If alpha, beta are the roots of the equation | vedclass

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

Question

(D) Given the equation: $x^{2}-\left(5+3^{\sqrt{\log _{3} 5}}-5^{\sqrt{\log _{5} 3}}\right)x+3\left(3^{\left(\log _{3} 5\right)^{\frac{1}{3}}}-5^{\lef

Vedclass · Accepted Answer

$3x^{2}-10x-4=0$

Vedclass · Answer

(D) Given the equation: $x^{2}-\left(5+3^{\sqrt{\log _{3} 5}}-5^{\sqrt{\log _{5} 3}}\right)x+3\left(3^{\left(\log _{3} 5\right)^{\frac{1}{3}}}-5^{\left(\log _{5} 3\right)^{\frac{2}{3}}}-1\right)=0$. Simplifying the coefficients: $3^{\sqrt{\log _{3} 5}} = 3^{\sqrt{\log _{3} 5} \cdot \frac{\sqrt{\log _{3} 5}}{\sqrt{\log _{3} 5}}} = 3^{\log _{3} 5 \cdot \sqrt{\log _{5} 3}} = (3^{\log _{3} 5})^{\sqrt{\log _{5} 3}} = 5^{\sqrt{\log _{5} 3}}$. Thus,the coefficient of $x$ is $-(5 + 5^{\sqrt{\log _{5} 3}} - 5^{\sqrt{\log _{5} 3}}) = -5$. Similarly,$3^{\left(\log _{3} 5\right)^{\frac{1}{3}}} = 5^{\left(\log _{5} 3\right)^{\frac{2}{3}}}$. Thus,the constant term is $3(5^{\left(\log _{5} 3\right)^{\frac{2}{3}}} - 5^{\left(\log _{5} 3\right)^{\frac{2}{3}}} - 1) = -3$. The equation simplifies to $x^{2}-5x-3=0$. For this equation,$\alpha+\beta=5$ and $\alpha\beta=-3$. The new roots are $\alpha+\frac{1}{\beta} = \frac{\alpha\beta+1}{\beta} = \frac{-3+1}{\beta} = \frac{-2}{\beta}$ and $\beta+\frac{1}{\alpha} = \frac{-2}{\alpha}$. Let $t = \frac{-2}{\alpha}$,then $\alpha = \frac{-2}{t}$. Substituting into $x^{2}-5x-3=0$: $(\frac{-2}{t})^{2} - 5(\frac{-2}{t}) - 3 = 0 \Rightarrow \frac{4}{t^{2}} + \frac{10}{t} - 3 = 0$. Multiplying by $t^{2}$: $4 + 10t - 3t^{2} = 0 \Rightarrow 3t^{2}-10t-4=0$. Therefore,the required equation is $3x^{2}-10x-4=0$.

Similar Questions

Find $\alpha^4+\beta^4$ if $\alpha, \beta$ are the roots of the equation $x^2+x+1=0$.

Let $f(n)=A(-2)^n+B(-3)^n$ for all $A, B \in \mathbb{R}$ and $n \in \mathbb{N}-\{1, 2\}$. If $f(n)+a f(n-1)+b f(n-2)=0$, then $(a+b)(b-a)=$

If the ratio of the roots of the equation $px^2+qx+r=0$ is $a:b$, then $\frac{ab}{(a+b)^2}=$

If the equation $x^{2}-cx+d=0$ has roots equal to the fourth powers of the roots of $x^{2}+ax+b=0,$ where $a^{2}>4b,$ then the roots of $x^{2}-4bx+2b^{2}-c=0$ will be

If $\alpha$ and $\beta$ are roots of the equation $x^2 - 4\sqrt{2}kx + 2e^{4\ln k} - 1 = 0$ for some $k$,and $\alpha^2 + \beta^2 = 66$,then $\alpha^3 + \beta^3$ is equal to: (in $\sqrt{2}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module