Let a, b in mathbb{R}, a neq 0 be such that t | vedclass

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

Question

(B) For the equation $a x^{2}-2 b x+5=0$,the roots are $\alpha, \alpha$. Thus,the sum of roots $2\alpha = \frac{2b}{a} \Rightarrow \alpha = \frac{b}{a

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(B) For the equation $a x^{2}-2 b x+5=0$,the roots are $\alpha, \alpha$. Thus,the sum of roots $2\alpha = \frac{2b}{a} \Rightarrow \alpha = \frac{b}{a}$ and the product of roots $\alpha^{2} = \frac{5}{a}$. From these,$b = a\alpha$ and $a = \frac{5}{\alpha^{2}}$. Substituting $a$,we get $b = \frac{5}{\alpha}$. Since $\alpha$ is also a root of $x^{2}-2 b x-10=0$,we have $\alpha^{2}-2 b \alpha-10=0$. Substituting $b = \frac{5}{\alpha}$ into this equation: $\alpha^{2}-2(\frac{5}{\alpha})\alpha-10=0$ $\Rightarrow \alpha^{2}-10-10=0$ $\Rightarrow \alpha^{2}=20$. Now,for the equation $x^{2}-2 b x-10=0$,the product of roots $\alpha \beta = -10$. Since $\alpha^{2} = 20$,we have $\alpha = \pm \sqrt{20}$. Then $\beta = \frac{-10}{\alpha}$. Thus,$\beta^{2} = \frac{100}{\alpha^{2}} = \frac{100}{20} = 5$. Therefore,$\alpha^{2}+\beta^{2} = 20 + 5 = 25$.

Let $a, b \in \mathbb{R}, a \neq 0$ be such that the equation $a x^{2}-2 b x+5=0$ has a repeated root $\alpha,$ which is also a root of the equation $x^{2}-2 b x-10=0$. If $\beta$ is the other root of this equation,then $\alpha^{2}+\beta^{2}$ is equal to:

Similar Questions

If $mn=3$ and $\frac{1}{m}+\frac{1}{n}=\frac{4}{3}$,then the value of $0.1+0.1^{\frac{1}{m}}+0.1^{\frac{1}{n}}$ is

If $\alpha$ and $\beta$ are the roots of ${x^2} + px + q = 0$ and $\alpha + h$ and $\beta + h$ are the roots of ${x^2} + rx + s = 0$,then

Let $f(x)$ be a polynomial and $a, b$ be distinct real numbers. Then the remainder in the division of $f(x)$ by $(x-a)(x-b)$ is

Let $z$ be a complex number such that the imaginary part of $z$ is non-zero and $a = z^2 + z + 1$ is real. Then $a$ cannot take the value

The roots of the equation $(x-a)(x-a-1)+(x-a-1)(x-a-2)+(x-a)(x-a-2)=0$ for $a \in R$ are always:

Vedclass Test Series

Exam Paper Generator

Online Exam Module