Question

Solve the quadratic equation by completing the square. What are the missing pieces to the steps?
-27 = 4x²- 24x
-27 = 4(x²- 6x + 9)
9 =4(x²- 6x + 9)
9/4 = (x -3)
+__= x - 3
__ = x​

Answer 1

Answer:

The missing pieces should be after line 3.

9/4 = (x - 3)^2

<=> (3/2)^2 = (x - 3)^2

<=> x - 3 = 3/2 or x - 3 = -3/2

<=> x = 3 + 3/2 or x = 3 - 3/2

<=> x = 9/2 or x = 3/2

Hope this helps!

:)

Answer 2

Answer:

I think the missing steps are:

$\frac{3}{2}=x-3$

and

$\frac{9}{2}=x$

Step-by-step explanation:

$-27=4x^2-24x$

First, factor 4 to make isolate $x^2$

$-27=4(x^2-6x)$

Now, divide by 4.

$-\frac{27}{4}=x^2-6x$

Complete the square by adding $(\frac{b}{2})^2$

$(\frac{b}{2})^2-\frac{27}{4}=x^2-6x+ (\frac{b}{2})^2$

$(\frac{6}{2})^2-\frac{27}{4}=x^2-6x+ (\frac{6}{2})^2$

$(3)^2-\frac{27}{4}=x^2-6x+ (3)^2$

$9-\frac{27}{4}=x^2-6x+9$

Solve the difference.

$\frac{9*4-27}{4}=x^2-6x+9$

$\frac{36-27}{4}=x^2-6x+9 \\\frac{9}{4}=x^2-6x+9$

Factor the equation.

$\frac{9}{4}=(x-3)(x-3)$

Rewrite it;

$\frac{9}{4}=(x-3)^2$

Extract the square root.

$\sqrt{\frac{9}{4} }=x-3$

Take the square root of 9 and 4.

$\frac{3}{2}=x-3$

Add 3.

$\frac{3}{2}+3=x$

Do the sum

$\frac{3+2*3}{2}=x\\\frac{3+6}{2}=x\\\frac{9}{2}=x$