Question

Find the value(s) 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

Step-by-step explanation:

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Answer 2

Given:-

• The quadratic equation (2p+1)x^2-(7p+2)x+(7p-3)=0 has equal roots.

To find:-

• Find the value of p..?

Solutions:-

• The given quadratic equation is (2p + 1)x² - (7p + 2)x + (7p - 3) = 0, and roots are real and equal.

Here,

• a = 2p + 1
• b = -7p - 2
• c = 7p - 3

We know that;

• D => b² - 4ac

Putting the value of a = 2p + 1, b = -7p - 2 and c = 7p - 3

• D => b² - 4ac

=> [-(7p + 2)]² - 4(2p + 1) (7p - 3)

=> (49p² + 28p + 4) - 4(2p + 1) (7p - 3)

=> 49p² + 28p + 4 - 56p² - 4p + 12

=> -7p² + 24p + 16

The given equation will have and equation roots.

• D = 0

Thus,

=> -7p² + 24p + 16 = 0

=> 7p² - 24p - 16 = 0

=> 7p² - 28p + 4p - 16 = 0

=> 7p(p - 4) + 4(p - 4) = 0

=> (7p + 4) (p -4) = 0

=> 7p + 4 = 0 or p - 4 = 0

=> 7p = -4 or p = 4

=> p = -4/7 or p = 4

Therefore,

the value of p is 4 or -4/7.

Now,

• For p = 4 the equation.

=> 9x² - 30x + 25 = 0

=> 9x² - 15x - 15x + 25 = 0

=> 3x(3x - 5) - 5(3x + 5) = 0

=> (3x - 5) (3x - 5)

=> (3x - 5)² = 0

=> x = 5/3 , 5/3

• For p = -4/7 the equation.

=> (-8/7 + 1)x² - (-4 + 2)x + (-4 - 3) = 0

=> (-8 + 1 / 7)x² + 2x - 7 = 0

=> -1/7 x² + 2x - 7 = 0

=> - x² + 14x - 49 = 0

=> x - 14x + 49 = 0

=> x - 7x - 7x + 49 = 0

=> x(x - 7) - 7(x - 7) = 0

=> (x - 7)² = 0

=> x = 7 , 7

Hence, the value of the equation are 5/3 and 7.