# Which value must be added to the expression x^{2 }+ 16x to make it a perfect-square trinomial?

8, 32, 64, 256

**Solution:**

Given the expression x^{2 }+ 16x.

If the quadratic equation is of the form ax^{2 }+ bx + c then to complete square

**Step1:** Take coefficient of x^{2} common from ax^{2 }+ bx + c,

⇒ a[x^{2} + (b/a)x + (c/a)]

**Step2:** Add and subtract (1/2 coefficient x)^{2} to quadratic term then,

⇒ a [(x + 1/2 coefficient x)^{2}) + (c/a) - (1/2 coefficient x)^{2}

⇒ a [(x + b/2a)^{2 }+ c/a - (b/2a)^{2}]

Note that if the coefficient of x^{2} is 1 then we have to add (1/2 coefficient x)^{2} to convert it into perfect square expression.

Thus, in the given problem x^{2 }+ 16x.

1/2 coefficient of x = (1/2) × 16 = 8

We have to add 8^{2} = 64, to convert it into a perfect square.

Therefore, 64 must be added to the expression to make it a perfect-square trinomial.

## Which value must be added to the expression x^{2} + 16x to make it a perfect-square trinomial?

**Summary:**

To make it a perfect-square trinomial, we have to add 8^{2} = 64 to the expression x^{2} + 16x.

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