solve math?

Question

Find all values of n such that x^2 + bx + (b/2)^2 = n
A. one real root B. Two real root C. two imaginary roots

Answer 1

Hey there!

Recall the formula (a+b)^2=a^2+2ab+b^2.

So we can rewrite x^2+bx+(b/2)^2=n as x^2+2(b/2)x+(b/2)^2=n. Factoring it, we get (x+b/2)^2=n. Take the square root on both sides of the equation, we will get x+b/2=sqrt(n).

So, our new equation is x+b/2=sqrt(n).

The answer depends on the value of b.

Hope it helps!

Answer 2

x^2 + bx + b^2/4 - n = 0
a=1; b=b, c= (b^2-4n)/4

the discriminant = b2-4ac
=b^2 - (b^2-4n) = 4n

A. one real root 4n=0 n=0
B. two real root 4n>0 n>0
C. two imag root 4n<0 n<0