AP EAMCET202022 Sep 2020Morning ShiftMathematicsQuadratic EquationActual
If the roots of the equation x^3-6 x^2+11 x-6=0 are , and . Then the equation whose roots are ^2, ^2, ^2 among the following is
Options
- Ax^3+14 x^2+49 x-36=0
- Bx^3-14 x^2+49 x-36=0
- Cx^3-14 x^2-49 x+36=0
- Dx^3-14 x^2-49 x-36=0
Correct answer
B. x^3-14 x^2+49 x-36=0
Step-by-step solution
It is given that roots of the equation x^3-6 x^2+11 x-6=0 are , , . Now, to find the equation whose roots are ^2, ^2, ^2 , put ^2=x = x . Since, is the root of the given equation, so aligned & x^ 3 / 2 -6 x+11 x^ 1 / 2 -6=0 . & x^ 1 / 2 (x+11)=6(x+1) aligned On squaring both sides, we get aligned & & x(x+11)^2=36(x+1)^2 & & x [x^2+22 x+121 ]=36 [x^2+2 x+1 ] & & x^3+22 x^2+121 x=36 x^2+72 x+36 & x^3-14 x^2+49 x-36 & =0 aligned