find the values of p if the roots of the equation (2p-1) ×2+6×+(p-+) =0 are real and equal
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Answer:
Given Eqn is
a
(2p+1)x
2
−
b
(7p+2)x
+
c
7p−3
=02
If this eqn has equal roots Discriminant
D=0
b
2
−4ac=0
⇒(7p+2)
2
−4(2p+1)(7p−3)=0
⇒49p
2
+4+28p−4(149
2
−6p−3+7p)=0
⇒49p
2
+4+28p−(56p
2
−24p−12+28p)=0
⇒49p
2
+4+28p−56p
2
−4p+12=0
⇒−7p
2
+24p+16=0
⇒7p
2
−24p−16=0⇒P=4,
7
−4
Now Roots At
i)P=4 ii)P=
7
−4
Eqn is 9x
2
−30x+25=0 Eqn is
7
−1
x
2
+2x−7=0
3x−5=0 x
2
−14x+49=0
x=
3
5
x=7
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