Math SAT Level I - Chapter 6 Equations and Inequalities

What are the roots of the equation x²-2x-15 = 0

x = 5 or -3

What are the roots of the equation x² = 10x-25

x = 5 or x = 5

For what value(s) of x is (3/4)-(5/x) = (7/x)

(3/4)-(5/x) = (7/x) => 4x(3/4)-4x(5/x) = 4x(7/x) => 3x-20 = 28 => 3x = 48 => x = 16

If ax²+bx+c = 0, then the sum of the two roots is ____ and the product of the two roots is ____

-b/a, c/a

If ax-b = c-dx, what is the value of x in terms of a, b, c, and d? A. (b+c)/(a+d) B. (c-b)/(a-d) C. (b+c-d)/(a) D. (c-b)/(a+d) E. (c/b)-(d/a)

A. (b+c)/(a+d) Treat a, b, c, and d as constants, and use the six-step method to solve for x: ax-b = c-dx => ax+dx = c+b => (b+c)/(a+d)

If v is an integer and the average (arithmetic mean) of 1, 2, 3, and v is less than 20, what is the greatest possible value of v?

Write the inequality (1+2+3+v)/4 < 20 => (6+v)/4 < 20 => 6+v < 80 => v < 74. Since v is an integer, the greatest possible value of v is 73.

If r₁ and r₂ are the two roots of the equation x²-2x-1 = 0 and r₁ > r₂, what is r₁ - r₂? A. -2√2 B. -2 C. 0 D. 2√2 E. 2

D. 2√2 Use the quadratic formula to find r₁ and r₂. Since a = 1, b = -2, and c = -1: x = 1(+/-)√2 Then r₁ = 1+√2 and r₂ = 1-√2, and so: r₁ - r₂ = (1+√2)-(1-√2) = 2√2

Solve the system using substitution 2x+y = 13 3x - y =12

y = 13 - 2x 3x - (13-2x) = 12 x = 5, y = 3

For what value of x is 2^(x-3) = 16?

2^(x-3) = 16 => 2^(x-3) = 2^(4) => x-3 = 4 => x = 7

Quadratic formula

((-b(+/-)√(b²-4ac))/(2a)

For what value(s) of x is (3/4)-(5/x) > (7/x)

(3/4)-(5/x) > (7/x) => add 5/x to both sides (combine like terms) => (3/4) > (7/x)+(5/x) => (3/4) > (12/x) Case I: x>0 (3/4)>(12/x) => 3x > 48 => x>16 So the POSITIVE solutions to the inequality are all numbers greater than 16 Case II: x<0 (3/4)>(12/x) => 3x < 48 => x < 16 So the NEGATIVE solutions the the inequality are all negative numbers less than 16. Since every negative number is less than 16, every negative number is a solution. Combining the two cases, the solution set is { x | x < 0 or x > 16}

For what value of x is 2^(x-3) = 15?

2^(x-3) = 15 => log(10)2^(x-3) = log(10)15 => (x-3)log(10)2 = log(10)15 => x-3 = log(10)15/log(10)2 => x = 3+ log(10)15/log(10)2 => x = 6.91

If 2^x = 4^y, what is the ratio of x to y?

2^x = 4^y => 2^x = (2)^2y => x = 2y. x/y = 2

For what value(s) of x is 3|x+5|-5 < 7

3|x+5| -5 = 7 => 3|x+5| < 12 => |x+5| < 4 -4 < x+5 < 4 => -9 < x < -1

For what value(s) of x is 3|x+5| -5 = 7

3|x+5| -5 = 7 => 3|x+5| = 12 => |x+5| = 4 => x+5 =4 or x+5 = -4 => x = -1 or x = -9

Solve for x 2x²+18

2x²+18 => 2x² =-18 => x² = -9 => x = (+/-)√-9 = (+/-)3i x = 3i or x = -3i

4^(x+3) = 7^(x-1)

4^(x+3) = 7^(x-1) => log4^(x+3) = log7^(x-1) => (x+3)log4 = (x-1)log7 => (x+3)(0.602) = (x-1)(0.845) => 0.602x+1.806 = 0.845x - 0.845 => 0.243x = 2.651 => x = 10.9

4√x - 5 = 7

4√x - 5 = 7 => 4√x = 12 => √x = 3 => (√x)² = 3² => x = 9

4√x+15 = 7

4√x+15 = 7 => 4√x = -8 => √x =-2 Since √x cannot be negative, no real number x satisfies the equation 4√x+15=7

For what value x is 5(x-10) = x +10?

5(x-10) = x +10 => 5x-50 = x+10 => 4x = 60 => x = 15

Find a quadratic equation for which the sum of the roots is 5 and the product of the roots is 5.

For simplicity, let a = 1. Then -b/a = 5 => -b/1 = 5 => b = -5, and c/a = 5 => c/1 = 5> c = 5. So the equation x²-5x+5 = 0 satisfies the given conditions.

Solve for x 2x²-5x = 0

Recall that if a product is equal to zero, one of the factors must be 0. So: 2x²-5x = 0 => x(2x-5) = 0 => x = 0 or x =5/2

Solve for x 3x² = 4x

Recall that if a product is equal to zero, one of the factors must be 0. So: 3x² = 4x => 3x²-4x => x(3x-4) = 0 => x = 0 or x = 4/3

Solve the system y = 2x-1 y = x²-2x+2

Solve by substitution, replacing the y in the second equation with 2x-1. 2x-1 = x²-2x+2 => x²-4x+3 = 0 => (x-3)(x-1) = 0 => x = 3 or 1

Solve the equation 2x²-4x-1 = 0

pg.89 x = 1+ (1/2)√6 or x = 1- (1/2)√6

Solve the equation x²-2x+2 = 0

pg.89 x = 1+i x = 1-i

If 4x+13 = 7-2x, what is the value of x? A. -10/3 B. -3 C. -1 D. 1 E. 10/3

C. -1 4x+13 = 7-2x => 6x+13 = 7 => 6x = -6 => x = -1

4^(x+3) = 8^x-1

(2²)^(x+3) = (2³)^(x-1) => 2^(2x+6) = 2^(3x-3) => 2x+6 = 3x-3 => x = 9

If 2^x = 5^y, what is the ratio of x to y?

2^x = 5^y => log2^x = log5^y => xlog2 = ylog5 => x/y = log5/log2 = 2.32

If the sum of two numbers is 13 and the difference of the numbers is 23, what is the product of the numbers? A. -90 B. -23 C. 0 D. 18 E. 36

A. -90 Represent the numbers by x and y. Then x+y = 13 and x-y = 23. Now add the equations: x+y = 13 x-y = 23 2x = 36 => x = 18

Solve the system 2x+y = 13 3x-y = 12

Add the two together as you would a basic addition problem. 2x+y = 13 3x-y = 12 => 5x = 25 => x = 5. Now plug in x and find the value for y in any of the equations. 2(5)+y=13 => 10+y=13 => y = 3 x = 5, y = 3

For what values of x is | 2x-3 |-4 < 7? A. x < 7 B. 0 < x < 3 C. -4 < x < 7 D. x < 0 or x > 3 E. x < -4 or x > 7

C. -4 < x < 7 | 2x-3 |-4 < 7 => | 2x-3 | < 11 => -11 < 2x-3 < 11=> -8 < 2x < 14 => -4 < x < 7

If 4^a = 8^b, what is the ratio of a to b? A. 1/2 B. 2/3 C. 3/2 D. 2/1 E. 4/1

C. 3/2 4^a = 8^b => 2^2a = 2^3b. Therefore, 4^a = 8^b => 2a = 3b => a/b = 3/2

If (x+3)/(2) + 3x = 5(x-3) + (x+23)/(5), what is the value of x? A. 3 B. 5 C. 7 D. 9 E. 11

C. 7 1) Multiply both sides by 10 to get rid of the fractions. (x+3)/(2) + 3x = 5(x-3) + (x+23)/(5) => 5(x+3)+30x = 50(x-3)+2(x+23) => 35x+15 = 52x-104 x = 7

If 4^a = 5^b, what is the ratio of a to b? A. 0.8 B. 0.86 C. 1.08 D. 1.16 E. 1.25

D. 1.16 4^a = 5^b => log4^a = log5^b => alog4 = blog5 => log5/log4 = 1.16

Solve the system 2x+3y = 31 3x+2y = 29

In this example we solve for x first. The top equation is multiplied by 3, the bottom by 2. y = 7 x = 5

Solve the system 2x+y = 13 5x-2y = 10

You must first cancel out a term. In this example we multiple the top equation by 2 so that the y's cancel: 4x+2y = 26 5x-2y = 10 => 9x = 36 => x = 4. Now substitute an x in any original equation. 2(4)+y = 13 => 8+y =13 => y = 5 x = 4, y = 5

Solve for x 1/2x+3(x-2) = 2(x+1)+1

1/2x+3(x-2) = 2(x+1)+1 => multiply both sides by a common denominator 2. => x+6(x-2) = 4(x+1)+2 => 7x-12 = 4x+6 => 3x = 18 => x = 6

If 2/3x = 3y-4z, what is the value of y in terms of x and z?

2/3x = 3y-4z => multiply both sides by 3 => 2x = 3(3y-4z) => 9y-12z => 2x+12z = 9y => y = (2x+12z)/9

If x²+5x+c = 0 has exactly one solution, what is the value of c? A. -6.25 B. -2.25 C. 0 D. 3.75 E. 6.23

E. 6.23 If a quadratic equation has exactly one solution, then its discriminant is 0. So: b²-4ac = 25-4(1)c = 0 => 25-4c = 0 => 4c = 25 => c = 6.25

Which of the following is a solution to the equation √(2x) + 5 = 3 A. 2 B. 4 C. 16 D. 32 E. This equation has no solution

E. This equation has no solution √(2x) + 5 = 3 => √(2x) = -2. Since by definition a square root cannot be negative, this equation has no solution.