The equation 16x^4 + 16x^3 - 4x - 1 = 0 has a | vedclass

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

Question

(D) Given the equation $f(x) = 16x^4 + 16x^3 - 4x - 1 = 0$. Since it has a multiple root,let $\alpha$ be the multiple root. Then $f(\alpha) = 0$ and $

Vedclass · Accepted Answer

$64$

Vedclass · Answer

(D) Given the equation $f(x) = 16x^4 + 16x^3 - 4x - 1 = 0$. Since it has a multiple root,let $\alpha$ be the multiple root. Then $f(\alpha) = 0$ and $f'(\alpha) = 0$. $f'(x) = 64x^3 + 48x^2 - 4$. Setting $f'(\alpha) = 0$ gives $16\alpha^3 + 12\alpha^2 - 1 = 0$. From $f(\alpha) = 16\alpha^4 + 16\alpha^3 - 4\alpha - 1 = 0$,we substitute $16\alpha^3 = 1 - 12\alpha^2$ to get $16\alpha^4 + (1 - 12\alpha^2) - 4\alpha - 1 = 0$,which simplifies to $16\alpha^4 - 12\alpha^2 - 4\alpha = 0$. Factoring gives $4\alpha(4\alpha^3 - 3\alpha - 1) = 0$. Since $\alpha 
eq 0$,we solve $4\alpha^3 - 3\alpha - 1 = 0$,which factors as $(\alpha - 1)(2\alpha + 1)^2 = 0$. Testing $\alpha = 1$ in $f(x)$ gives $16+16-4-1 
eq 0$. Thus,$\alpha = -\frac{1}{2}$ is the multiple root. Dividing $f(x)$ by $(x + \frac{1}{2})^2 = (x^2 + x + \frac{1}{4})$,we get $16x^2 - 4 = 0$,so $x = \pm \frac{1}{2}$. The roots are $\alpha = -\frac{1}{2}, \beta = -\frac{1}{2}, \gamma = -\frac{1}{2}, \delta = \frac{1}{2}$. Then $\frac{1}{\alpha^4} + \frac{1}{\beta^4} + \frac{1}{\gamma^4} + \frac{1}{\delta^4} = (-2)^4 + (-2)^4 + (-2)^4 + (2)^4 = 16 + 16 + 16 + 16 = 64$.

The equation $16x^4 + 16x^3 - 4x - 1 = 0$ has a multiple root. If $\alpha, \beta, \gamma, \delta$ are the roots of this equation,then $\frac{1}{\alpha^4} + \frac{1}{\beta^4} + \frac{1}{\gamma^4} + \frac{1}{\delta^4} =$

Similar Questions

Let $a$ be an integer such that all the real roots of the polynomial $2x^{5}+5x^{4}+10x^{3}+10x^{2}+10x+10$ lie in the interval $(a, a+1)$. Then,$|a|$ is equal to ...... .

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

If $x = \sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}$,then $x^2(x-4)^2$ is equal to :

If $l, m, n$ are real and $l \neq m$,then the roots of the equation $(l - m)x^2 - 5(l + m)x - 2(l - m) = 0$ are

If each root of the equation $2x^3 + ax^2 - 8x + b = 0$ is reduced by $1$,then in the transformed equation thus formed,the term containing $x^2$ and the constant term vanish. The roots of the original equation are

Vedclass Test Series

Exam Paper Generator

Online Exam Module