b2 = 16. This would've been sufficient, but you need ab to leave only the required a2 + b2. Alone, stat. (2)->IS->CE. Stat. (1) + (2) combined: Now you have all the pieces of the puzzle: a2 - 2ab + b2 = 16. Plug in ab=77 (stat. (1) into the expanded form to get: a2 - 2·77 + b2 = 16 Therefore, a2+ b2 = 16 + 2·77 = 170. Stat. (1)+(2)->S->C.

What is the value of a2+b2? (1) ab=30 (2) (a+b)2=121

Correct. The question stem itself should make you think of the issue of recycled quadratics - a2 + b2 are the main pieces of the puzzle of recycled quad I. The third piece of the puzzle, ab, is given in stat. (1); If you missed all this, stat (2) is a dead giveaway, as it contains the recycled quad in factor form. Stat. (1): at this point, all you have is a single equation with two unknowns - ab. No way to find the individual values of a and b, or to reach a recycled quad. Therefore, stat (1)->IS->BCE. Stat. (2): change the form to expanded form: (a+b)2 = a2 + 2ab + b2 = 121. This would've been sufficient, but you need ab to leave only the required a2 + b2. Alone, stat. (2)->IS->CE. Stat. (1) + (2) combined: Now you have all the pieces of the puzzle: a2 + 2ab + b2 = 121. plug in ab=30 (stat. (1) into the expanded form to get: a2 + 2·30 + b2 = 121 Therefore, a2+ b2 = 121 - 2·30 = 61. Stat. (1)+(2)->S->C.

The discriminant is what distinguishes x1 from x2: If the discriminant is positive, (b2-4ac > 0) then x1 and x2 will have different values - 2 different solutions. If the discriminant is equal to zero (b2-4ac = 0), then x1=x2 =-b / 2a.The quadratic has only one solution. If the discriminant is negative (b2-4ac < 0), then the quadratic has no real solutions, since you can't take the square root of a negative number without resorting to imaginary numbers (which are thankfully not tested on the GMAT). Therefore, if a question asks for the number of solutions a quadratic has, plug in the values of a, b and c into the discriminant b2-4ac and find out whether it is positive, zero or negative.

Example: How many solutions does the equation x2-4x+4=0 have? 1) Find out the coefficients: a=1, b=-4, c=4. 2) Plug into the discriminant b2-4ac = (-4)2 - 4⋅1⋅4 = 16-16 = 0. 3) Since the discriminant equals zero, the quadratic above has only one solution.

If x2-7x-30=0, what is x? *groan* Yes, we know. You hated quadratics back in high-school, and you hoped you'd never ever see them again the minute you finished your exams, and now they're back to haunt you. The GMAT does have questions involving quadratics, and we're here to revisit the material needed to deal with them. We promise to make this as painless as possible. A quadratic equation (or "quadratic") is an equation involving x2. It is usually of the form of ax2 + bx + c=0. a is the coefficient for x2, b is the coefficient for x, and c is a constant - just a number.

If a problem presents this equation: 3x2+2x=10 What's the first thing you do? Correct - two points. In this case, subtract 10 to get 3x2+2x-10=0. A quadratic equation will have up to 2 possible solutions (also called roots). These solutions are values of x which satisfy the equation. For the equation we started with, both x=10 and x=-3 are solutions for x2-7x-30=0. We'll revisit two ways of finding those solutions soon: The solutions formula, and the technique for factoring quadratics. We'll also look at some recycled quadratics, how to recognize them, and what to do when you do.

What is the reciprocal of (3−√8)? 1/(√8 − 3) √8 − 3 3/(√8) 3 + √8 3·√8

Incorrect. The reciprocal of x is 1/x. Thus, the reciprocal of (3−√8) is 1/(3−√8). Unfortunately, there is no such answer among the answer choices. A different approach is needed: Recall that if a·b=1 then a and b are reciprocal. Eliminate any answer choice whose product with (3−√8) is not one. Correct. Answer choices A and B can be eliminated immediately for being negative (√8 < √9=3). The product of a negative number with positive (3−√8) cannot be equal to one. Check C: 3/(√8) × (3−√8) = 9/(√8) − 3. This is not equal to one, so eliminate C. Checking D you get (3+√8) × (3−√8). That should ring a bell. The numbers fit in the quadratic formula (a−b)(a+b)=a2−b2, so (3+√8) × (3−√8) = 32−(√8)2 = 9−8 = 1. The product of (3−√8) and (3+√8) is equal to 1, therefore they are each other's reciprocal.

There are several major ways of finding the solutions of a quadratic. In this lesson we'll refresh your memory regarding The Solutions formula. You might recall seeing this in high-school. Finding the solutions of a quadratic of the form of ax2+bx+c=0 is a matter of plugging in a, b and c into the following formula: x1,2 = -b± √b2-4ac / 2a The two solutions of the quadratic are x1 and x2. The ± sign denotes that x1 = -b+ √b2-4ac / 2a and x2 = -b-√b2-4ac / 2a

Let's practice finding the solutions on the following quadratic equation: x2+3x-10=0 1) First, identify your coefficients: a is the coefficient for x2 - in this case a=1. b is the coefficient for x - in this case, b=3 c is the constant (the number in the end) - in this case c=-10. x1 = -5 and x2 = 2 The above example shows a classic quadratic with two solutions. However, some quadratics have only one solution, or even no solutions at all (i.e. no values of x that satisfy the equation ax2+bx+c=0). Some GMAT questions will ask how many solutions does a specific quadratic have, rather than what those solutions are. A quick way of finding the number of solutions without going to trouble of finding the actual solutions is to focus on the part under the square root (b2-4ac), which is called the discriminant.

GMAT problems involving quadratic equations often require you to build a quadratic equation from factored form. Doing so is relatively easy using a simple multiplication process called FOIL, an acronym for the four steps of the process: First, Outer, Inside, Last. Let's change the expression (x+4)(x-5) into a quadratic expression using FOIL: First: multiply the first term of each factor together: x·x = x2 Outer: Multiply the outer terms of the expressions together: x·(-5) = -5x Inside: Multiply the inside terms of the expressions together: 4·x = 4x Last: Multiply the last term of each factor together: 4·(-5) = -20. Finally, add the 4 terms: x2 - 5x + 4x - 20. Combine the two middle terms to reach our quadratic expression x2 - x - 20.

To sum up: FOIL: an acronym for a process for building a quadratic equation from factored form (x+e)(x+f): First, Outer, Inside, Last. First: multiply the first term of each factor together. Outer: Multiply the outer terms of the expressions together. Inside: Multiply the inside terms of the expressions together. Last: Multiply the last term of each factor together.

If (x−2√3) is one factor of the equation x2+(4√3)·x−36=0, what is the other factor of the equation? (x+2√3) (x+4√3) (x+6√3) (x−2√3) (x−6√3)

Very good! You took 1 minutes and 18 seconds to answer this question. Correct. The issue is factoring a quadratic equation.Write the quadratic in expanded form: If (x+e)(x+f) = x2+(4√3)·x−36, then e·f=-36. This, together with the known factor, (x−2√3), makes it possible to find the other factor. If (x−2√3) is a factor of x2+(4√3)·x−36, then e=−2√3, and you need to find f. Since e·f=−36, it follows that f = (−36) / (−2√3) = 18 / √3 The answer is of the form (x+f), so clearly D and E can be eliminated. The problem with 18/√3 is that nasty square root in the bottom. Multiply the top and bottom by √3 / √3 to get rid of the root in the bottom, as √3√3 = 3. (18 / √3) ⋅ (√3 / √3) = --> 18⋅√3 / √3⋅√3 Since √3⋅√3 = 3, the result of the fraction above is: --> 18⋅√3 / 3 Reduce 18 and 3 to get: --> 6⋅√3 / 1 = 6⋅√3

As we've mentioned, many GMAT problems involving quadratics will test your understanding of meaning of factoring quadratics, rather than your ability to employ the technique. Take a look at this one: If x=5 is a solution of the equation x2-nx+10=0, which of the following is the value of n? -7 -2 2 7 25

Very good. What does it mean that x=5 is the solution? That's right, it means that x=5 satisfies the equation x2-nx+10=0. In other words, if you plug in x=5 into the equation, the result has to be zero!. Once you've reached this insight, the question is easy - plug in x=5 and find a simple equation with n: x2-nx+10 = 52-n·5+10 = 0 Now solve for n: 25-5n+10=0 --> 5n=35 --> n=7 Remember this about solutions of an equation - they are the values that satisfy the equation. In the case of a quadratic, plugging the solutions x into the equation ax2+bx+c will give the result of zero.