The number of distinct real roots of the equa | vedclass

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

Question

(A) We analyze the equation $|x||x+2|-5|x+1|-1=0$ by considering different intervals for $x$:Case $1$: $x \geq 0$The equation becomes $x(x+2)-5(x+1)-1

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) We analyze the equation $|x||x+2|-5|x+1|-1=0$ by considering different intervals for $x$:Case $1$: $x \geq 0$The equation becomes $x(x+2)-5(x+1)-1=0 \implies x^2+2x-5x-5-1=0 \implies x^2-3x-6=0$.The roots are $x = \frac{3 \pm \sqrt{9+24}}{2} = \frac{3 \pm \sqrt{33}}{2}$. Since $x \geq 0$,we take $x = \frac{3+\sqrt{33}}{2}$. (One root)Case $2$: $-1 \leq x The equation becomes $(-x)(x+2)-5(x+1)-1=0 \implies -x^2-2x-5x-5-1=0 \implies x^2+7x+6=0$.$(x+6)(x+1)=0 \implies x=-6, -1$. In the range $[-1, 0)$,only $x=-1$ is valid. (One root)Case $3$: $-2 \leq x The equation becomes $(-x)(x+2)-5(-(x+1))-1=0 \implies -x^2-2x+5x+5-1=0 \implies x^2-3x-4=0$.$(x-4)(x+1)=0 \implies x=4, -1$. Neither value lies in the interval $[-2, -1)$. (No root)Case $4$: $x The equation becomes $(-x)(-(x+2))-5(-(x+1))-1=0 \implies x^2+2x+5x+5-1=0 \implies x^2+7x+4=0$.The roots are $x = \frac{-7 \pm \sqrt{49-16}}{2} = \frac{-7 \pm \sqrt{33}}{2}$.Both $\frac{-7+\sqrt{33}}{2} \approx -0.63$ and $\frac{-7-\sqrt{33}}{2} \approx -6.37$ are candidates. Since $x Total distinct real roots: $1+1+0+1 = 3$.

The number of distinct real roots of the equation $|x||x+2|-5|x+1|-1=0$ is:

Similar Questions

If one root of the equation $x(x + 2) = 3 - ax^2$ approaches infinity,then the value of $a$ approaches which of the following?

Let $[x]$ be the greatest integer less than or equal to $x$ for a real number $x$. Then the equation $[x^2] = x + 1$ has:

If the equation $x^2 + \lambda x + \mu = 0$ has equal roots and one root of the equation $x^2 + \lambda x - 12 = 0$ is $2$,then $(\lambda, \mu) = $

If $a, b$ and $c$ are in arithmetic progression, then the roots of the equation $a x^{2}-2 b x+c=0$ are

If the roots of $x^4-10x^3+37x^2-60x+36=0$ are $\alpha, \alpha, \beta, \beta$ with $\alpha < \beta$,then find the value of $2\alpha+3\beta-2\alpha\beta$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module