X^2+(p-3)x+p=0
Pls give me the answer fast in need this answer for my unit exams
Answers
Given quadratic equation is= (p + 1)x² - 6(p + 1) x + 3 (p + 9) = 0.
On comparing with standard form of quadratic equation i.e ax² + bx + c =0, a≠ 0
Here, a = p+1 , b= -6(p+1) , c= 3(p+9)
D(discriminant)= b²-4ac
=-6 (p+1)² - 4× (p+1)× 3(p+9)
= 36(p+1)² - 12(p+1)(p+9)
= 36(p² + 1 + 2p) - 12 (p²+9p+p+9)
= 36p² + 36 + 72p - 12p²- 108p-12p- 108
= 36p²-12p² +72p - 108p-12p +36- 108
= 24p² - 48p - 72
= 24(p² - 2p - 3)
Since, roots of given equation are equal. D= 0
0 = 24(p² - 2p - 3)
(p² - 2p - 3)= 0
p² - 3p +p - 3 = 0
p(p -3) +1(p-3)= 0
(p + 1) (p +3)= 0
(p + 1) (p +3)= 0
p ≠ 0 or p = 3
Hence, required value of p is 3
Put the value of p = 3 in given equation
(p + 1)x² - 6(p + 1) x + 3 (p + 9) = 0.
(3+1)x² -6(3+1)x + 3(3+9)= 0
4x² -24x +36= 0
4(x² -6x +9)=0
x² -6x +9 = 0
x² -3x -3x +9= 0
x(x -3) -3(x -3)= 0
(x-3)(x-3)= 0
(x-3)= 0 or (x-3)= 0
x = 3 or x= 3
Hence , the roots are x =3, 3.