Algebra
Quadratic Functions
MMTS_Full_Test_12
Grade 12
Question:
Let $f(x)=x^2+bx+c$, minimum value of $f(x)$ is $-5$, then absolute value of the difference of the roots of $f(x)$ is
5
$\sqrt{20}$
$\sqrt{15}$
Can't be determined
Step-by-Step Solution
Key Concept: Minimum = $c - b^2/4 = -5$; difference of roots = $\sqrt{b^2-4c}/1$
Min $= c-b^2/4=-5\Rightarrow b^2-4c=20$. $|\alpha-\beta|=\sqrt{b^2-4c}=\sqrt{20}$.
Correct Answer: 2