Question

The equation 2x2 – 5x = –12 is rewritten in the form of 2(x – p)2 + q = 0. what is the value of q?

Answer 1

Answer:

$q=\dfrac{71}{8}$

Step-by-step explanation:

The given equation is

$2x^2-5x=-12$

Rewrite the given equation in the form of  

$2(x-p)^2+q=0$       ... (i)

We need to find the value of q.

The given equation can be rewritten as

$2(x^2-\dfrac{5}{2}x)=-12$

Now, add and subtract square of half of coefficient of x inside the parenthesis.

$2(x^2-\dfrac{5}{2}x+(\dfrac{5}{4})^2-(\dfrac{5}{4})^2)=-12$

$2(x^2-\dfrac{5}{2}x+(\dfrac{5}{4})^2)-2(\dfrac{5}{4})^2)=-12$

$2(x-\dfrac{5}{4})^2-2(\dfrac{25}{16})=-12$

Add 12 on both sides.

$2(x-\dfrac{5}{4})^2-\dfrac{25}{8}+12=0$

$2(x-\dfrac{5}{4})^2+\dfrac{96-25}{8}=0$

$2(x-\dfrac{5}{4})^2+\dfrac{71}{8}=0$      ... (ii)

From (i) and (ii), we get

$q=\dfrac{71}{8}$

Therefore, $q=\dfrac{71}{8}$.

Answer 2

2x^2 - 5x + 12 = 0
Factoring out the leading coefficient:
2(x^2 - 5/2 x) + 12 = 0
Completing the square:
2[(x - 5/4)^2 - 25/16] + 12 = 0
2(x - 5/4)^2 - 25/8 + 12 = 0
2(x - 5/4)^2 + 71/8 = 0
q = 71/8