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Grade: 8


If x 2 +6x+a=0,ax 2 +1x+b=0,have a common root and first equation has equal roots then show that 2a 2 +b=0

5 months ago

Answers : (2)

24737 Points
For common roots
(6a -1)x +a²-b = 0
x = (b-a²)/(6a -1)
36 -4a² = 0
a² = 9
a = +- 3
Now out this and solve
5 months ago
Vikas TU
12119 Points
Dear student
so the question says there is one common root.
now we know for the general equation ax^2+bx+c=0,
x = [-b±√(b^2 -4ac)]/2a
so assume that x=[-b+√(b^2–4ac)]/2a is the common root.
So, for the roots of the given equation (eq.1, x^2+2x+3=0) will be
x= [-2±√(2^2–4(1)(3))]/2(1)
But we have selected [-b+√(b^2–4ac)]/2a as the common root.
so as per the question,
x= [-b+√(b^2–4ac)]/2a = [-2+√(2^2–4(1)(3))]/2(1)
so now you have the values for a,b, and c
Hence a=1,b=2,c=3
5 months ago
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