Question

Find the value of p for which the quadratic equation (2p + 1)x² − (7p + 2)x + (7p − 3) = 0 has equal roots. Also find these roots.

Answer 1

Given quadratic equation:
(2p+1)x²−(7p+2)x+(7p−3)=0

To have equal roots, the discriminant should be zero.

∴ D = 0
⇒ (7p+2)²−4×(2p+1)×(7p−3)=0
⇒49p²+28p+4−4×(14p²−6p+7p−3)=0
⇒49p²+28p+4−56p²−4p+12=0
⇒−7p²+24p+16=0

⇒7p²−24p−16=0
⇒7p²−28p+4p−16=0

⇒7p(p−4)+4(p−4)=0

⇒(p−4)(7p+4)=0
⇒p=4 or −4/7
Therefore, the values of p for which the given equation has equal roots are 4, −4/7.

For p = 4:
(2p+1)x²−(7p+2)x+(7p−3)=0
⇒9x²−30x+25=0⇒(3x−5)2=0
⇒x=5/3, 5/3

For p = −4/7:
(2p+1)x²−(7p+2)x+(7p−3)=0
⇒(2(−47)+1)x²−(7(−47)+2)x+(7(−47)−3)=0
⇒(−17)x²+2x−7=0⇒17x²−2x+7=0
⇒x²−14x+49=0⇒(x−7)2=0
⇒x=7, 7

Thus, the equal roots are 7 or 5/3.