What are the roots of x squared plus 8x plus 25?

Answer 1

We will use a tool called a quadratic formula. First, I will prove it.

For all degree two function f(x) = ax^2 + bx + c where a, b, c are constants, claim the roots are -b +/- sqrt(b^2 - 4ac))/2a IF a is not 0 and b^2 - 4ac >= 0.

Proof. Suppose the conditions are satisfied, finding the roots of the function is the same as solving the following

ax^2 + bx + c = 0 we will simplify, since a != 0

a(x^2 + (b/a)x) + c = 0 we will do a trick, a trick that you see once, you will never forget, we add then subtract the same thing

a(x^2 + (b/a)x + b^2/(4a^2) - b^2/(4a^2)) + c = 0

Now notice b/a = 2 . b/2a, we will do that, also, we will take -b^2/(4a^2) out, notice, b^2/4a^2 = (b/2a)^2

Then we get

a (-b^2/(4a^2)) + a( x^2 + 2 (b/2a)x + (b/2a)^2) + c = 0 we use the complete square identity

-b^2/(4a) + a(x + b/2a)^2 + c = 0

Isolate x

a(x + b/2a)^2 = -c + b^2/4a = (-4ac + b^2)/4a

(x + b/2a)^2 = (b^2 - 4ac)/(4a^2)

(x + b/2a) = +/- sqrt (b^2 - 4ac) / 2a

x = +/- sqrt (b^2 - 4ac) / 2a - b/2a

x = -b +/- sqrt(b^2 - 4ac))/2a

Q.E.D

Now apply the quadratic formula to your question: x^2 + 8x + 25. First check if a = 0, a = 1 != 0, pass. Now check if b^2 - 4ac is non-negative. 8^2 - 4 x 1 x 25 = 64 - 100 = - 36 < 0. Done, this function have no roots.