Question

# If $$\displaystyle \frac{1}{2}, \displaystyle \frac{1}{3}, \displaystyle \frac{1}{4}$$ are the roots of $$ax^{3}+bx^{2}+cx+d=0$$, then the roots of $$a(x+1)^{3}+b(x+1)^{2}+c(x+1)+d=0$$ are

A
121314
B
32,43,54
C
12,23,34
D
12,23,34

Solution

## The correct option is A $$\displaystyle -\frac{1}{2},-\frac{2}{3},-\frac{3}{4}$$Let the equation whose roots are $$\alpha ,\beta$$ and $$\gamma$$ is $$f(x)=0$$ Then,The equation whose roots are $$\alpha -1,\beta -1$$ and $$\gamma -1$$ is $$f(x+1)=0$$Thus the roots of above equation [which is $$f(x+1)=0$$] will be $$-1+\dfrac12, -1+\dfrac13, -1+\dfrac14$$Maths

Suggest Corrections

0

Similar questions
View More

People also searched for
View More