Find c, if the quadratic equation $x^{2} - 2 (c + 1) x + c^{2} = 0$ has real and equal roots.

c = \frac{1}{2}

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

c = - \frac{1}{2}

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

c = 2

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

c = - 2

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

Open in App

Solution

The correct option is B $c = - \frac{1}{2}$
Since, the roots are real and equal, $D = 0$
$D = \sqrt{b^{2} - 4 a c = 0}$
$(- 2 (c + 1))^{2} - 4 c^{2} = 0$
$4 (c^{2} + 2 c + 1) - 4 c^{2} = 0$
$8 c + 4 = 0$
$∴ c = \frac{- 1}{2}$

Suggest Corrections

Similar questions

For every real number c $\geq 0$ , find all the complex number z which satisfy the equation

$| z |^{2}$ - 2iz + 2c(1 + i) = 0

(1 + m^{2}) x^{2} + 2 m c x + c^{2} - a^{2} = 0

c^{2} = a^{2} (1 + m^{2})

p

(3 p + 1) c^{2} + 2 (p + 1) c + p = 0

x^{2} + b_{1} x + c_{1} = 0

x^{2} + b_{2} x + c_{2} = 0

b_{1} b_{2} = 2 (c_{1} + c_{2})

b_{1}, b_{2}, c_{1}, c_{2} \in R,

x^{2} - m x - 4 x + 1 = 0

m

(- \infty, - 6)

(- \infty, - 3)

(- 2, \infty)

(2, \infty)

Join BYJU'S Learning Program