Verified answer

There is no need to use the quadratic formula as many of the answerers have. Solve the following quadratics by completing the square.

1.

3x² = -15x + 21

3x² + 15x - 21 = 0

3(x² + 5x - 7) = 0

x² + 5x - 7 = 0

4x² + 20x - 28 = 0

(2x + 5)² - 53 = 0

(2x + 5)² = 53

2x + 5 = ±√(53)

2x = -5 ± √(53)

x = (-5 ± √(53)) / 2

Choice D

2.

x² = 8x - 11

x² - 8x + 11 = 0

(x - 4)² - 5 = 0

(x - 4)² = 5

x - 4 = ±√5

x = 4 ± √5

The quadratic formula should be used here since both quadratic equations are not factorable.

The formula to find (x) is: [-b+or-√(b^2-4ac)]/2a

To get all the variables in the right order, and put them in the quadratic formula, you have to rearrange the equation you have, into

ax^2+bx+c=0,

So you need to turn that

3x^2=-15x+21

into

3x^2+15x-21=0

and same with the other one to find the answers.

The correct answers are:

1.) D D. (-5± sqrt53/2)

2.) x = 4 ± sqrt 5

Hope this helps!

(Make sure you know how to solve the question, because that is always more important than the answer itself)

1. 3x^2 = –15x + 21

3x^2 + 15x - 21 = 0

Real solutions:

Root 1: -6.14

Root 2: 1.14

B

2. x^2 = 8x – 11

x^2 - 8x + 11 = 0

Real solutions:

Root 1: 1.764

Root 2: 6.236

C

3x^2 + 15x -21 = 0

dividing by 3,we get

x^2 + 5x - 7 = 0

roots of the eqn. ax^2+bx+c=0 is -b±√(b^2-4ac)/2a

=(-5±√53)/2

-------------------------------------

2)x^2-8x+11 = 0

=(8±√64-44)/2

=4±√5