Find The value of 'p' for which quadratic equation (2p+1)x²–(7p+2)+(7p–3)=0 has equal roots. find the roots
Answers
D =[ -(7p+2)]2 - 4 (2p+1)(7p- 3)
since D =
Given : (2p + 1)x² - (7p + 2)x + (7p - 3) = 0 ………(1)
On comparing the given equation with ax² + bx + c = 0
Here, a = 2p + 1 , b = - (7p + 2) , c = 7p - 3
D(discriminant) = b² – 4ac
D = [- (7p + 2)]² - 4 × (2p + 1) × (7p - 3)
D = [(7p)² + 2²+ 2× 7p× 2) - 4(14p² + 6p + 7p - 3)
[(a + b)² = a² + b² + 2ab]
D = 49p² + 4 + 28p - 4(14p² + p - 3)
D = 49p² + 4 + 28p - 56 p² - 4 p + 12
D = 49p² - 56 p² + 28p - 4p + 12 + 4
D = - 7p² + 24p + 16
Given : Equal roots
Therefore , D = 0
- 7p² + 24p + 16 = 0
7p² - 24p - 16 = 0
7p² - 28p + 4p - 16 = 0
[By middle term splitting]
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
Hence, the value of p is - 4/7 & 4 .
On putting p = 4 in eq 1 ,
(2p + 1)x² - (7p + 2)x + (7p - 3) = 0
(2 × 4 + 1)x² - (7 × 4 + 2)x + (7 × 4 - 3) = 0
(8 + 1)x² - (28 + 2)x + (28 - 3) = 0
9x² - 30x + 25 = 0
9x² - 15x - 15x + 25 = 0
[By middle term splitting]
3x(3x - 5) - 5(3x - 5) = 0
(3x - 5) = 0 or (3x - 5) = 0
3x = 5 or 3x = 5
x = 5/3 or x = 5/3
Roots are 5/3 & 5/3
On putting p = - 4/7 in eq 1 ,
(2p + 1)x² - (7p + 2)x + (7p - 3) = 0
(2(- 4/7) + 1)x² - (7(- 4/7) + 2)x + (7( - 4/7) - 3) = 0
(- 8/7 + 1)x² - (- 4 + 2)x + ( - 4 - 3) = 0
[(- 8 + 7)]/7x² - (- 2)x + ( - 7) = 0
[By taking LCM]
- 1/7x² + 2x - 7 = 0
(- x² + 2x × 7 - 7 × 7)/7 = 0
[By taking LCM]
-x² + 14x - 49 = 0
x² - 14x + 49 = 0
x² - 7x - 7x + 49 = 0
[By middle term splitting]
x(x - 7) - 7( x - 7) = 0
( x - 7) = 0 or ( x - 7) = 0
x = 7 or x = 7
Roots are 7 & 7
Hence, the roots of the equation (2p + 1)x² - (7p + 2)x + (7p - 3) = 0 are 5/3 & 7 .
★★ NATURE OF THE ROOTS
If D = 0 roots are real and equal
If D > 0 roots are real and distinct
If D < 0 No real roots