Question

Quadratic and Ratios...​

Answer 1

Answer:

x=0

Step-by-step explanation:

in the quadratic formula to only have one solution

$b^{2} -4ac=0$ for $ax^2+bx+c=0$

so we have

$ax^2+b=c$

$ax^2+(b-c)=0$

if we insert in the formula we have

$0^2-4(a)(b-c)=0\\\\-4ab+4ac=0\\\\4ac=4ab\\\\\frac{4ac}{4a} = \frac{4ab}{4a}\\\\c=b$

$ax^2+b=c\\\\ax^2=b-b\\\\ax^2=0$ since a≠0

$\frac{ax^2}{a}=\frac{0}{a} \\\\x^2=0\\\\x=0$