Question

Find the value of p for which the quadratic equation (2p + 1 )x² - (7p + 2)x + (7p -3) = 0 . has equal roots. Also find these roots.​

Answer 1

Answer:

quadratic equation (-2p + 1)x²2 + (p - 2)x + 6p = 0

a = (-2p + 1), b = (p - 2), c = 6p

a) and b)

discriminant

Δ = b²2 – 4ac

Sub for a, b, c

Δ = (p - 2)²2 - 4(-2p + 1) 6p

= (p - 2)²2 – 24p(-2p + 1)

= p²2 - 4p + 4 + 48p²2 - 24p

= 49p²2 -28p + 4

= (7p - 2)²2 which is a perfect square

The significance of the discriminant being a perfect square is that the roots of the quadratic equation will be real, unequal and RATIONAL

c) use the quadratic formula

x = (-b ± √ Δ)/2a

= (-(p - 2) ± √(7p - 2)²2)/2(-2p + 1)

= (-p + 2) ± (7p - 2))/2(-2p + 1)

= (-p + 2) + (7p - 2))/2(-2p + 1) and (-p + 2) - (7p - 2))/2(-2p + 1)

= (6p )/2(-2p + 1) and (- 8p)/2(-2p + 1)

= (3p )/(-2p + 1) and (- 4p)/(-2p + 1)

= 3p /(1 - 2p) and - 4p/(1 - 2p) provided p ≠ ½

Step-by-step explanation:

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