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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