If a root of the equation ax^2 + bx + c = 0 i | vedclass

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

Question

(A) Let $\alpha$ be a root of the first equation $ax^2 + bx + c = 0$. Then,$\frac{1}{\alpha}$ is a root of the second equation $a'x^2 + b'x + c' = 0$.

Vedclass · Accepted Answer

$(cc' - aa')^2 = (ba' - cb')(ab' - bc')$

Vedclass · Answer

(A) Let $\alpha$ be a root of the first equation $ax^2 + bx + c = 0$. Then,$\frac{1}{\alpha}$ is a root of the second equation $a'x^2 + b'x + c' = 0$.For the first equation: $a\alpha^2 + b\alpha + c = 0$.For the second equation: $a'(\frac{1}{\alpha})^2 + b'(\frac{1}{\alpha}) + c' = 0$,which simplifies to $c'\alpha^2 + b'\alpha + a' = 0$.Using the method of cross-multiplication for the system of equations:$\frac{\alpha^2}{ba' - b'c} = \frac{\alpha}{cc' - aa'} = \frac{1}{ab' - bc'}$.From the second and third terms: $\alpha = \frac{cc' - aa'}{ab' - bc'}$.From the first and third terms: $\alpha^2 = \frac{ba' - b'c}{ab' - bc'}$.Equating $(\frac{cc' - aa'}{ab' - bc'})^2 = \frac{ba' - b'c}{ab' - bc'}$,we get:$(cc' - aa')^2 = (ba' - b'c)(ab' - bc')$.Thus,option $A$ is correct.

If a root of the equation $ax^2 + bx + c = 0$ is the reciprocal of a root of the equation $a'x^2 + b'x + c' = 0$,then:

Similar Questions

Solve the given two equations and select the correct answer from the given options.
$I. x^{2}-2x-15=0$
$II. y^{2}+5y+6=0$

Let $f(x) = x^2 - x + k - 2$,$k \in R$. Then the complete set of values of $k$ for which $y = |f(|x|)|$ is non-derivable at $5$ distinct points is:

The roots of the equation $(a^2 + b^2)t^2 - 2(ac + bd)t + (c^2 + d^2) = 0$ are equal,then

Solve the given two equations and provide the correct answer from the given options.
$I.$ $x^{2}-82x+781=0$
$II.$ $y^{2}-5041=0$

Find the value of $(a+b+c)^{3}-3(b+c)(c+a)(a+b)$ if $a=5, b=3$ and $c=2$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module