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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
For common roots (6a -1)x +a²-b = 0 x = (b-a²)/(6a -1) Now36 -4a² = 0 a² = 9a = +- 3 Now out this and solve
Dear studentso 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)]/2aso 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 bex= [-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 cHence a=1,b=2,c=3∴a:b:c=1:2:3
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