If the sum of the roots of the equation lambd | vedclass

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

Question

(D) For a quadratic equation $ax^2 + bx + c = 0$,the sum of the roots is given by $-\frac{b}{a}$ and the product of the roots is given by $\frac{c}{a}

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) For a quadratic equation $ax^2 + bx + c = 0$,the sum of the roots is given by $-\frac{b}{a}$ and the product of the roots is given by $\frac{c}{a}$.Given the equation $\lambda x^2 + 2x + 3\lambda = 0$,we have $a = \lambda$,$b = 2$,and $c = 3\lambda$.Sum of the roots $= -\frac{2}{\lambda}$.Product of the roots $= \frac{3\lambda}{\lambda} = 3$.According to the problem,the sum of the roots is equal to their product:$-\frac{2}{\lambda} = 3$.Multiplying both sides by $\lambda$,we get $-2 = 3\lambda$.Therefore,$\lambda = -\frac{2}{3}$.Since $-\frac{2}{3}$ is not among the options $A, B,$ or $C$,the correct option is $D$.

If the sum of the roots of the equation $\lambda x^2 + 2x + 3\lambda = 0$ is equal to their product,then $\lambda = $

Similar Questions

If $\alpha, \beta$ are the roots of the quadratic equation $x^{2}-8x+k=0$,find the value of $k$ such that $\alpha^{2}+\beta^{2}=40$.

Solve the given two equations and select the correct answer from the given options.
$I.$ $7x - 3y = 13$
$II.$ $5x + 4y = 40$

Solve the given two equations and choose the correct answer from the given options.
$I.$ $x^{2} = 729$
$II.$ $y = \sqrt{729}$

If $x+\frac{1}{x}=3,$ then the value of $x^{8}+\frac{1}{x^{8}}$ will be

Find the value of $\frac{(a-b)^{2}}{(b-c)(c-a)}+\frac{(b-c)^{2}}{(a-b)(c-a)}+\frac{(a-c)^{2}}{(a-b)(b-c)}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module