solve x^3-7x^2+36=0 given one root being...

solve `x^3-7x^2+36=0` given one root being twice the other .

Verified by Experts

The correct Answer is:

`therefore` The roots are 3,6,-2.

THEORY OF EQUATIONS
VIKRAM PUBLICATION ( ANDHRA PUBLICATION)|Exercise EXERCISE -4(a)|23 Videos
THEORY OF EQUATIONS
VIKRAM PUBLICATION ( ANDHRA PUBLICATION)|Exercise EXERCISE -4(b)|24 Videos
SYSTEM OF CIRCLE
VIKRAM PUBLICATION ( ANDHRA PUBLICATION)|Exercise EXERXCISE - 2(b)|23 Videos

Similar Questions

Explore conceptually related problems

solve x^3-7x^2+14x-8=0 given that the roots are in geometric progression.

Solve 18x^3+81x^2+121x+60=0 given that one roots is equal to half the sum of the remainging roots .

solve x^3 -9x^2 + 14x+ 24 =0 given that two of the roots are in the ratio 3:2

Solve 6x^4-13x^3-35x^2-x+3=0 given that one of its roots is 2+sqrt(3)

Solve the equation x^3 -6x^2 +7x +2=0 one root being 2 + sqrt(5)

solve the equations x^3 -3x^2 - 6x +8=0 the roots being in A.P

Solve the equation 6x^3 -11x^2 +6x -1=0 the roots being in H.P

Solve x^3-9x^2+14x +24-0 , given that two of the roots are in the ratio 3:2 .

Solve the x^3-3x^2-6x+8=0 equation , given that the roots of each are in A.P .