If ax^2 + bx + c = 0 and bx^2 + cx + a = 0 ha | vedclass

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

Question

(B) Let $\alpha$ be the common root of the equations $ax^2 + bx + c = 0$ and $bx^2 + cx + a = 0$.Then,$a\alpha^2 + b\alpha + c = 0$ and $b\alpha^2 + c

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) Let $\alpha$ be the common root of the equations $ax^2 + bx + c = 0$ and $bx^2 + cx + a = 0$.Then,$a\alpha^2 + b\alpha + c = 0$ and $b\alpha^2 + c\alpha + a = 0$.Using the method of cross-multiplication for these two equations:$\frac{\alpha^2}{b(a) - c(c)} = \frac{\alpha}{c(b) - a(a)} = \frac{1}{a(c) - b(b)}$$\frac{\alpha^2}{ab - c^2} = \frac{\alpha}{bc - a^2} = \frac{1}{ac - b^2}$From the first and third parts: $\alpha^2 = \frac{ab - c^2}{ac - b^2}$From the second and third parts: $\alpha = \frac{bc - a^2}{ac - b^2}$Since $\alpha^2 = (\alpha)^2$,we have:$\frac{ab - c^2}{ac - b^2} = \left( \frac{bc - a^2}{ac - b^2} ight)^2$$(ab - c^2)(ac - b^2) = (bc - a^2)^2$$a^2bc - ab^3 - ac^3 + b^2c^2 = b^2c^2 - 2a^2bc + a^4$$a^4 + ab^3 + ac^3 = 3a^2bc$Dividing both sides by $abc$ (since $a, b, c 
eq 0$):$\frac{a^3}{bc} + \frac{b^2}{c} + \frac{c^2}{b} = 3$ (This is not the target expression).Actually,dividing $a^4 + ab^3 + ac^3 = 3a^2bc$ by $abc$ gives $\frac{a^3}{bc} + \frac{b^2}{c} + \frac{c^2}{b} = 3$. Wait,let's re-evaluate: $a(a^3 + b^3 + c^3) = 3a^2bc \Rightarrow \frac{a^3 + b^3 + c^3}{abc} = 3$.

If $ax^2 + bx + c = 0$ and $bx^2 + cx + a = 0$ have a common root and $a, b, c$ are non-zero real numbers,then $\frac{a^3 + b^3 + c^3}{abc} = $

Similar Questions

If $ax^2 + bx + c = 0$,then $x =$

If $x+y=2z$,find the value of $\frac{x}{x-z}+\frac{z}{y-z}$.

Two candidates attempt to solve the equation $x^2 + px + q = 0$. One starts with a wrong value of $p$ and finds the roots to be $2$ and $6$,and the other starts with a wrong value of $q$ and finds the roots to be $2$ and $-9$. The roots of the original equation are

Solve the given two equations and select the correct option.
$I.$ $9x^{2} - 114x + 361 = 0$
$II.$ $y^{2} = 36$

If the roots of the equation $x^2 + x + 1 = 0$ are $\alpha$ and $\beta$,and the roots of the equation $x^2 + px + q = 0$ are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$,then $p$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module