Given: \(\displaystyle{a}{x}^{{2}}+{b}{x}+{c}={0}\) with two real roots, \(\displaystyle{x}_{{{1}}}{>}{x}_{{{2}}}\),

Jackson Floyd

Jackson Floyd

Answered question


ax2+bx+c=0 with two real roots, x1>x2, find a quadratic equation whose roots are x1+1 and x21 without solving the first equation

Answer & Explanation



Beginner2022-03-27Added 20 answers

f(x)=ax2+bx+c=a(xx1)(xx2) is given to you. Consider
g(x) =a(x(x1+1))(x(x21))  =a(xx11)(xx2+1)  =a((xx1)1)((xx2)+1)  =a(xx1)(xx2)+a(xx1)a(xx2)a  =f(x)+a(x2x1)given >0a.g(x) =a(x(x1+1))(x(x21))  =a(xx11)(xx2+1)  =a((xx1)1)((xx2)+1)  =a(xx1)(xx2)+a(xx1)a(xx2)a  =f(x)+a(x2x1)given >0a.
Now x2x2=(x1+x2)24x1x2=(b2a2)4ca=b24aca
Hence, g(x)=ax2+bx+c+b24aca

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