The number of real solutions of the equation | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) To find the number of real solutions for $x|x+3|+|x-1|-2=0$,we analyze the equation in three intervals based on the critical points $x=-3$ and $x=

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) To find the number of real solutions for $x|x+3|+|x-1|-2=0$,we analyze the equation in three intervals based on the critical points $x=-3$ and $x=1$: $I$. Case $x \ge 1$: $x(x+3) + (x-1) - 2 = 0 \implies x^2+3x+x-1-2=0 \implies x^2+4x-3=0$. Using the quadratic formula,$x = \frac{-4 \pm \sqrt{16 - 4(1)(-3)}}{2} = -2 \pm \sqrt{7}$. Since $x \ge 1$,both $-2+\sqrt{7} \approx 0.646$ and $-2-\sqrt{7} \approx -4.646$ are rejected. $II$. Case $-3 \le x $x(x+3) - (x-1) - 2 = 0 \implies x^2+3x-x+1-2=0 \implies x^2+2x-1=0$. Using the quadratic formula,$x = \frac{-2 \pm \sqrt{4 - 4(1)(-1)}}{2} = -1 \pm \sqrt{2}$. Since $-3 \le x $III$. Case $x $x(-(x+3)) - (x-1) - 2 = 0 \implies -x^2-3x-x+1-2=0 \implies -x^2-4x-1=0 \implies x^2+4x+1=0$. Using the quadratic formula,$x = \frac{-4 \pm \sqrt{16 - 4(1)(1)}}{2} = -2 \pm \sqrt{3}$. Since $x Thus,the valid solutions are $x = -1+\sqrt{2}$,$x = -1-\sqrt{2}$,and $x = -2-\sqrt{3}$. The total number of real solutions is $3$.

The number of real solutions of the equation $x|x+3|+|x-1|-2=0$ is

Similar Questions

If the roots of the equation $(p-3)x^2 + 2(p-3)x + 2p-5 = 0$ are real and distinct for $\alpha < p < \beta$ and $(\beta - \alpha)$ is maximum,then the extreme value of the quadratic expression $-(\alpha + \beta)x^2 + \alpha \beta x + (\alpha - \beta)$ is

The product of real roots of the equation $|x|^{6/5} - 26|x|^{3/5} - 27 = 0$ is:

Which of the following is/are always false?

The roots of the equation $|x - 2|^2 + |x - 2| - 6 = 0$ are:

The product of the roots of the equation $9x^{2}-18|x|+5=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module