Question

Find The value of P for a which the quadiratic equation (2p - 1)x^2 - (7p + 2)x + (7p -3) =0 has real and equal roots

Answer 1

Answer:

0.1, 11.33

Step-by-step explanation:

equal roots: b² - 4ac = 0

b = -(7p + 2), a=(2p - 1), c= (7p-3)

(7p + 2)² - 4(2p - 1)(7p - 3) = 0

49p² + 28p + 4 - 56p² + 52p - 12 = 0

-7p² + 80p - 8 = 0

p$p_{1} =\frac{-80+\sqrt{80^{2} -4(-7)(-8)} }{2 (-7)} =\frac{40-2\sqrt{386} }{7} =0.1\\p_{2} =\frac{-80-\sqrt{80^{2} -4(-7)(-8)} }{2 (-7)} =\frac{40+2\sqrt{386} }{7} =11.33$