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

3 Answers
Apr 28, 2018

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)"

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+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!

Apr 28, 2018

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. :)