If 1 is a root of the quadratic equation 3x²+ax-2=0 and the quadratic equation a(x²+6x)-b=0 has equal roots, find the value of b.
Answers
SOLUTION :
Given : 1 is the root of quadratic equation 3x² + ax - 2 = 0………….(1)
& a(x² + 6x) - b = 0 has equal roots ………….(2)
On putting the value of given root i.e x = 1 in eq 1 .
3x² + ax - 2 = 0
3(1)² + a(1) − 2 = 0
3 × 1 + a - 2 = 0
3 + a − 2 = 0
1 + a = 0
a = - 1
Hence the value of a is - 1.
On putting the value of a = - 1 in eq 2,
a(x² + 6x) - b = 0
-1 (x² + 6x) - b = 0
- x² - 6x - b = 0
x² + 6x + b = 0
On comparing the given equation with Ax² + Bx + C = 0
Here, A = 1, B = 6 , and C = b
D(discriminant) = B² – 4AC
Given : Quadratic equation has equal roots i.e D = 0
B² – 4AC = 0
6² – 4(1)(b) = 0
36 – 4b = 0
36 = 4b
b = 36/4 = 9
b = 9
Hence, the value of b is 9 .
★★ 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…
3×1^2 + a×1 - 2 = 0
a = -1.
Thus second equation is
-1×(-1x^2+6x)-b=0
i.e., x^2 - 6x -b = 0
This equation has equal roots, so discriminant = 0.
(-6)^2-4×1×-b = 0
b = -36/4 = -9
hope it's help you.
please mark it as brainliest.