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

#### Solution

Given quadratic equation:

(2p+1)x^{2}−(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

⇒49p^{2}+28p+4−4×(14p^{2}−6p+7p−3)=0

⇒49p^{2}+28p+4−56p^{2}−4p+12=0

⇒−7p^{2}+24p+16=0

⇒7p^{2}−24p−16=0

⇒7p^{2}−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)x^{2}−(7p+2)x+(7p−3)=0

⇒9x^{2}−30x+25=0⇒(3x−5)2=0

⇒x=5/3, 5/3

For p = −4/7:

(2p+1)x^{2}−(7p+2)x+(7p−3)=0

⇒(2(−47)+1)x^{2}−(7(−47)+2)x+(7(−47)−3)=0

⇒(−17)x^{2}+2x−7=0⇒17x^{2}−2x+7=0

⇒x^{2}−14x+49=0⇒(x−7)2=0

⇒x=7, 7

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