Identical roots of a quadratic equation Let \(\displaystyle{f{{\left({x}\right)}}}={a}{x}^{{2}}+{b}{x}+{c}\)

Jakobe Norton

Jakobe Norton

Answered question

2022-04-09

Identical roots of a quadratic equation
Let f(x)=ax2+bx+c and g(x)=x2+d.
Now consider f(x)+λg(x)=0

Answer & Explanation

ysnlm8eut

ysnlm8eut

Beginner2022-04-10Added 14 answers

For the second part, the solution in (1) holds good till the line (*) below,
(adc)2+b2d=0  ()  (caD)2b2D=0 (where D=d>0)  (c+aD)2(bD)2=0  (c+aDbD)(c+aD+bD)=0  (1)  f(D)f(D)=0  at least one of D and D is a root of f(x)

but D,D are precisely the solutions of the second quadratic
g(x)=x2+d=0x2D=0(xD)(x+D)=0
Thus one of the numbers D,D is a root of both the quadratics f(x),g(x), and by the Factor Theorem, one of the terms
(xD)    or    (x+D)
divides both f and g, hence, is a common factor of both.

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