Question

Find the value of p if the equation $3x^2-2x+p=0$ and $6x^2-17x+12=0$ have a common root.

• A. $-8/3,-15/4$
• B. $-7/3,-14/3$
• C. $2,4$
• D. none of these

Answer 1

The correct option is A $-8/3,-15/4$

Let $a$ be the common root of the equations.

Then $a$ will satisfy both the equations.

Thus, $3a^2-2a+p=0$

$6a^2-17a+12=0$

By cross multiplication, we have,

$\frac{a^2}{-24+17p}=\frac{a}{6p-36}=\frac{-1}{39}$

From 1 and 2, we have,

$a=\frac{17p-24}{6p-36}$ ......(1)

From 2 and 3, we have,

$a=\frac{2p-12}{-13}$ ......(2)

From 1 and 2, we get,

$\frac{17p-24}{6p-36}=\frac{2p-12}{-13}$

$12p^2+77p+120=0$

$(3p+8)(4p+15)=0$

$p=-\frac{8}{3},-\frac{15}{4}$