If a, b, c ∈ R are such that 4 a+2 b+ c0 and a x2+b x+ c=0 has no real roots, then the value of (c+ a)(c +b) is

Question

If a, b, c ∈ R are such that 4 a+2 b+ c>0 and a x2+b x+ c=0 has no real roots, then the value of (c+ a)(c +b) is (A) greater than ab (B) less than ab (C) greater than ca (D) less than ab + bc + ca

Tardigrade Admin · Accepted Answer

Since, the equation a x2+b x +c=0 have no real roots and 4 a+2 b+ c>0 ∴ a +b +c >0 ∵ a x2+b x+ c>0, ∀ x ∈ R So, c+ a>-b and c+b>-a ⇒ (c +a)(c +b) >a b

Q. If $a, b, c \in R$ are such that $4 a + 2 b + c > 0$ and $a x^{2} + b x + c = 0$ has no real roots, then the value of $(c + a) (c + b)$ is

greater than ab

less than ab

greater than ca

less than ab + bc + ca

Since, the equation $a x^{2} + b x + c = 0$
have no real roots and $4 a + 2 b + c > 0$
$∴ a + b + c > 0$
${∵ a x^{2} + b x + c > 0, \forall x \in R}$
So, $c + a > - b$ and $c + b > - a$
$\Rightarrow (c + a) (c + b) > ab$

Q. If a,b,c∈R are such that 4a+2b+c>0 and ax2+bx+c=0 has no real roots, then the value of (c+a)(c+b) is