If (3+2 sqrt{2})^{x^2-4}+(3-2 sqrt{2})^{x^2-4 | vedclass

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

Question

(D) Given $(3+2 \sqrt{2}) \cdot (3-2 \sqrt{2}) = 9-8 = 1$. Thus,$(3-2 \sqrt{2}) = \frac{1}{3+2 \sqrt{2}}$. Let $y = (3+2 \sqrt{2})^{x^2-4}$. The equat

Vedclass · Accepted Answer

$35$

Vedclass · Answer

(D) Given $(3+2 \sqrt{2}) \cdot (3-2 \sqrt{2}) = 9-8 = 1$. Thus,$(3-2 \sqrt{2}) = \frac{1}{3+2 \sqrt{2}}$. Let $y = (3+2 \sqrt{2})^{x^2-4}$. The equation becomes $y + \frac{1}{y} = 6$,which simplifies to $y^2 - 6y + 1 = 0$. Solving for $y$ using the quadratic formula,$y = \frac{6 \pm \sqrt{36-4}}{2} = 3 \pm 2 \sqrt{2}$. Case $1$: $(3+2 \sqrt{2})^{x^2-4} = 3+2 \sqrt{2}$ $\Rightarrow x^2-4 = 1$ $\Rightarrow x^2 = 5$. Case $2$: $(3+2 \sqrt{2})^{x^2-4} = 3-2 \sqrt{2} = (3+2 \sqrt{2})^{-1}$ $\Rightarrow x^2-4 = -1$ $\Rightarrow x^2 = 3$. If $x^2 = 5$,then $x^4+x^2+5 = (5)^2 + 5 + 5 = 25+5+5 = 35$. If $x^2 = 3$,then $x^4+x^2+5 = (3)^2 + 3 + 5 = 9+3+5 = 17$. Since $35$ is the only option provided,the correct answer is $35$.

If $(3+2 \sqrt{2})^{x^2-4}+(3-2 \sqrt{2})^{x^2-4}=6$,then $x^4+x^2+5=$

Similar Questions

If $b_{1} b_{2} = 2(c_{1} + c_{2})$ and $b_{1}, b_{2}, c_{1}, c_{2}$ are all real numbers, then at least one of the equations $x^{2} + b_{1} x + c_{1} = 0$ and $x^{2} + b_{2} x + c_{2} = 0$ has

Consider a quadratic equation $ax^2 + 2bx + c = 0$ where $a, b, c$ are positive real numbers. If the equation has no real root, then which of the following is true?

If $\sin A, \sin B, \cos A$ are in $G.P.$,then the roots of $x^2 + 2x \cot B + 1 = 0$ are always ......

The value of $\lambda$ such that the sum of the squares of the roots of the quadratic equation $x^2 + (3 - \lambda)x + 2 = \lambda$ has the least value is

Let $\alpha, \beta$ be two roots of the equation $x^{2}+(20)^{\frac{1}{4}} x+(5)^{\frac{1}{2}}=0$. Then $\alpha^{8}+\beta^{8}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module