If x = sqrt{7} + sqrt{3} and xy = 4,then find | vedclass

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

Question

(B) Given $x = \sqrt{7} + \sqrt{3}$ and $xy = 4$. First,find $y$: $y = \frac{4}{x} = \frac{4}{\sqrt{7} + \sqrt{3}}$. Rationalizing the denominator: $y

Vedclass · Accepted Answer

$368$

Vedclass · Answer

(B) Given $x = \sqrt{7} + \sqrt{3}$ and $xy = 4$. First,find $y$: $y = \frac{4}{x} = \frac{4}{\sqrt{7} + \sqrt{3}}$. Rationalizing the denominator: $y = \frac{4(\sqrt{7} - \sqrt{3})}{7 - 3} = \frac{4(\sqrt{7} - \sqrt{3})}{4} = \sqrt{7} - \sqrt{3}$. Now,$x + y = (\sqrt{7} + \sqrt{3}) + (\sqrt{7} - \sqrt{3}) = 2\sqrt{7}$. We need to find $x^4 + y^4$. Note that $x^2 + y^2 = (x + y)^2 - 2xy = (2\sqrt{7})^2 - 2(4) = 28 - 8 = 20$. Then,$x^4 + y^4 = (x^2 + y^2)^2 - 2(xy)^2 = (20)^2 - 2(4)^2 = 400 - 2(16) = 400 - 32 = 368$.

If $x = \sqrt{7} + \sqrt{3}$ and $xy = 4$,then find the value of $x^4 + y^4$.

Similar Questions

Let $f(x) = ax^{2} + bx + c$ be such that $f(1) = 3$,$f(-2) = \lambda$,and $f(3) = 4$. If $f(0) + f(1) + f(-2) + f(3) = 14$,then $\lambda$ is equal to...

The roots of the equation $(p - q)x^2 + (q - r)x + (r - p) = 0$ are:

If $a, b \in \{1, 2, 3\}$ and the equation $ax^{2} + bx + 1 = 0$ has real roots,then

$\alpha$ and $\beta$ are the real roots of the equation $12 x^{1/3} - 25 x^{1/6} + 12 = 0$. If $\alpha > \beta$,then $\sqrt[6]{\frac{\alpha}{\beta}} =$

The quadratic equation with real coefficients whose one root is $7 + 5i$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module