Q. 134.5( 13 Votes )

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Answer :

Given: equation (2p+1) x2 - (7p+2) x + (7p-3) =0 has equal roots.

To find: The value of p

The roots of the equation.

Formula Used:

For a quadratic equation, ax2 + bx + c = 0,

D = b2 – 4ac

If D = 0, roots are equal

Explanation:

Consider (2p+1) x2 - (7p+2) x + (7p-3) = 0,

Here a = 2p+1, b = 7p+2 and c = 7p-3

As roots are real.

So,

D = 0

b2 – 4ac = 0

(7p + 2)2 – 4(7p – 3) (2p + 1) = 0

(7p + 2)2 – 4(14p2+ 7p – 6p – 3) = 0

(7p + 2)2 – 4(14p2+ p – 3) = 0

49p2 + 28p + 4 – 56p2 + 12 – 4p = 0

7p2 – 24p – 16 = 0

Split the middle terms.

7p2 – 28p + 4p – 16 = 0

7p (p – 4) + 4(p – 4) = 0

(7p + 4) (p – 4) = 0

p = -4/7, 4

When p = -4/7

The equation is:     -x2 + 14 x - 49 = 0

x2 - 14 x + 49 = 0

Split the middle terms.

x2 – 7x – 7x + 49 = 0

x(x-7) - 7(x-7) = 0

(x-7) (x-7) =0

x = 7

When p = 4

The equation is:

(2(4) +1) x2 - (7(4) +2) x + (7(4)-3) = 0

(8 +1) x2 - (28 +2) x + (28-3) = 0

9x2 - 30x + 25 = 0

Split the middle terms.

9x2 - 15x – 15x + 25 = 0

3x(3x-5) – 5(3x-5) = 0

(3x-5) (3x-5) = 0

x = 5/3

Hence the value of p is -4/7, 4 and x is 7, 5/3.

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