Question

The equation whose roots exceed by $\frac{1}{2}$ than those of $8x^3-4x^2+6x-1=0$, is:

• A. $8x^3-16x^2+8x-3=0$
• B. $8x^3-16x^2-8x-3=0$
• C. $8x^3-8x^2+8x-3=0$
• D. $4x^3-8x^2+8x-3=0$

Answer 1

Answer: (D)

$4x^3-8x^2+8x-3=0$

Suppose the roots of the given equation is ${\alpha }_1$ ,${\beta }_1$ ,${\gamma }_1$

And the roots of the equation whose roots is exceed by $\frac{1}{2}$ are ${\alpha }_2$ ,${\beta }_2$ ,${\gamma }_2$

So we can get the relation

${\alpha }_2$ $={\alpha }_1+\frac{1}{2}$, ${\beta }_2=$ ${\beta }_1+$$\frac{1}{2}$, ${\gamma }_2=$ ${\gamma }_1+$$\frac{1}{2}$

Now we know that ${\alpha }_1$, ${\beta }_1$, ${\gamma }_1$ will satisfy the equation

So replace ${\alpha }_1$ by ${\alpha }_2-$ $\frac{1}{2}$, and then put it in the given equation

$\Rightarrow$ $8{\alpha }_1^3-4{\alpha }_1^2+6{\alpha }_1-1=0$, ${\alpha }_1$$\to$${\alpha }_2-$ $\frac{1}{2}$

Then we will get $4{\alpha }_2^3-8{\alpha }_2^2+8{\alpha }_2-3=0$

This is the equation which have roots ${\alpha }_2$, ${\beta }_2$, and ${\gamma }_2$ and which is exceed $\frac{1}{2}$ from ${\alpha }_1$ ,${\beta }_1$ ,${\gamma }_1$

Now final equation i is $4x^3-8x^2+8x-3=0$

Hence, option D is correct.