The number of real roots of the equation x |x | vedclass

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

Question

(B) We analyze the equation $x|x| - 5|x + 2| + 6 = 0$ by considering different intervals for $x$.Case $1$: $x \ge 0$.The equation becomes $x^2 - 5(x +

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) We analyze the equation $x|x| - 5|x + 2| + 6 = 0$ by considering different intervals for $x$.Case $1$: $x \ge 0$.The equation becomes $x^2 - 5(x + 2) + 6 = 0$,which simplifies to $x^2 - 5x - 4 = 0$.The roots are $x = \frac{5 \pm \sqrt{25 - 4(1)(-4)}}{2} = \frac{5 \pm \sqrt{41}}{2}$.Since $x \ge 0$,we accept $x = \frac{5 + \sqrt{41}}{2}$. ($1$ root)Case $2$: $-2 \le x The equation becomes $-x^2 - 5(x + 2) + 6 = 0$,which simplifies to $-x^2 - 5x - 4 = 0$,or $x^2 + 5x + 4 = 0$.Factoring gives $(x + 1)(x + 4) = 0$,so $x = -1$ or $x = -4$.Since $-2 \le x Case $3$: $x The equation becomes $-x^2 - 5(-(x + 2)) + 6 = 0$,which simplifies to $-x^2 + 5x + 10 + 6 = 0$,or $x^2 - 5x - 16 = 0$.The roots are $x = \frac{5 \pm \sqrt{25 - 4(1)(-16)}}{2} = \frac{5 \pm \sqrt{89}}{2}$.Since $x Total number of real roots is $1 + 1 + 1 = 3$.

The number of real roots of the equation $x |x| - 5|x + 2| + 6 = 0$ is:

Similar Questions

The roots of the equation $x^4 - 2x^3 + x = 380$ are

The number of integral solutions of $2(x^2 + \frac{1}{x^2}) - 7(x + \frac{1}{x}) + 9 = 0$ for $x \neq 0$ is:

The sum of all the real values of $x$ satisfying the equation $(x^2-7x+11)^{x^2-6x-7}=1$ is

$A$ building construction work can be completed by two masons $A$ and $B$ together in $22.5$ days. Mason $A$ alone can complete the work in $24$ days less than mason $B$ alone. Then mason $A$ alone will complete the work in: (in $days$)

Solve the equation,$3^{x^2-x}=25-4^{x^2-x}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module