Question

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

Answer 1

`x=-14" (repeated"`

Explanation:

`x^2+28x+196" is a "color(blue)"perfect square"`

`"note that "(a+b)^2=a^2+2ab+b^2`

`"with "a=x" and "b=14`

`rArrx^2+28x+196=(x+14)^2`

`"solve "(x+14)^2=0`

`rArrx+14=0rArrx=-14" (repeated)"`

Answer 2

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+28x+196`

`ax^2+bx+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 `sqrt196 = 14`, does `14+14 = 28`? Yes!

Then factor out the quadratic into 2 binomials...

`(x + 14)(x+14)`
.. and rearrange each individually to equal `x`

`x+14=0 => x = -14`

Quadratic formula

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

`ax^2+bx+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!

Answer 3

The solution of this given quadratic equation is `(x+14)^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 `28x` 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+28x+196`
= `x^2+14x+14x+196`
= `x(x+14)+14(x+14)`
= `(x+14)(x+14)`
= `(x+14)^2`
I hope this was helpful. :)