Using quadratic formula solve the given quadratic equation
$9 x^{2} - 9 (a + b) + (2 a^{2} + 5 a b + 2 b^{2}) = 0$

Open in App

Solution

We have $9 x^{2} - 9 (a + b) + (2 a^{2} + 5 a b + 2 b^{2}) = 0$

Comparing this equation with $A x^{2} + B x + C = 0$ we have

$A = 9, B = - 9 (a + b), C = 2 a^{2} + 5 a b + 2 b^{2}$

$D = B^{2} - 4 A C \Rightarrow D = 81 {(a + b)}^{2} - 36 (2 a^{2} + 5 a b + 2 b^{2}) \Rightarrow D = 9 {(a - b)}^{2} \geq 0$

So, the roots of the given equation are real and are given by

$α = - \frac{B - \sqrt{D}}{2 A} = - \frac{9 (a + b) + 3 (a - b)}{18} = \frac{2 a + b}{3}$

$β = \frac{B - \sqrt{D}}{2 A} = - \frac{9 (a + b) - 3 (a - b)}{18} = \frac{a + 2 b}{3}$

Suggest Corrections

Similar questions

Solve the following quadratic equation using quadratic formula .

$9 x^{2} - 9 (a + b) x + (2 a^{2} + 5 a b + 2 b^{2}) = 0$

9 x^{2} - 9 (a + b) x + (2 a^{2} + 5 a b + 2 b^{2}) = 0

9 x^{2} - 9 (a + b) x + (2 a^{2} + 5 a b + 2 b^{2}) = 0

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

9 x^{2} - 9 (a + b) x + (2 a^{2} + 5 a + 2 b^{2})

9 x^{2} - 9 (a + b) x + (2 a 62 + 5 a b + 2 b^{2}) = 0

Join BYJU'S Learning Program