If 4a - frac{4}{a} + 3 = 0,then the value of | vedclass

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

Question

(B) Given equation: $4a - \frac{4}{a} + 3 = 0$.Divide the entire equation by $4$: $a - \frac{1}{a} + \frac{3}{4} = 0$,which implies $a - \frac{1}{a} =

Vedclass · Accepted Answer

$\frac{21}{64}$

Vedclass · Answer

(B) Given equation: $4a - \frac{4}{a} + 3 = 0$.Divide the entire equation by $4$: $a - \frac{1}{a} + \frac{3}{4} = 0$,which implies $a - \frac{1}{a} = -\frac{3}{4}$.We know the identity $x^{3} - y^{3} = (x - y)^{3} + 3xy(x - y)$.Let $x = a$ and $y = \frac{1}{a}$. Then $a^{3} - \frac{1}{a^{3}} = (a - \frac{1}{a})^{3} + 3(a)(\frac{1}{a})(a - \frac{1}{a})$.Substitute $a - \frac{1}{a} = -\frac{3}{4}$:$a^{3} - \frac{1}{a^{3}} = (-\frac{3}{4})^{3} + 3(-\frac{3}{4}) = -\frac{27}{64} - \frac{9}{4}$.To subtract,find a common denominator: $-\frac{27}{64} - \frac{144}{64} = -\frac{171}{64}$.Now,calculate $a^{3} - \frac{1}{a^{3}} + 3 = -\frac{171}{64} + 3 = \frac{-171 + 192}{64} = \frac{21}{64}$.

If $4a - \frac{4}{a} + 3 = 0$,then the value of $a^{3} - \frac{1}{a^{3}} + 3$ is equal to?

Similar Questions

The set of values of $a$ such that $x^2 - 2ax + a^2 - 6a \leqslant 0$ for all $x \in [1, 2]$ is:

The positive value of $m$ for which the roots of the equation $12x^2 + mx + 5 = 0$ are in the ratio $3:2$ is

Let $\alpha$ and $\beta$ be the roots of the equation $5x^{2} + 6x - 2 = 0$. If $S_{n} = \alpha^{n} + \beta^{n}$ for $n = 1, 2, 3, \ldots$,then:

The value of $p$ for which one root of the equation ${x^2} - 30x + p = 0$ is the square of the other,are

Vedclass Test Series

Exam Paper Generator

Online Exam Module