If x is a rational number and frac{(x+1)^{3}- | vedclass

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

Question

(B) Given the equation: $\frac{(x+1)^{3}-(x-1)^{3}}{(x+1)^{2}-(x-1)^{2}}=2$Using algebraic identities $(a+b)^3 - (a-b)^3 = 2b^3 + 6a^2b$ and $(a+b)^2

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Given the equation: $\frac{(x+1)^{3}-(x-1)^{3}}{(x+1)^{2}-(x-1)^{2}}=2$Using algebraic identities $(a+b)^3 - (a-b)^3 = 2b^3 + 6a^2b$ and $(a+b)^2 - (a-b)^2 = 4ab$:Numerator: $(x+1)^3 - (x-1)^3 = 2(1)^3 + 6(x)^2(1) = 2 + 6x^2$Denominator: $(x+1)^2 - (x-1)^2 = 4(x)(1) = 4x$Substituting these into the equation:$\frac{6x^2 + 2}{4x} = 2$Simplify the fraction by dividing by $2$:$\frac{3x^2 + 1}{2x} = 2$Multiply both sides by $2x$:$3x^2 + 1 = 4x$Rearrange into a standard quadratic equation:$3x^2 - 4x + 1 = 0$Factor the quadratic equation:$(3x - 1)(x - 1) = 0$This gives two possible values for $x$: $x = 1/3$ or $x = 1$.If $x = 1/3$,the sum of the numerator and denominator is $1 + 3 = 4$.If $x = 1$,the sum of the numerator and denominator is $1 + 1 = 2$ (not in options).Thus,the correct value is $4$.

If $x$ is a rational number and $\frac{(x+1)^{3}-(x-1)^{3}}{(x+1)^{2}-(x-1)^{2}}=2$,then the sum of the numerator and denominator of $x$ is

Similar Questions

Let $p, q \in \{1, 2, 3, 4\}$. The number of equations of the form $px^2 + qx + 1 = 0$ having real roots is

Sum of roots is $-1$ and sum of their reciprocals is $\frac{1}{6}$,then the equation is

If the sum of the roots of the quadratic equation $ax^2 + bx + c = 0$ is equal to the sum of the squares of their reciprocals,then $a/c, b/a, c/b$ are in

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2}-16x+63=0$
$II.$ $y^{2}-2y-35=0$

The sum of the solutions of the equation $|\sqrt{x} - 2| + \sqrt{x}(\sqrt{x} - 4) + 2 = 0$ $(x > 0)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module