Algebra Final
What does 2n^(2)-7n+5 look like after been factored?
2n^(2)−2n−5n+5
In 2n^(2)-7n+5 , What is factoring Steps?
For a polynomial of the form ax^(2)+bx+c, rewrite the middle term as a sum of a) two terms whose product is a⋅c=2⋅5=10, and b) whose sum is b=−7b=-7.
What is the quadratic formula
− b ± √( b^(2) − 4 ( a c )) / 2 a
Solving: 2x+5 − 3x _______ _________ = x 2 x-3 nets x=-15. What are the steps?
1) Cross Multiply the denominators, divide into themselves. 2x + 5 * (x-3) - 3x * 2 ______ ______ - ___ __ __ = x 2 (x-3) x-3 2 2) Simplify, Combine (2x + 5)(x-3) - 2(3x) ------------------------------ = x 2(x-3) 3) Expand (2x+5)(x−3)(2x+5)(x-3) using the FOIL Method. Simplify 2(3x), then add it to the final Foil Solution 2x^(2)−7x−15 ------------------------ = x 2(x-3) 5) Multiply both sides of the equation by the denominator, removing the fraction on the left side and combining on the right 2x^(2)−7x−15 = 2x^(2)-6x 6) Simplify x = -15
factoring 2n^(2)-7n+5=0 Answer is n=2.5,1 What are the Steps?
1) Factor by grouping. (2n−5)(n−1)=0 2) Set 2n−5 equal to 0 and solve for n. n=2.5 3 Set n−1 equal to 0 and solve for n. n=1
Using the Quadratic Formula to solve, x^(2)=6x−34 The answer is 3±5i What are the steps?
1) Move all the expressions to the left side of the equation. x^(2) -6x+34 = 0 2) Use the quadratic formula to find the solutions. − b ± √( b^(2) − 4 ( a c )) / 2 a 3) Substitute the values a=1, b=−6, and c=34 into the quadratic formula and solve for x. − (-6) ± √( (-6)^(2) − 4 ( (1* 34 )) / 2 (1) 4) Simplify the numerator. x=6±10i/2⋅1 5) Simplify again x=3±5i
Solved equation via the square root property. (2x−5)^2=30 is 5 (±) √30 __________________ 2 What are the steps?
1) Take the square root of each side of the equation to set up the solution for x ( 2 x − 5 )^2 ⋅ 1/2 = ± √ 30 2) Remove the perfect root factor 2 x − 5 under the radical to solve for x . ( 2 x − 5 ) = ± √ 30 3 Dividing the equation into positive and negative, add 5 to both sides, place in front of the math symbols, then divide each side by 2 and simplify x= 5 (+) √ 30/2 , x= 5 (-) √ 30/2
Using the Quadratic Formula to solve, r^(2)+3r−5=0 solution is r = − 3 ± √ 29 _______________ 2 What are the steps?
1) Use the quadratic formula to find the solutions. − b ± √( b^(2) − 4 ( a c )) / 2 a 2) Substitute the values a = 1 , b = 3 , and c = − 5 into the quadratic formula and solve for r . − 3 ± √ (3^(2) − 4 ⋅ ( 1 ⋅ − 5 ) 2 ⋅ 1 3) Simplify. r = − 3 ± √ 29 /2
