The roots of $(p^{2} + q^{2}) x^{2} + 2 (q^{2} + r^{2}) x + (q^{2} + r^{2}) = 0$ are equal. Which of the following must be true?

q = r = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

r = p

No worries! We‘ve got your back. Try BYJU‘S free classes today!

r = -p

No worries! We‘ve got your back. Try BYJU‘S free classes today!

At least one of (1), (2) and (3)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D At least one of (1), (2) and (3)
For the equation $a x^{2} + b x + c = 0$ , to have equal roots, the discriminant $b^{2} - 4 a c$ should be $= 0$
So, ${(2 (q^{2} + r^{2}))}^{2} - 4 \times (p^{2} + q^{2}) \times (q^{2} + r^{2}) = 0$
$=> 4 {(q^{2} + r^{2})}^{2} - 4 \times (p^{2} + q^{2}) \times (q^{2} + r^{2}) = 0$
$=> 4 (q^{2} + r^{2}) [(q^{2} + r^{2}) - (p^{2} + q^{2})] = 0$
$=> 4 (q^{2} + r^{2}) [r^{2} - p^{2}] = 0$
$=> q^{2} + r^{2} = 0$ or $r^{2} - p^{2} = 0$
When $q^{2} + r^{2} = 0$ then $q = r = 0$
When $r^{2} - p^{2} = 0$ , then $(r + p) = 0 o r (r - p) = 0$
$=> r = - p o r r = p$
So option D is correct.

Suggest Corrections

Similar questions

(p^{2} - q^{2}) x^{2} + (q^{2} - r^{2}) x + r^{2} - p^{2} = 0

(p^{2} - q^{2}) y^{2} + (r^{2} - p^{2}) y + q^{2} - r^{2} = 0

(p^{2} + q^{2}) x^{2} - 2 (p r + q s) x + r^{2} + s^{2} = 0

p, q a n d r

(p \neq q, r \neq 0)

\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}

Using quadratic formula solve the following quadratic equation:

$p^{2} x^{2} + (p^{2} - q^{2}) x - q^{2} = 0, p \neq 0$

p^{2} + q^{2} + r^{2} = - 2

p, q, r \in C

g (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + p^{2} x & (1 + q^{2}) x & (1 + r^{2} x) (1 + p^{2}) x & 1 + q^{2} x & (1 + r^{2}) x (1 + p^{2}) x & (1 + q^{2}) x & 1 + r^{2} x \end{matrix} ∣ ∣ ∣ ∣ ∣

x \in R,

4 \int 1 g (x) d x

Join BYJU'S Learning Program