Let alpha and beta be the roots of the quadra | vedclass

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

Question

(C) Given that $a, b, c$ are in $A.P.$,we have $2b = a + c$.Dividing by $a$,we get $\frac{2b}{a} = 1 + \frac{c}{a}$ ... $(i)$.From the properties of r

Vedclass · Accepted Answer

$-31$

Vedclass · Answer

(C) Given that $a, b, c$ are in $A.P.$,we have $2b = a + c$.Dividing by $a$,we get $\frac{2b}{a} = 1 + \frac{c}{a}$ ... $(i)$.From the properties of roots,$\alpha + \beta = -\frac{b}{a} = 15$,which implies $\frac{b}{a} = -15$.Substituting $\frac{b}{a} = -15$ into equation $(i)$:$2(-15) = 1 + \frac{c}{a} \Rightarrow -30 = 1 + \frac{c}{a} \Rightarrow \frac{c}{a} = -31$.Since $\alpha\beta = \frac{c}{a}$,we have $\alpha\beta = -31$.

Let $\alpha$ and $\beta$ be the roots of the quadratic equation $ax^2 + bx + c = 0$. If $a, b, c$ are in $A.P.$ and $\alpha + \beta = 15$,then $\alpha\beta$ equals:

Similar Questions

The reciprocal of $x+\frac{1}{x}$ is

If the equation $\frac{x^2 + 5}{2} = x - 2\cos(ax + b)$ has at least one solution,then $(b + a)$ can be equal to

If $x$ is real and $k = \frac{x^2 - x + 1}{x^2 + x + 1}$,then

Solve the given two equations and select the correct answer from the given options.
$I.$ $\frac{12 \times 4}{x^{4/7}} - \frac{3 \times 4}{x^{4/7}} = x^{10/7}$
$II.$ $y^3 + 783 = 999$

Construct a quadratic equation whose roots have the sum $= 6$ and product $= -16$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module