Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$x^{2} - 4 x - 1 = 0$

Open in App

Solution

$Given : x^{2} - 4 x - 1 = 0 On c omparing it with a x^{2} + b x + c = 0, we get : a = 1, b = - 4 and c = - 1 Discriminant D is given by : D = (b^{2} - 4 a c) = {(- 4)}^{2} - 4 \times 1 \times (- 1) = 16 + 4 = 20 = 20 > 0 Hence, the roots of the equation are real. Roots α and β are given by : α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 4) + \sqrt{20}}{2 \times 1} = \frac{4 + 2 \sqrt{5}}{2} = \frac{2 (2 + \sqrt{5})}{2} = (2 + \sqrt{5}) β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 4) - \sqrt{20}}{2} = \frac{4 - 2 \sqrt{5}}{2} = \frac{2 (2 - \sqrt{5})}{2} = (2 - \sqrt{5}) Thus, the roots of the equation are (2 + \sqrt{5}) and (2 - \sqrt{5}) .$

Suggest Corrections

Similar questions

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:
$x^{2} + 5 x - (a^{2} + a - 6) = 0$

x^{2} + x + 2 = 0

x^{2} - (\sqrt{3} + 1) x + \sqrt{3} = 0

x^{2} - 2 a x - (4 b^{2} - a^{2}) = 0

x^{2} - 4 a x - b^{2} + 4 a^{2} = 0

Join BYJU'S Learning Program