find the value of p for which the quadratic equation (p+1)x^2-6(p+1)x+3(p+q)=0,p not equal to 1. hence , find the roots of the equation
Answers
SOLUTION :
Given : (p + 1)x² - 6(p + 1)x + 3(p + 9) = 0, p ≠ -1 has equal roots ………(1)
On comparing the given equation with ax² + bx + c = 0
Here, a = p + 1 , b = - 6(p +1) , c = 3(p +9)
D(discriminant) = b² – 4ac
D = [- 6(p +1)² - 4 × (p + 1) × 3(p + 9)
D = [36((p)² + 1²+ 2× p× 1)) - 12(p² + 9p + p + 9)
[(a + b)² = a² + b² + 2ab]
D = 36(p² + 1 + 2p - 12(p² + 10p + 9)
D = 36p² + 36 + 72p - 12 p² - 120 p - 108
D = 36p² - 12 p² + 72p - 120p + 36 - 108
D = 24p² - 48p - 72
Given : Equal roots
Therefore , D = 0
24p² - 48p - 72 = 0
24(p² - 2p - 3) = 0
p² - 2p - 3 = 0
p² - 3p + p - 3 = 0
[By middle term splitting]
p(p - 3) + 1 (p - 3) = 0
(p + 1) (p - 3) = 0
p + 1 = 0 or (p - 3) = 0
p = - 1 or p = 3
The value of p is - 1 & 3 .
It is given that p ≠ - 1 , so p = 3
Hence , the value of p is 3 only.
On putting p = 3 in eq 1 ,
(p + 1)x² - 6(p + 1)x + 3(p + 9) = 0
(3 + 1)x² - 6(3 + 1)x + 3(3 + 9) = 0
4x² - 6(4)x + 3(12) = 0
4x² - 24x + 36 = 0
4(x² - 6x + 9) = 0
x² - 6x + 9 = 0
x² - 3x - 3x + 9 = 0
[By middle term splitting]
x(x - 3) -3(x - 3) = 0
(x - 3) = 0 or (x - 3) = 0
x = 3 or x = 3
Roots are 3 & 3
Hence ,the roots of the equation (p + 1)x² - 6(p + 1)x + 3(p + 9) = 0 is 3 .
★★ 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
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