The number of values of x such that $x + 1, 2 x + 5, 3 x + 2$ are the sides of a right angled triangle is

1

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

3

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

4

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

Open in App

Solution

The correct option is A $2$
We have, 3 side of triangle
$(x + 1), (2 x + 5), (3 x + 2)$

Therefore, $(2 + 5)^{2} = (3 x + 2)^{2} + (x + 1)^{2}$

$\Rightarrow 4 x^{2} + 25 + 20 x = 9 x^{2} + 4 + 12 x + x^{2} + 1 + 2 x$

$4 x^{2} + 25 + 20 x = 10 x^{2} + 14 x + 5$

$6 x^{2} - 6 x - 20 = 0$

$3 x^{2} - 3 x - 10 = 0$

$x = + 3 \pm \frac{\sqrt{9 - 4 \times 3 \times (- 10)}}{6}$

$x = \frac{3 \pm \sqrt{9 + 120}}{6}$

$x = \frac{3 \pm \sqrt{129}}{6}$

Hence, 2 values of x.

Suggest Corrections

Similar questions

x^{2} + x + 1, 2 x + 1

x^{2} - 1

x^{2} + x + 1, 2 x + 1, x^{2} - 1

x^{2} + x + 1, 2 x + 1

x^{2} - 1

120^{\circ}

A B C,

C,

x

{sin}^{- 1} (x) = {sin}^{- 1} (\frac{a x}{c}) + {sin}^{- 1} (\frac{b x}{c})

a, b, c

x, x + 1, 2 x - 1

x \sqrt{10}

x

Join BYJU'S Learning Program