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If x2+6x+a=0,ax2+1x+b=0,have a common root and first equation has equal roots then show that 2a2+b=0

Dama sriram , 4 Years ago
Grade 8
anser 2 Answers
Arun

Last Activity: 4 Years ago

For common roots
 
(6a -1)x +a²-b = 0
 
x = (b-a²)/(6a -1)
 
Now
36 -4a² = 0
 
a² = 9
a = +- 3
 
Now out this and solve

Vikas TU

Last Activity: 4 Years ago

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
a:b:c=1:2:3

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