# What should be added to the polynomial so that it becomes perfect square 4X square plus 8X?

## $4 {x}^{2} + 8 x$

Jan 11, 2018

4

#### Explanation:

By adding four we can make it perfect square
$4 {x}^{2} + 8 x + 4$
${\left(2 x\right)}^{2} + 2 \cdot 2 x \cdot 2 + {\left(2\right)}^{2}$
${\left(2 x + 2\right)}^{2}$
Thus it becomes a perfect square

But how do we get to this answer without any guess , here's the role of discriminant.
Any quadratic equation is of form
$a {x}^{2} + b x + c = 0$
Here , discriminant $\left(d\right) = {b}^{2} - 4 a c$
For perfect squares it is $0$
So , we get
0=8^2-4×4×c
$64 = 16 c$
$c = 4$

Jan 11, 2018

$4 {x}^{2} + 8 x + 4$ so you add 4

#### Explanation:

A perfect square is of form: ${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$

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Let the unknown value be $t$ ( you have to write something for it!). $t$ could be just a number and or a coefficient of x->?x

Set $4 {x}^{2} + 8 x + t \textcolor{w h i t e}{\text{ddd")=color(white)("ddd}} {a}^{2} + 2 a b + {b}^{2}$

Compare a^2=4x^2->2^2x^2color(white)("d")"so "color(blue)(a =sqrt(2^2x^2)=2x)

Compare $2 a b = 8 x$

Substituting for $\textcolor{red}{a}$ we have:

Compare $\textcolor{g r e e n}{2 \textcolor{red}{a} b = 8 x \textcolor{w h i t e}{\text{d")->color(white)("d")2(color(red)(2x))b=8x color(white)("d")->color(white)("d}} 4 x b = 8 x}$

$\textcolor{b l u e}{\text{so } b = 2}$

Thus ${a}^{2} + 2 a b + {b}^{2} \textcolor{w h i t e}{\text{d")->color(white)("d}} 4 {x}^{2} + \left(2\right) \left(2 x\right) \left(2\right) + {2}^{2}$

Giving:

$\textcolor{w h i t e}{\text{ddddddddd}} \textcolor{b l u e}{\underline{\overline{| \textcolor{w h i t e}{\frac{2}{2}} 4 {x}^{2} + 8 x + 4 \textcolor{w h i t e}{2} |}}}$