Question

If roots of the equation $f(x+2)=a{x}^2+bx+c=0$ and $\alpha ,\beta$ are such that $\alpha <-2<\beta$, then for the equation $f\left(x\right)=A{x}^2+Bx+C$.

• A. $AC<0$
• B. $AC>0$
• C. $B^2-4AC>0$
• D. $B^2-4AC<0$

Answer 1

The correct option is C $B^2-4AC>0$

We are given

$f(x+2)=a{x}^2+bx+c=0\to \left(1\right)$

have roots $\alpha$ & $\beta$ & $\alpha <-2<\beta$ it show thta $\alpha$ & $\beta$ are real and distinct roots So, $b^2-4ac>0$

Now, we also given,

$f\left(x\right)=A{x}^2+Bx+C\to \left(2\right)$

In the equation $\left(1\right)$ let put $x+2\to$

So in the left hand side we have to put $x\to x-2$

$\therefore f\left(x\right)=a(x-2{)}^2+b(x-2)+c$

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

$=a{x}^2-4ax+4a+bx-2b+c$

$f\left(x\right)=a{x}^2+(-4a+b)+x+4a-2b+1---\left(3\right)$

Now, compare equation $\left(2\right)$ and $\left(3\right)$ we get

$A=a;B=-4a+b$ & $C=4a-2b+c$

If both $\left(2\right)$ & $\left(3\right)$ equation are equal then the roots of the equation $\left(3\right)$ must be real & distinct

So, $B^2-4AC>0$

$\therefore (-4a+b{)}^2-4(a\left)\right(4a-2b+c)>0$

$\therefore +16a^2-8ab+b^2-16a^2+8ab-4ac>0$

$(16-16)a^2+(-8+8)ab+(b^2-4ac)>0$

$\therefore b^2-4ac>0$

Hence it proof that for equation $f\left(x\right)=A{x}^2+Bx+C$

$B^2-4AC>0$