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

The value of $k$ for which the equation $(k - 2)x^2 + 8x + (k + 4) = 0$ has both roots real,distinct,and negative is

Solve the given two equations and select the correct answer from the given options.
$I.$ $(x+y)^{2} = 3136$
$II.$ $y+2513 = 2569$

If $5x + 9y = 5$ and $125x^3 + 729y^3 = 120$,then the value of the product of $x$ and $y$ is:

Number of rational roots of the equation $x^{2016} - x^{2015} + x^{1008} + x^{1003} + 1 = 0$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module