The roots of the quadratic equation $x^{2} - 2 (a + b) x + 2 (a^{2} + b^{2}) = 0 (a \neq b)$ are

real and distinct

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

rational

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

real and equal

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

imaginary

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

Open in App

Solution

The correct option is D imaginary
$x^{2} - 2 (a + b) x + 2 (a^{2} + b^{2}) = 0$

$Δ = b^{2} - 4 a c = 4 (a + b)^{2} - 8 (a^{2} + b^{2})$
$= 4 a^{2} + 4 b^{2} + 8 a b - 8 a^{2} - 8 b^{2}$
$= - 4 a^{2} - 4 b^{2} + 8 a b$
$= - 4 (a - b)^{2} < 0$

Hence, roots are imaginary.

Suggest Corrections

Similar questions

Q. Show that the roots of the equation

x^{2} + 2 (a + b) x + 2 (a^{2} + b^{2}) = 0

are unreal.

The roots of a quadratic equation $a x^{2} + b x + c = 0$ are given by $\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ , provided $b^{2} - 4 a c \geq 0$ .

Q. If a and b are distinct real numbers, show that the quadratic equation

2 (a^{2} + b^{2}) x^{2} + 2 (a + b) x + 1 = 0

has no real roots.

If b² – 4ac = 0 then The roots of the Quadratic equation ax² + bx + c = 0 are given by :

Q. Find the discriminant for the following quadratic equation. Also, determine the nature of the roots.

2 (a^{2} + b^{2}) x^{2} + 2 (a + b) x + 1 = 0; a \neq b

Join BYJU'S Learning Program

The roots of the quadratic equation x2−2(a+b)x+2(a2+b2)=0 (a≠b) are

The correct option is D imaginaryx2−2(a+b)x+2(a2+b2)=0 Δ=b2−4ac=4(a+b)2−8(a2+b2) =4a2+4b2+8ab−8a2−8b2 =−4a2−4b2+8ab =−4(a−b)2<0 Hence, roots are imaginary.

The roots of the quadratic equation $x^{2} - 2 (a + b) x + 2 (a^{2} + b^{2}) = 0 (a \neq b)$ are

The correct option is D imaginary
$x^{2} - 2 (a + b) x + 2 (a^{2} + b^{2}) = 0$

$Δ = b^{2} - 4 a c = 4 (a + b)^{2} - 8 (a^{2} + b^{2})$
$= 4 a^{2} + 4 b^{2} + 8 a b - 8 a^{2} - 8 b^{2}$
$= - 4 a^{2} - 4 b^{2} + 8 a b$
$= - 4 (a - b)^{2} < 0$

Hence, roots are imaginary.