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

(C) We analyze the equation $x|x-2|+3|x-3|+1=0$ in three cases:Case $(I): x < 2$The equation becomes $x(2-x) + 3(3-x) + 1 = 0$$-x^2 + 2x + 9 - 3x + 1

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) We analyze the equation $x|x-2|+3|x-3|+1=0$ in three cases:Case $(I): x The equation becomes $x(2-x) + 3(3-x) + 1 = 0$$-x^2 + 2x + 9 - 3x + 1 = 0$$-x^2 - x + 10 = 0 \Rightarrow x^2 + x - 10 = 0$The roots are $x = \frac{-1 \pm \sqrt{1 - 4(1)(-10)}}{2} = \frac{-1 \pm \sqrt{41}}{2}$.Since $\sqrt{41} \approx 6.4$,$x_1 = \frac{5.4}{2} = 2.7$ (not in range) and $x_2 = \frac{-7.4}{2} = -3.7$ (in range).So,$1$ root here.Case $(II): 2 \leq x The equation becomes $x(x-2) + 3(3-x) + 1 = 0$$x^2 - 2x + 9 - 3x + 1 = 0$$x^2 - 5x + 10 = 0$The discriminant $D = (-5)^2 - 4(1)(10) = 25 - 40 = -15 No real roots in this interval.Case $(III): x \geq 3$The equation becomes $x(x-2) + 3(x-3) + 1 = 0$$x^2 - 2x + 3x - 9 + 1 = 0$$x^2 + x - 8 = 0$The roots are $x = \frac{-1 \pm \sqrt{1 - 4(1)(-8)}}{2} = \frac{-1 \pm \sqrt{33}}{2}$.Since $\sqrt{33} \approx 5.74$,$x_3 = \frac{4.74}{2} = 2.37$ (not in range) and $x_4 = \frac{-6.74}{2} = -3.37$ (not in range).No real roots in this interval.Thus,there is only $1$ real root.

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

Similar Questions

For which interval of values of $a$ will the equation $3x^2 + 2(a^2 + 1)x + (a^2 - 3a + 2) = 0$ have roots of opposite signs?

The roots of the equation $\sqrt{3x + 1} + 1 = \sqrt{x}$ are

Number of equations of the form $ax^2 + bx + 1 = 0$ having real roots,where $a, b \in \{1, 2, 3, 4\}$ is-

If the roots of the equation $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}$ are equal in magnitude but opposite in sign,then the product of the roots will be

The solution of the equation $2x^3 - x^2 - 22x - 24 = 0$,given that two of the roots are in the ratio $3:4$,is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module