If alpha and beta are the roots of ax^2 + bx | vedclass

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

Question

(D) Given that $\alpha$ and $\beta$ are roots of $ax^2 + bx + c = 0$,we have $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.Let the n

Vedclass · Accepted Answer

$ax^2 + x(b - 4a) + 4a - 2b + c = 0$

Vedclass · Answer

(D) Given that $\alpha$ and $\beta$ are roots of $ax^2 + bx + c = 0$,we have $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.Let the new roots be $\alpha' = 2 + \alpha$ and $\beta' = 2 + \beta$.The sum of the new roots is $\alpha' + \beta' = (2 + \alpha) + (2 + \beta) = 4 + (\alpha + \beta) = 4 - \frac{b}{a} = \frac{4a - b}{a}$.The product of the new roots is $\alpha'\beta' = (2 + \alpha)(2 + \beta) = 4 + 2(\alpha + \beta) + \alpha\beta = 4 + 2(-\frac{b}{a}) + \frac{c}{a} = \frac{4a - 2b + c}{a}$.The required quadratic equation is $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$.Substituting the values,we get $x^2 - (\frac{4a - b}{a})x + \frac{4a - 2b + c}{a} = 0$.Multiplying by $a$,we obtain $ax^2 - (4a - b)x + 4a - 2b + c = 0$,which simplifies to $ax^2 + x(b - 4a) + 4a - 2b + c = 0$.

If $\alpha$ and $\beta$ are the roots of $ax^2 + bx + c = 0$,then the equation whose roots are $2 + \alpha$ and $2 + \beta$ is:

Similar Questions

If for the function $f(x) = \frac{1}{4}x^2 + bx + 10$; $f(12 - x) = f(x)$ for all $x \in R$,then the value of $b$ is:

Let $\alpha$ and $\beta$ be the roots of the equation $5x^{2} + 6x - 2 = 0$. If $S_{n} = \alpha^{n} + \beta^{n}$ for $n = 1, 2, 3, \dots$,then:

The two roots of the equation $x^3 - 9x^2 + 14x + 24 = 0$ are in the ratio $3 : 2$. Find the roots.

For the equation $\frac{1}{x + a} - \frac{1}{x + b} = \frac{1}{x + c}$,if the product of the roots is zero,what is the sum of the roots?

If the $A.M.$ and $G.M.$ of the roots of a quadratic equation are $8$ and $5$ respectively,then the quadratic equation is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module