Question

The roots of the equation $x^2+ax-4=0$ are, where $a\in R$

• A. real distinct
• B. equal
• C. imaginary
• D. real

Answer 1

Answer: (A)

real distinct

$x^2+ax-4=0$

$A=1,B=a,C=-4$

$D=b^2-4ac$

$\Rightarrow D=a^2-4\times (-4)\times 1$

$\Rightarrow D=a^2+16>0\left(\text{always}\right)$

Since $D$ is greater than $0$ for any value of $a$, the roots of the given equation are real and distinct.