In the equation x^3 + 3Hx + G = 0,if G and H | vedclass

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

Question

(C) The given cubic equation is $x^3 + 3Hx + G = 0$,where $G$ and $H$ are real constants.For a cubic equation of the form $x^3 + px + q = 0$,the discr

Vedclass · Accepted Answer

One real and two imaginary

Vedclass · Answer

(C) The given cubic equation is $x^3 + 3Hx + G = 0$,where $G$ and $H$ are real constants.For a cubic equation of the form $x^3 + px + q = 0$,the discriminant is given by $\Delta = - (4p^3 + 27q^2)$.In our case,$p = 3H$ and $q = G$.Substituting these,we get $\Delta = - (4(3H)^3 + 27G^2) = - (108H^3 + 27G^2) = -27(4H^3 + G^2)$.Given that $G^2 + 4H^3 > 0$,it follows that $\Delta = -27(G^2 + 4H^3) For a cubic equation with real coefficients,if the discriminant $\Delta < 0$,the equation has one real root and two complex conjugate (imaginary) roots.

In the equation $x^3 + 3Hx + G = 0$,if $G$ and $H$ are real and $G^2 + 4H^3 > 0$,then the roots are

Similar Questions

Let $\alpha \neq \beta$ satisfy $\alpha^2+1=6 \alpha$ and $\beta^2+1=6 \beta$. Then,the quadratic equation whose roots are $\frac{\alpha}{\alpha+1}$ and $\frac{\beta}{\beta+1}$ is

If $x$ is real,then the value of $x^2 - 6x + 13$ will not be less than

The number of distinct real solutions of the equation $x|x+4|+3|x+2|+10=0$ is

The roots of the equations $2x^2 - 5x + 1 = 0$ and $x^2 + 5x + 2 = 0$ are:

The midpoint of the diagonal of a rectangle formed by $x^2+5x-6=0$ and $y^2-8y-20=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module