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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