Name: Wavy Curve Method for Solving Quadratic Inequalities
Uploaded: 2019-07-19
Duration: 5 min 36 s
Description: In this interesting video, we will learn how to solve quadratic inequalities using a wavy curve method with interesting examples. Come, let's learn!

Question

Solve

|x^{2} + 4 x + 3| + 2 x + 5 = 0

.

Accepted Answer

Given $|x^{2} + 4 x + 3| + 2 x + 5 = 0$ $. . . (i)$

As we know that $|x| = \{\begin{cases} x : x \geq 0 \\ - x : x < 0 \end{cases}$

Case I:

Let, $x^{2} + 4 x + 3 \geq 0$ .

$(x + 1) (x + 3) \geq 0$

$x \in (- \infty, - 3] \cup [- 1, \infty)$ $. . . (ii)$

So for this interval, $x^{2} + 4 x + 3 \geq 0$ . Now using equation $(i)$ we get,

$|x^{2} + 4 x + 3| + 2 x + 5 = 0$

$(x^{2} + 4 x + 3) + 2 x + 5 = 0$

$x^{2} + 6 x + 8 = 0$

$(x + 2) (x + 4) = 0$

$x = - 2, - 4$ $. . . (iii)$

But for this case, our solution set will be the intersection of equation $(ii)$ and equation $(iii)$ .

Hence, $x \in \{- 4\}$ $. . . (iv)$

Case II:

Let, $x^{2} + 4 x + 3 < 0$ .

$(x + 1) (x + 3) < 0$

$x \in (- 3, - 1)$ $. . . (v)$

So for this interval, $x^{2} + 4 x + 3 < 0$ . Now using equation $(i)$ we get,

$|x^{2} + 4 x + 3| + 2 x + 5 = 0$

$- (x^{2} + 4 x + 3) + 2 x + 5 = 0$

$- x^{2} - 4 x - 3 + 2 x + 5 = 0$

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

$x = - 1 \pm \sqrt{3}$ $. . . (v i)$

But for this case, our solution set will be the intersection of equation $(v)$ and equation $(vi)$ . Hence, $x \in \{- 1 - \sqrt{3}\}$ $. . . (vii)$

Now, the final solution set of equation $|x^{2} + 4 x + 3| + 2 x + 5 = 0$ will be the union of equation $(iv)$ i.e., $x \in \{- 4\}$ and equation $(vii)$ i.e., $x \in \{- 1 - \sqrt{3}\}$ .

Hence, $x \in \{- 4, - 1 - \sqrt{3}\}$ .

Solve $|x^{2} + 4 x + 3| + 2 x + 5 = 0$ .

Important Questions on Essential Mathematical Tools

Learn and Prepare for any exam you want !

Solve x2+4x+3+2x+5=0.

Important Questions on Essential Mathematical Tools

Learn and Prepare for any exam you want !

Solve $|x^{2} + 4 x + 3| + 2 x + 5 = 0$ .