The condition that {x^3} - 3px + 2q may be di | vedclass

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

Question

(c) Given that ${x^2} + 2ax + {a^2}$ is a factor of

${x^3} - 3px + 2q = 0$

Let ${x^3} - 3px + 2q = ({x^2} + 2ax + {a^2})(x + \lambda )$,

wher

Vedclass · Accepted Answer

${p^3} = {q^2}$

Vedclass · Answer

(c) Given that ${x^2} + 2ax + {a^2}$ is a factor of

${x^3} - 3px + 2q = 0$

Let ${x^3} - 3px + 2q = ({x^2} + 2ax + {a^2})(x + \lambda )$,

where $\lambda $ is a constant.

Then equating the coefficients of like powers of x on both sides,

${x^3} - 3px + 2q = {x^3} + (2a + \lambda ){x^2} + ({a^2} + 2a\lambda )x + \lambda {a^2}$

==> $2a + \lambda = 0$$ \Rightarrow \lambda = - 2a$ &hellip;..$(i)$

and $ - 3p = {a^2} + 2a\lambda $ &hellip;..$(ii)$

and $2q = \lambda {a^2}$ &hellip;..$(iii)$

Put the value of $\lambda $ in $(iii),$

==> $2q = - 2{a^3} \Rightarrow q = - {a^3}$ ..&hellip;$(iv)$

Put the value of $\lambda $ in $(ii),$

==>$ - 3p = {a^2} + 2a( - 2a) = {a^2} - 4{a^2} = - 3{a^2}$

==> $ - 3p = - 3{a^2} \Rightarrow p = {a^2}$

==>$p = {( - q)^{2/3}}$

==> ${p^3} = {q^2}$.

The condition that ${x^3} - 3px + 2q$ may be divisible by a factor of the form ${x^2} + 2ax + {a^2}$ is

Similar Questions

In the real number system, the equation $\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1$ has

Let $a, b, c$ be non-zero real roots of the equation $x^3+a x^2+b x+c=0$. Then,

In a cubic equation coefficient of $x^2$ is zero and remaining coefficient are real has one root $\alpha = 3 + 4\, i$ and remaining roots are $\beta$ and $\gamma$ then $\alpha \beta \gamma$ is :-

If $x$ is real , the maximum value of $\frac{{3{x^2} + 9x + 17}}{{3{x^2} + 9x + 7}}$ is

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