1.

Find the values of p for which the quadratic equation (2p + 1)x2 – (7p + 2)x + (7p – 3) = 0 has equal roots. Also, find the roots.

Answer»

The given equation (2p + 1)x2 – (7p + 2)x + (7p – 3) = 0 is in the form of ax+ bx + c = 0

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

For the equation to have real and equal roots, the condition is

D = b– 4ac = 0

⇒ (-(7p + 2))– 4(2p +1)( 7p – 3) = 0

⇒ (7p + 2)– 4(14p2 + p – 3) = 0

⇒ 49p2 + 28p + 4 – 56p2 – 4p + 12 = 0

⇒ -7p+ 24p + 16 = 0

Solving for p by factorization,

⇒ -7p+ 28p – 4p + 16 = 0

⇒ -7p(p – 4) -4(p – 4) = 0

⇒ (p – 4) (-7p – 4) = 0

Either p – 4 = 0 ⇒ p = 4 Or, 7p + 4 = 0 ⇒ p = -4/7,

So, the value of k can either be 4 or -4/7

Now, using k = 4 in the given quadratic equation we get

(2(4) + 1)x2 – (7(4) + 2)x + (7(4) – 3) = 0

9x2 – 30x + 25 = 0

⇒ (3x – 5)2 = 0

Thus, x = 5/3 is the root of the given quadratic equation.

Next, on using k = 1 in the given quadratic equation we get

(2(-4/7 ) + 1)x2 – (7(-4/7 ) + 2)x + (7(-4/7 ) – 3) = 0

x2 – 14x + 49 = 0

⇒ (x – 7)2 = 0

Thus, x – 7 = 0 ⇒ x = 7 is the root of the given quadratic equation.



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