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find the value of p such that the quadratic equation x²-(2p+1)x+(3p+7)=0 has sum of the roots as half of their product.​

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Answer:

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

239
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