Let alpha, beta be the roots of the equation | vedclass

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

Question

(D) Given that $\alpha$ and $\beta$ are the roots of $x^2-px+r=0$. From the relation between roots and coefficients,we have $\alpha+\beta=p$ $(i)$ and

Vedclass · Accepted Answer

$\frac{2}{9}(2p-q)(2q-p)$

Vedclass · Answer

(D) Given that $\alpha$ and $\beta$ are the roots of $x^2-px+r=0$. From the relation between roots and coefficients,we have $\alpha+\beta=p$ $(i)$ and $\alpha\beta=r$. Given that $\frac{\alpha}{2}$ and $2\beta$ are the roots of $x^2-qx+r=0$. From the relation between roots and coefficients,we have $\frac{\alpha}{2}+2\beta=q$,which implies $\alpha+4\beta=2q$ $(ii)$. Also,the product of roots is $\frac{\alpha}{2} 	imes 2\beta = r$,so $\alpha\beta=r$. Subtracting $(i)$ from $(ii)$,we get $(\alpha+4\beta)-(\alpha+\beta)=2q-p$,which simplifies to $3\beta=2q-p$,so $\beta=\frac{2q-p}{3}$. Substituting $\beta$ into $(i)$,we get $\alpha=p-\frac{2q-p}{3} = \frac{3p-2q+p}{3} = \frac{4p-2q}{3} = \frac{2(2p-q)}{3}$. Since $r=\alpha\beta$,we have $r = \left(\frac{2(2p-q)}{3}ight) \left(\frac{2q-p}{3}ight) = \frac{2}{9}(2p-q)(2q-p)$.

Let $\alpha, \beta$ be the roots of the equation $x^2-px+r=0$ and $\frac{\alpha}{2}, 2\beta$ be the roots of the equation $x^2-qx+r=0$. Then the value of $r$ is

Similar Questions

For what value of $a$ is one root of the quadratic equation $(a^2 - 5a + 3) x^2 + (3a - 1) x + 2 = 0$ twice the other (in $/3$)?

Suppose that the equation $ax^2+bx+c=0$ has roots $\alpha$ and $\beta$,both of which are different from $\frac{1}{3}$. Then,an equation whose roots are $\frac{1}{3\alpha-1}$ and $\frac{1}{3\beta-1}$ is

If $\alpha$ and $\beta$ are the roots of the quadratic equation $ax^{2}+bx+c=0$ and $3b^{2}=16ac$, then:

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3-9x^2+23x-15=0$,then $\alpha^3+\beta^3+\gamma^3=$

The difference between two roots of the equation $x^3-13x^2+15x+189=0$ is $2$. Then the roots of the equation are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module