If P(x)=a x2+b x+c and Q(x)=-a x2+b x+c, where a c ≠ 0, then P(x) Q(x) has atleast two real roots.

Question

If P(x)=a x2+b x+c and Q(x)=-a x2+b x+c, where a c ≠ 0, then P(x) Q(x) has atleast two real roots. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

P(x) ⋅ Q(x)=(a x2+b x+c)(-a x2+b x+c) Now, D1=b2-4 a c and D2=b2+4 a c Clearly, D1+D2=2 b2 ≥ 0 ∴ Atleast one of D1 and D2 is (+ ve). Hence, atleast two real roots. Hence, statement is true.

Q. If $P (x) = a x^{2} + b x + c$ and $Q (x) = - a x^{2} + b x + c$ , where $a c \neq = 0$ , then $P (x) Q (x)$ has atleast two real roots.

$P (x) \cdot Q (x) = (a x^{2} + b x + c) (- a x^{2} + b x + c)$
Now, $D_{1} = b^{2} - 4 a c$ and $D_{2} = b^{2} + 4 a c$
Clearly, $D_{1} + D_{2} = 2 b^{2} \geq 0$
$∴$ Atleast one of $D_{1}$ and $D_{2}$ is (+ ve). Hence, atleast two real roots.
Hence, statement is true.

Q. If P(x)=ax2+bx+c and Q(x)=−ax2+bx+c, where ac=0, then P(x)Q(x) has atleast two real roots.

P(x)⋅Q(x)=(ax2+bx+c)(−ax2+bx+c) Now, D1​=b2−4ac and D2​=b2+4ac Clearly, D1​+D2​=2b2≥0 ∴ Atleast one of D1​ and D2​ is (+ ve). Hence, atleast two real roots. Hence, statement is true.

Q. If $P (x) = a x^{2} + b x + c$ and $Q (x) = - a x^{2} + b x + c$ , where $a c \neq = 0$ , then $P (x) Q (x)$ has atleast two real roots.

$P (x) \cdot Q (x) = (a x^{2} + b x + c) (- a x^{2} + b x + c)$
Now, $D_{1} = b^{2} - 4 a c$ and $D_{2} = b^{2} + 4 a c$
Clearly, $D_{1} + D_{2} = 2 b^{2} \geq 0$
$∴$ Atleast one of $D_{1}$ and $D_{2}$ is (+ ve). Hence, atleast two real roots.
Hence, statement is true.