Question

The equation whose roots exceed by $ \dfrac{1}{2} $ than those of $ 8{x^3} - 4{x^2} + 6x - 1 = 0 $ , is:
A. $ 8{x^3} - 16{x^2} + 8x - 3 = 0 $
B. $ 8{x^3} - 16{x^2} - 8x - 3 = 0 $
C. $ 8{x^3} - 8{x^2} + 8x - 3 = 0 $
D. $ 4{x^3} - 8{x^2} + 8x - 3 = 0 $

Answer 1

Hint: Roots of an equation are those values of the variable at which the function has zero value. The sum of the roots of the equation, product of the roots, etc. is related with the coefficient of variables in the equation. An equation can also be formed when the roots of that equation are known. Use this approach to find out the correct answer.

Complete step-by-step answer:
We are given an equation $ 8{x^3} - 4{x^2} + 6x - 1 = 0 $ . Let its roots be $ \alpha ,\beta \,and\,\gamma $ .
Sum of roots of a cubic equation is $ \alpha + \beta + \gamma = \dfrac{{ - b}}{a} $ where $ b $ is the coefficient of $ {x^2} $ and $ a $ is the coefficient of $ {x^3} $
Product of roots of a cubic equation is $ \alpha \beta \gamma = \dfrac{{ - d}}{a} $ where $ d $ is the constant term and $ a $ is the coefficient of $ {x^3} $
The sum of the product of the roots of a cubic equation is $ \alpha \beta + \beta \gamma + \alpha \gamma = \dfrac{c}{a} $ where $ c $ is the coefficient of $ x $ and $ a $ is the coefficient of $ {x^3} $
So, in the case of the given equation,
 $
\Rightarrow \alpha + \beta + \gamma = \dfrac{{ - ( - 4)}}{8} = \dfrac{1}{2} \\
\Rightarrow \alpha \beta + \beta \gamma + \alpha \gamma = \dfrac{6}{8} = \dfrac{3}{4} \\
\Rightarrow \alpha \beta \gamma = \dfrac{{ - ( - 1)}}{8} = \dfrac{1}{8} \;
  $
Now, we have to find the equation whose roots exceed the roots of the given equation by $ \dfrac{1}{2} $ so its roots will be $ \alpha + \dfrac{1}{2},\beta + \dfrac{1}{2}\,and\,\gamma + \dfrac{1}{2} $ .
We know that the cubic equation having roots $ \alpha ,\beta \,and\,\gamma $ is given as –
$\Rightarrow {x^3} - (\alpha + \beta + \gamma ){x^2} + (\alpha \beta + \beta \gamma + \alpha \gamma )x - \alpha \beta \gamma = 0 $
So the equation having roots $ \alpha + \dfrac{1}{2},\beta + \dfrac{1}{2}\,and\,\gamma + \dfrac{1}{2} $ is -
 $
\Rightarrow {x^3} - (\alpha + \dfrac{1}{2} + \beta + \dfrac{1}{2} + \gamma + \dfrac{1}{2}){x^2} + [(\alpha + \dfrac{1}{2})(\beta + \dfrac{1}{2}) + (\beta + \dfrac{1}{2})(\gamma + \dfrac{1}{2}) + (\alpha + \dfrac{1}{2})(\gamma + \dfrac{1}{2})]x - (\alpha + \dfrac{1}{2})(\beta + \dfrac{1}{2})(\gamma + \dfrac{1}{2}) = 0 \\
\Rightarrow {x^3} - (\alpha + \beta + \gamma + \dfrac{3}{2}){x^2} + (\alpha \beta + \dfrac{{\alpha + \beta }}{2} + \dfrac{1}{4} + \beta \gamma + \dfrac{{\beta + \gamma }}{2} + \dfrac{1}{4} + \alpha \gamma + \dfrac{{\alpha + \gamma }}{2} + \dfrac{1}{4})x - (\alpha \beta + \dfrac{{\alpha + \beta }}{2} + \dfrac{1}{4})(\gamma + \dfrac{1}{2}) = 0 \\
\Rightarrow {x^3} - (\alpha + \beta + \gamma + \dfrac{3}{2}){x^2} + (\alpha \beta + \beta \gamma + \alpha \gamma + \alpha + \beta + \gamma + \dfrac{3}{4})x - (\alpha \beta \gamma + \dfrac{{\alpha \beta + \beta \gamma + \alpha \gamma }}{2} + \dfrac{{\alpha + \beta + \gamma }}{4} + \dfrac{1}{8}) = 0 \\
  $
Now, put the known values in the equation obtained above –
 $
\Rightarrow {x^3} - (\dfrac{1}{2} + \dfrac{3}{2}){x^2} + (\dfrac{3}{4} + \dfrac{1}{2} + \dfrac{3}{4})x - (\dfrac{1}{8} + \dfrac{1}{2}(\dfrac{3}{4}) + \dfrac{1}{8} + \dfrac{1}{8}) = 0 \\
  {x^3} - 2{x^2} + 2x - \dfrac{3}{4} = 0 \\
  4{x^3} - 8{x^2} + 8x - 3 = 0 \;
  $
So, the correct answer is “Option D”.

Note: Usually, the polynomial equations have x as the variable, for example $ f(x) = a{x^2} + bx + c $ , on putting different values of x, we get different values of ā $ f(x) $ that is the y-coordinate on the graph. So the roots of an equation can also be defined as the x-intercepts as y-coordinate of the points lying on the x-axis is zero.