Question

find the values of p for which the quadratic equation: (2p+1)x^2 - (7p+2)x + (7p-3) =0

Answer 1

Answer:

2p + 1)x² + (7p+2)x + (7p-3) = 0

If they have equal roots,

b²-4ac=0

Let,

a=2p+1

b=7p+2

c=7p-3

Now,

(7p+2)² - 4(2p + 1) (7p-3)=0 

49p² +4+ 28p – 4(14p² +p-3)=0

After simplification,

7p²-24p+16=0 

The roots are,

4 and -4/7.

Read more on Brainly.in - https://brainly.in/question/214175#readmore-by-step explanation:

Answer 2

Answer:

Here is your answer

Step-by-step explanation:

Given Eqn is

a=(2p+1)x2

b=(7p+2)x

c=7p−3=02

If this eqn has equal roots Discriminant

D=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

Eqn is 9x2−30x+25=0

-1/7x 2 +2x−7=0.

3x−5=0

x=5/3

ii)P=7/-4

Eqn is 9x2-30x+25=0

x2-14x+49=0

x=7

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