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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

User Nemeth
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2 Answers

4 votes

Answer:


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

Explanation:

Given the equation


x^2-(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-(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 }(-(3)/(4))


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


x^2-2(x)((3)/(8))+(9)/(64)=5+(9)/(64)


x^2-2(x)((3)/(8))+(-(3)/(8))^2=5+(9)/(64)


(x-(3)/(8))^2=5+(9)/(64)

which makes LHS a perfect square trinomial.


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

User Kirakun
by
7.5k points
2 votes

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

... ((1/2)·(-3/4))² = (-3/8)² =

... A) 9/64

User Dodomorandi
by
9.0k points

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