Find the values of p for which the quadratic equation (2p+1)x^(2)-(7p+2)x+7p-3=0 has equal roots.
Answers
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Answer:
Discriminant = b² - 4ac
Here, b = -(7p+2) , a = (2p+1) and c = 7p - 3
Substitute these value in Formula!
Discriminant = b² - 4ac
✏ [{-(7p+2)² - 4(2p+1)(7p-3)}] = 0
✏ [{-(49p²+4+28p - 4(14p²-6p+7p-3)}] = 0
✏ [{-(49p²+28p+4 -56p² + 24p -28p + 12)}] = 0
✏ [{-(49p²-56p²+28p+24p-28p+12+4)}] = 0
✏ [{-(-7p² -24p+16)}] = 0
✏ [7p²-24p-16] = 0
Solving By Splitting Method!
✏ 7p²-24p-16 = 0
✏ 7p²-28p+4p-16 = 0
✏ 7p(p-4)+4(p-4) = 0
✏ (7p+4)(p-4) = 0
p = -4/7 or 4
Therefore, Value of P is 4 ,
Now, Substitute the Value of P in the Given Equation!
(2p+1)x²-(7p+2)x+(7p-3)=0
✏ [(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
✏ 9x² - 15x - 15x + 25 = 0
✏ 3x(3x-5)-5(3x-5) = 0
✏ (3x-5)(3x-5) = 0
•°•
(3x - 5) = 0
✏ 3x = 5
✏ x = 5/3 ----(1)
For other root Substitute the value of x ( -4/7) in The Given Equation!
✏ (2p+1)x²-(7p+2)x+(7p-3)=0
✏ (2.-4/7 + 1 ) x² - (7.-4/7 + 2)x + 7.-4/7 - 3 = 0
✏ ( -8/7 + 1 )x² -(-4+2)x + (-4-3) = 0
✏ (-1/7)x² -(-2) + (-7) = 0
✏ (-1/7)x² + 2 - 7 = 0
✏-x² + 14x - 49 = 0
✏ x² -14x + 49 = 0
✏ x² - 7x - 7x + 49 = 0
✏ x(x-7)-7(x-7) = 0
✏ (x-7)(x-7) = 0
✏ (x-7) = 0 or (x-7) = 0
✏ x = 7 0r 7 -----(2)
Hence, Value of x is 7
From Equation (1) and (2) , Required Roots of this Equation are 5/3 and 7.
Therefore, Required Roots of this Equation are 5/3 and 7