If the equations 3x2 + px + 1 = 0 and 2x2 + qx + 1 = 0 have a common root, then

The given equations are3x2+px+1=0 and 2x2+qx+1=0 Multiplying Eq. (i) by 2 and Eq. (ii) by 3 and subtracting, we get2px+2−3qx−3=0⇒(2p−3q)x=1⇒x=12p−3qPut the value of x in Eq. (i), we get312p−3q2+p12p−3q+1=0⇒3+p(2p−3q)+(2p−3q)2=0⇒3+2p2−3pq+4p2+9q2−12pq=0⇒6p2+9q2−15pq+3=0⇒2p2+3q2−5pq+1=0

If the equations 3x2 + px + 1 = 0 and 2x2 + qx + 1 = 0 have a common root, then

The given equations are3x2+px+1=0 and 2x2+qx+1=0 Multiplying Eq. (i) by 2 and Eq. (ii) by 3 and subtracting, we get2px+2−3qx−3=0⇒(2p−3q)x=1⇒x=12p−3qPut the value of x in Eq. (i), we get312p−3q2+p12p−3q+1=0⇒3+p(2p−3q)+(2p−3q)2=0⇒3+2p2−3pq+4p2+9q2−12pq=0⇒6p2+9q2−15pq+3=0⇒2p2+3q2−5pq+1=0

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If the equations 3x² + px + 1 = 0 and 2x² + qx + 1 = 0 have a common root, then

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$(2 p^{2} + 3 q^{2} - 51 p q + 1) = 0$

$(2 p^{2} + 3 q^{2} - 5 p q + 1) = 10$

$(2 p^{2} + 3 q^{2} - 5 p q + 5)$

$(2 p^{2} + 3 q^{2} - 5 p q + 1) = 0$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

The given equations are

$3 x^{2} + p x + 1 = 0$ and $2 x^{2} + q x + 1 = 0$ Multiplying Eq. (i) by 2 and Eq. (ii) by 3 and subtracting, we get

$\begin{array}{l} 2 p x + 2 - 3 q x - 3 = 0 \\ \Rightarrow (2 p - 3 q) x = 1 \\ \Rightarrow x = \frac{1}{2 p - 3 q} \end{array}$

Put the value of x in Eq. (i), we get

$\begin{array}{l} 3 {(\frac{1}{2 p - 3 q})}^{2} + p (\frac{1}{2 p - 3 q}) + 1 = 0 \\ \Rightarrow 3 + p (2 p - 3 q) + (2 p - 3 q)^{2} = 0 \\ \Rightarrow 3 + (2 p^{2} - 3 p q) + (4 p^{2} + 9 q^{2} - 12 p q) = 0 \\ \Rightarrow (6 p^{2} + 9 q^{2} - 15 p q + 3) = 0 \\ \Rightarrow (2 p^{2} + 3 q^{2} - 5 p q + 1) = 0 \end{array}$