Question

Find the value of $ \mu $ for which one root of the quadratic equation $ \mu {{\text{x}}^{\text{2}}} - {\text{14x}} + {\text{8 = 0}} $ is $ 6 $ times the other.

Answer 1

Hint: General form of quadratic equation is $ {\text{a}}{{\text{x}}^{\text{2}}} + {\text{bx}} + {\text{c}} = {\text{0}} $
If $ \alpha $ and $ \beta $ are the roots of the given quadratic equation, then
Sum of the roots of the equation is $ - \dfrac{{\text{b}}}{{\text{a}}} $
Product of the roots of the equation is $ \dfrac{{\text{c}}}{{\text{a}}} $
Follow the given conditions in the question and solve accordingly.

Complete step-by-step answer:
Given: The quadratic equation is $ \mu {{\text{x}}^{\text{2}}} - {\text{14x}} + {\text{8 = 0}} $
One root of the equation given is $ 6 $ times the other.
We need to find the value of $ \mu $
Let the roots of the given quadratic equation be $ \alpha $ and $ \beta $ respectively.
According to the question,
 $ \alpha $ is 6 times of $ \beta $
 $ \Rightarrow \alpha = 6 \times \beta $
Sum of the roots of the equation in general form $ {\text{a}}{{\text{x}}^{\text{2}}} + {\text{bx}} + {\text{c}} = {\text{0}} $ is $ - \dfrac{{\text{b}}}{{\text{a}}} $
Comparing the given equation $ \mu {{\text{x}}^{\text{2}}} - {\text{14x}} + {\text{8 = 0}} $ with the general form, we get
 $
  {\text{a}} = \mu \\
  {\text{b}} = - 14 \\
  {\text{c}} = 8 \;
  $
 $ \alpha + \beta = - \dfrac{{\text{b}}}{{\text{a}}} = - \left( {\dfrac{{ - {\text{14}}}}{\mu }} \right) $
 $ \Rightarrow \alpha + \beta = \dfrac{{14}}{\mu } $ ……..Equation (1)
Given that $ \alpha = 6 \times \beta $
Substituting $ \alpha = 6 \times \beta $ in Equation (1)
We get,
 $
  {\text{6}}\beta + \beta = \dfrac{{{\text{14}}}}{\mu } \\
   \Rightarrow {\text{7}}\beta = \dfrac{{{\text{14}}}}{\mu } \;
  $
 $ \Rightarrow \beta = \dfrac{2}{\mu } $ ……..Equation (2)
Product of the roots of the equation in general form $ {\text{a}}{{\text{x}}^{\text{2}}} + {\text{bx}} + {\text{c}} = {\text{0}} $ is $ \dfrac{{\text{c}}}{{\text{a}}} $
 $ \alpha \times \beta = \dfrac{c}{a} = \dfrac{8}{\mu } $ ……..Equation (3)
Substituting $ \alpha = 6 \times \beta $ in Equation (3)
We get,
 $ 6\beta \times \beta = \dfrac{8}{\mu } $
 $ \Rightarrow 6{\beta ^2} = \dfrac{8}{\mu } $ ……..Equation (4)
Substituting Equation (2) in Equation (4) , we get
 $
  {\text{6}} \times {\left( {\dfrac{{\text{2}}}{\mu }} \right)^{\text{2}}} = \dfrac{{\text{8}}}{\mu } \\
   \Rightarrow {\text{6}} \times \dfrac{{\text{4}}}{{{\mu ^{\text{2}}}}} = \dfrac{{\text{8}}}{\mu } \\
   \Rightarrow \mu = 3 \;
  $
Therefore, the value of $ \mu $ is $ 3 $ .
So, the correct answer is “3”.

Note: In this type of questions which involves the concept of quadratic equations, we need to have knowledge about the formulae related to sum of roots and product of roots of a general quadratic equation. Be careful with calculations as we come across many calculations. Follow the conditions given in the question and solve them accordingly to find the required answer.