The product of all real roots of the equation | vedclass

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

Question

(A) Given equation is ${x^2} - |x| - 6 = 0$.Case $1$: If $x \ge 0$,then $|x| = x$. The equation becomes ${x^2} - x - 6 = 0$.Factoring the quadratic: $

Vedclass · Accepted Answer

$-9$

Vedclass · Answer

(A) Given equation is ${x^2} - |x| - 6 = 0$.Case $1$: If $x \ge 0$,then $|x| = x$. The equation becomes ${x^2} - x - 6 = 0$.Factoring the quadratic: $(x - 3)(x + 2) = 0$.This gives $x = 3$ or $x = -2$. Since we assumed $x \ge 0$,we accept $x = 3$.Case $2$: If $x Factoring the quadratic: $(x + 3)(x - 2) = 0$.This gives $x = -3$ or $x = 2$. Since we assumed $x The real roots of the equation are $3$ and $-3$.The product of all real roots is $3 	imes (-3) = -9$.

The product of all real roots of the equation ${x^2} - |x| - 6 = 0$ is

Similar Questions

The number of real solutions of the equation $(\frac{3}{2})^x = -x^2 + 5x - 10$ is:

If $ax^2 + bx + c = 0$ and $bx^2 + cx + a = 0$ have a common root and $a, b, c$ are non-zero real numbers,then $\frac{a^3 + b^3 + c^3}{abc} = $

The roots of the equation $ix^2 - 4x - 4i = 0$ are

If $a, b, c$ are distinct real numbers and $a^3 + b^3 + c^3 = 3abc$,then the equation $ax^2 + bx + c = 0$ has two roots,out of which one root is

Solve the given two equations and select the correct answer from the given options.
$I.$ $x = \sqrt[3]{2197}$
$II.$ $2y^2 - 54y + 364 = 0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module