If the equations 2ax^2 - 3bx + 4c = 0 and 3x^ | vedclass

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

Question

(C) Given the equations $2ax^2 - 3bx + 4c = 0$ and $3x^2 - 4x + 5 = 0$.Since the second equation $3x^2 - 4x + 5 = 0$ has a discriminant $D = (-4)^2 -

Vedclass · Accepted Answer

$\frac{34}{15}$

Vedclass · Answer

(C) Given the equations $2ax^2 - 3bx + 4c = 0$ and $3x^2 - 4x + 5 = 0$.Since the second equation $3x^2 - 4x + 5 = 0$ has a discriminant $D = (-4)^2 - 4(3)(5) = 16 - 60 = -44 If two quadratic equations have a common root and the coefficients are real,and one equation has imaginary roots,then both equations must have the same roots.Thus,the ratios of the coefficients must be equal:$\frac{2a}{3} = \frac{-3b}{-4} = \frac{4c}{5} = k$$\frac{2a}{3} = \frac{3b}{4} = \frac{4c}{5} = k$From this,we get $a = \frac{3k}{2}$,$b = \frac{4k}{3}$,and $c = \frac{5k}{4}$.Now,calculate $\frac{a + b}{c}$:$\frac{a + b}{c} = \frac{\frac{3k}{2} + \frac{4k}{3}}{\frac{5k}{4}} = \frac{\frac{9k + 8k}{6}}{\frac{5k}{4}} = \frac{17k}{6} 	imes \frac{4}{5k} = \frac{17 	imes 2}{3 	imes 5} = \frac{34}{15}$.

If the equations $2ax^2 - 3bx + 4c = 0$ and $3x^2 - 4x + 5 = 0$ have a common root,then $\left( \frac{a + b}{c} \right)$ is equal to (where $a, b, c \in R$).

Similar Questions

Solve the given two equations and select the correct option.
$I.$ $2x^2 + 23x + 63 = 0$
$II.$ $4y^2 + 19y + 21 = 0$

The roots of $4x^2 + 6px + 1 = 0$ are equal,then the value of $p$ is

If $p$ and $q$ are the roots of the equation ${x^2} + pq = (p + 1)x$,then $q=$

Number of natural solutions of the equation $xyz = 2^5 \times 3^2 \times 5^2$ is equal to

For what values of $k$ will the equation $x^2 - 2(1 + 3k)x + 7(3 + 2k) = 0$ have equal roots?

Vedclass Test Series

Exam Paper Generator

Online Exam Module