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