Number of values of $x$ satisfying the equation $x^{2} - 2 x = 3$ .

0

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

1

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

3

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

2

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

Open in App

Solution

The correct option is B $2$
Consider the given equation.
$x^{2} - 2 x = 3$

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

$(x - 1)^{2} = 4$

$x - 1 = \pm 2$

$x = - 1, 3$

Hence, number of values of $x$ is $2$ .

Suggest Corrections

Similar questions

^{'} x^{'}

(x^{2} . 2^{x + 1} + 2^{| x - 3 | + 2}) = (x^{2} . 2^{| x - 3 | + 4} + 2^{x - 1})

[0, 3 π]

2 {sin}^{2} x + 5 sin x - 3 = 0

x ϵ [0, π]

3^{s i n 2 x + 2 c o s^{2} x} + 3^{1 - s i n 2 x + 2 s i n^{2} x} = 28

x

The number of polynomials P(x) satisfying the equation P(x²) + 2x² + 10x = 2x. P(x + 1) + 3 is

\int_{- 1}^{x} (8 t^{2} + \frac{28}{3} t + 4) d t = \frac{(\frac{3}{2}) x + 1}{{log}_{x + 1} \sqrt{x + 1}} .

Join BYJU'S Learning Program