Exact set of values of a for which {x^3} | vedclass

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

Question

Equation becomes $4 a^{2}+6 a x-\left(x^{4}+x^{3}-2 x^{2}\right)=0$

$a=\frac{-6 x \pm \sqrt{36 x^{2}+16\left(x^{4}+x^{3}-2 x^{2}\right)}}{8}$

$a

Vedclass · Accepted Answer

$\left[ { - \frac{1}{8},\frac{1}{2}} \right]$

Vedclass · Answer

Equation becomes $4 a^{2}+6 a x-\left(x^{4}+x^{3}-2 x^{2}\right)=0$

$a=\frac{-6 x \pm \sqrt{36 x^{2}+16\left(x^{4}+x^{3}-2 x^{2}\right)}}{8}$

$a=\frac{-x^{2}}{2}-x$ and $a=\frac{x^{2}}{2}-\frac{x}{2}$

$\mathrm{x}^{2}+2 \mathrm{x}+2 \mathrm{a}=0$ or $\mathrm{x}^{2}-\mathrm{x}-2 \mathrm{a}=0$

$\mathrm{D}_{1} \geq 0$  and $ \mathrm{D}_{2} \geq 0$

$\mathrm{a} \in\left[-\frac{1}{8}, \frac{1}{2}\right]$

Exact set of values of $a$ for which ${x^3}(x + 1) = 2(x + a)(x + 2a)$ is having four real solutions is

Similar Questions

If $\alpha , \beta , \gamma$ are roots of equation $x^3 + qx -r = 0$ then the equation, whose roots are

$\left( {\beta \gamma + \frac{1}{\alpha }} \right),\,\left( {\gamma \alpha + \frac{1}{\beta }} \right),\,\left( {\alpha \beta + \frac{1}{\gamma }} \right)$

Let $a, b, c, d$ be numbers in the set $\{1,2,3,4,5,6\}$ such that the curves $y=2 x^3+a x+b$ and $y=2 x^3+c x+d$ have no point in common. The maximum possible value of $(a-c)^2+b-d$ is

If $S$ is a set of $P(x)$ is polynomial of degree $ \le 2$ such that $P(0) = 0,$$P(1) = 1$,$P'(x) > 0{\rm{ }}\forall x \in (0,\,1)$, then

If $a, b, c, d$ and $p$ are distinct real numbers such that $(a^2 + b^2 + c^2)\,p^2 -2p\, (ab + bc + cd) + (b^2 + c^2 + d^2) \le 0$, then