Question

Brian is solving the equation x^2 -3/4 x =5 What value must be added to both sides of the equation to make the left side a perfect-square trinomial?
A)9/64

B)9/16

C)3/4

D)9/4

Answer 1

Answer:

$\frac{9}{64}\text{ is added to both sides of the equation to make the left side a perfect-square trinomial.}$

Step-by-step explanation:

Given the equation

$x^2-\frac{3}{4}x=5$

we have to find the value which must be added to both sides of the equation to make the left side a perfect-square trinomial.

$x^2-\frac{3}{4}x=5$

To form the perfect square we have to add the square of half the coefficient of x,

$\text{Here the coefficient of x is }(-\frac{3}{4})$

$\text{Now, the square of half of above is }(-\frac{3}{8})^2=\frac{9}{64}$

$x^2-2(x)(\frac{3}{8})+\frac{9}{64}=5+\frac{9}{64}$        

$x^2-2(x)(\frac{3}{8})+(-\frac{3}{8})^2=5+\frac{9}{64}$        

$(x-\frac{3}{8})^2=5+\frac{9}{64}$        

which makes LHS a perfect square trinomial.

$\frac{9}{64}\text{ is added to both sides of the equation to make the left side a perfect-square trinomial.}$

Answer 2

The square of half the x-coefficient must be added. That value is

... ((1/2)·(-3/4))² = (-3/8)² =

... A) 9/64