Find the value of p for which the quadratic equation (2p+1)x^2-(7p+2)x+(7p-3)=0 has equal roots. Also find these roots.
Answers
Answer:
p=4,-4/7
x=5/3,7
Step-by-step explanation:
b^2 - 4ac=0
(-7p-2)^2-4*(2p+1)*(7p-3)=0
49p^2+4+28p-4(14p^2+p-3)=0
-7p^2+16+24p=0
7p^2-24p-16=0
7p^2-28p+4p-16=0
7p(p-4)+4(p-4)=0
(7p+4)(p-4)=0
p=4,-4/7
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