# How do you solve x^2+28x+196=0?

Apr 28, 2018

$x = - 14 \text{ (repeated}$

#### Explanation:

${x}^{2} + 28 x + 196 \text{ is a "color(blue)"perfect square}$

$\text{note that } {\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$

$\text{with "a=x" and } b = 14$

$\Rightarrow {x}^{2} + 28 x + 196 = {\left(x + 14\right)}^{2}$

$\text{solve } {\left(x + 14\right)}^{2} = 0$

$\Rightarrow x + 14 = 0 \Rightarrow x = - 14 \text{ (repeated)}$

Apr 28, 2018

Either by inspection, graphing, or the quadratic formula.

#### Explanation:

Solve by Graphing

Every quadratic is a function on a graph. Just replace 0 with y= and graph the quadratic as a function. the x-coordinate(s) wherever the function intersects the x axis are your solution(s). in this instance, there's only 1 x-intercept and therefore only one solution.

Solve by Inspection

${x}^{2} + 28 x + 196$

$a {x}^{2} + b x + c$
$a = 1 , b = 28 , c = 196$

Ask yourself, what factors of $c$ multiplied by the factors of $a$ both multiply together to make $c$ and add to $b$. This sounds confusing, but when I lay it out, it will make more sense...

Think, if $\sqrt{196} = 14$, does $14 + 14 = 28$? Yes!

Then factor out the quadratic into 2 binomials...

$\left(x + 14\right) \left(x + 14\right)$
.. and rearrange each individually to equal $x$

$x + 14 = 0 \implies x = - 14$

If the quadratic doesn't look easy to factor, then pull out the quadratic formula!

$a {x}^{2} + b x + c = 0$
$a = 1 , b = 28 , c = 196$

x=(-b±sqrt(b^2-4ac))/(2a)

x=(-28±sqrt(28^2-4(28)(1)))/(2(1))
$x = - 14$

± is literally just a symbol for saying "plus or minus",

Hope it helps!

Apr 28, 2018

The solution of this given quadratic equation is ${\left(x + 14\right)}^{2}$

#### Explanation:

In the given expression we see the polynomials have the following numbers as their coefficients $1 , 28$ and $196$ so we can factorise this expression by middle term factorisation.
In middle term factorisation we have to split lower degree polynomial into two in our $28 x$ into the sum of two factors of $196$ which when multiplied with each other give $196$
Here those two factors are $14$ and $14$,as ${14}^{2} = 196$ and when you add $14$ twice you get $28$.
${x}^{2} + 28 x + 196$
= ${x}^{2} + 14 x + 14 x + 196$
= $x \left(x + 14\right) + 14 \left(x + 14\right)$
= $\left(x + 14\right) \left(x + 14\right)$
= ${\left(x + 14\right)}^{2}$
I hope this was helpful. :)