Name: Introduction to Algebra
Uploaded: 2019-07-19
Duration: 11 min 18 s
Description: In this video, we have discussed basic concepts of Algebra. Having a clear idea of the concept of this chapter will helps you to solve problems which are related to this topic.

Question

Solve.

|x^{2} + 2 x| = x + 2

Accepted Answer

To solve: $|x^{2} + 2 x| = x + 2 . . . . . (i)$

We know that $|a x + b| = c x + d$ means that $a x + b = c x + d$ or $a x + b = - (c x + d)$

Thus, from equation $(i)$ , we have

$x^{2} + 2 x = x + 2$ or $x^{2} + 2 x = - (x + 2)$

$\Rightarrow x^{2} + x - 2 = 0 . . . . (i i)$ or $x^{2} + 3 x + 2 = 0 . . . . (i i i)$

Now, we need to solve equation $(i i)$

Consider

$x^{2} + x - 2 = 0$

$\Rightarrow x = \frac{- 1 \pm \sqrt{1 - 4 (1) (- 2)}}{2} = \frac{- 1 \pm \sqrt{9}}{2} = \frac{- 1 \pm 3}{2}$

$\Rightarrow x = - 2$ or $x = 1$

If $x = - 2$ is a solution for the given equation then it must satisfy that equation i.e., $|x^{2} + 2 x| = x + 2$

L.H.S. $= |x^{2} + 2 x| = |{(- 2)}^{2} + 2 (- 2)| = |4 - 4| = 0$

R.H.S. $= - 2 + 2 = 0$

Since L.H.S. $=$ R.H.S.

Thus, $x = - 2$ is the solution for the given equation.

Similarly, let us check for $x = 1$

L.H.S. $= |x^{2} + 2 x| = |{(1)}^{2} + 2 (1)| = |3| = 3$

R.H.S. $= 1 + 2 = 3$

$\Rightarrow x = 1$ is also the solution for the given equation

Now, we need to solve equation $(i i i)$

Consider

$x^{2} + 3 x + 2 = 0$

$\Rightarrow x = \frac{- 3 \pm \sqrt{9 - 8}}{2} = \frac{- 3 \pm 1}{2}$

$\Rightarrow x = - 2$ or $x = - 1$

Checking whether $x = - 1$ satisfies the given equation or not

L.H.S. $= |x^{2} + 2 x| = |{(- 1)}^{2} + 2 (- 1)| = |- 1| = 1$

R.H.S. $= - 1 + 2 = 1$

Since L.H.S. $=$ R.H.S.

Thus, $x = - 1$ is also the solution for the given equation.

Hence, the solutions for the given equation are $- 2, \pm 1$

Solve.
$|x^{2} + 2 x| = x + 2$

Important Questions on Algebra

Learn and Prepare for any exam you want !

Solve. x2+2x=x+2

Important Questions on Algebra

Learn and Prepare for any exam you want !

Solve.
$|x^{2} + 2 x| = x + 2$