Question

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

Answer 1

Answer:

Let's solve for p.

(2p+1)(x2)−(7p+2)(x)+7p−3=0

Step 1: Add -x^2 to both sides.

2px2−7px+x2+7p−2x−3+−x2=0+−x2

2px2−7px+7p−2x−3=−x2

Step 2: Add 2x to both sides.

2px2−7px+7p−2x−3+2x=−x2+2x

2px2−7px+7p−3=−x2+2x

Step 3: Add 3 to both sides.

2px2−7px+7p−3+3=−x2+2x+3

2px2−7px+7p=−x2+2x+3

Step 4: Factor out variable p.

p(2x2−7x+7)=−x2+2x+3

Step 5: Divide both sides by 2x^2-7x+7.

p(2x2−7x+7)

2x2−7x+7

=

−x2+2x+3

2x2−7x+7

p=

−x2+2x+3

2x2−7x+7

Answer:

p=

−x2+2x+3

2x2−7x+7