Product of all real roots of equation {x^2} - |x|

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4-2.Quadratic Equations and Inequations

medium

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

A$-9$

B$6$

C$9$

D$36$

Solution

(a) Given equation ${x^2} – |x| – 6 = 0$
If $x > 0$, equation is ${x^2} – x – 6 = 0$
==> $(x – 3)(x + 2) = 0$
==> $x = 3,\,x = – 2$
==> $x = 3$
If $x < 0$, equation is ${x^2} + x – 6 = 0$
==> $(x + 3)(x – 2) = 0$
==> $x = – 3,\,x = 2$
==> $x = – 3$
Hence product of all possible real roots $= -9.$

Standard 11

Mathematics

If $x$ is real, the expression $\frac{{x + 2}}{{2{x^2} + 3x + 6}}$ takes all value in the interval

hard

(IIT-1969)

If $\alpha, \beta$ are the roots of the equation, $x^2-x-1=0$ and $S_n=2023 \alpha^n+2024 \beta^n$, then

hard

(JEE MAIN-2024)

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

normal

If $\alpha ,\beta$ are the roots of $x^2 -ax + b = 0$ and if $\alpha^n + \beta^n = V_n$, then –

normal

The sum of all the real roots of the equation $\left( e ^{2 x }-4\right)\left(6 e ^{2 x }-5 e ^{ x }+1\right)=0$ is

hard

(JEE MAIN-2022)

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

Solution

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