Find the values of p for which the quadratic equation (2p+1)x²-(7p+2)x+(7p-3)=0 has equal roots. Also, find these roots.
Answers
SOLUTION :
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
HOPE THIS ANSWER WILL HELP YOU…
Question :
Find the values of p for which the quadratic equation (2p+1)x²-(7p+2)x+(7p-3)=0 has equal roots. Also, find these roots.
Method of Solution :
In this Question, it is given that this Equation has Equal Roots , it's meant Discriminant must be 0.
Discriminant = b² - 4ac
Here, b = -(7p+2) , a = (2p+1) and c = 7p - 3
Substitute these value in Formula!
Discriminant = b² - 4ac
✏ [{-(7p+2)² - 4(2p+1)(7p-3)}] = 0
✏ [{-(49p²+4+28p - 4(14p²-6p+7p-3)}] = 0
✏ [{-(49p²+28p+4 -56p² + 24p -28p + 12)}] = 0
✏ [{-(49p²-56p²+28p+24p-28p+12+4)}] = 0
✏ [{-(-7p² -24p+16)}] = 0
✏ [7p²-24p-16] = 0
Solving By Splitting Method!
✏ 7p²-24p-16 = 0
✏ 7p²-28p+4p-16 = 0
✏ 7p(p-4)+4(p-4) = 0
✏ (7p+4)(p-4) = 0
p = -4/7 or 4
Therefore, Value of P is 4 ,
Now, Substitute the Value of P in the Given Equation!
(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
✏ 3x(3x-5)-5(3x-5) = 0
✏ (3x-5)(3x-5) = 0
•°•
(3x - 5) = 0
✏ 3x = 5
✏ x = 5/3 ----(1)
For other root Substitute the value of x ( -4/7) in The Given Equation!
✏ (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
✏ (-1/7)x² -(-2) + (-7) = 0
✏ (-1/7)x² + 2 - 7 = 0
✏-x² + 14x - 49 = 0
✏ x² -14x + 49 = 0
✏ x² - 7x - 7x + 49 = 0
✏ x(x-7)-7(x-7) = 0
✏ (x-7)(x-7) = 0
✏ (x-7) = 0 or (x-7) = 0
✏ x = 7 0r 7 -----(2)
Hence, Value of x is 7
From Equation (1) and (2) , Required Roots of this Equation are 5/3 and 7.
Therefore, Required Roots of this Equation are 5/3 and 7.