Question

The roots of the equation $ {x^2} - 3x - m\left( {m + 3} \right) = 0 $ , where m is a constant are
a. m, m+3
b. –m, m+3
c. m, -(m+3)
d. –m, -(m+3)

Answer 1

Hint:In this question consider the roots of the equation be $ \alpha $ and $ \beta $ , use the relationship between the sum of the roots and the coefficients of the quadratic equation that is sum of the roots is given as the ratio of the negative times the coefficient of x to the coefficient of x2 and the products of the roots with coefficients that is product of the roots is given as the ratio of the constant term to the coefficient of $ {x^2} $ . This will help approaching the problem.

Complete step-by-step answer:
Given quadratic equation
 $ {x^2} - 3x - m\left( {m + 3} \right) = 0 $ .................... (1)
Where m is a constant.
Let the roots of this equation be $ \alpha $ and $ \beta $
Now as we know that in a quadratic equation the sum of the roots is given as the ratio of the negative times the coefficient of x to the coefficient of x2.
So the sum of the roots is $ \alpha + \beta = - \dfrac{{{\text{coefficient of }}x}}{{{\text{coefficient of }}{x^2}}} $
Now from equation (1) the coefficient of x is -3 and the coefficient of x2 is 1.
Therefore, $ \alpha + \beta = - \left( {\dfrac{{ - 3}}{1}} \right) = 3 $ ............................. (2)
Now as we know that in a quadratic equation the product of the roots is given as the ratio of the constant term to the coefficient of x2.
Now from equation (1) constant term is –m (m + 3) and the coefficient of x2 is 1.
So the product of the roots, $ \alpha \times \beta = \dfrac{{{\text{constant term}}}}{{{\text{coefficient of }}{x^2}}} $
Therefore, $ \alpha \times \beta = \dfrac{{ - m\left( {m + 3} \right)}}{1} = - m\left( {m + 3} \right) $ ............................. (3)
Now it is a known fact that $ {\left( {a - b} \right)^2} = {\left( {a + b} \right)^2} - 4ab $ so use this property we have,
 $ \Rightarrow {\left( {\alpha - \beta } \right)^2} = {\left( {\alpha + \beta } \right)^2} - 4\alpha \beta $
Now from equation (2) and (3) we have,
 $ \Rightarrow {\left( {\alpha - \beta } \right)^2} = {\left( 3 \right)^2} - 4\left[ { - m\left( {m + 3} \right)} \right] = 9 + 4m\left( {m + 3} \right) $
\[ \Rightarrow {\left( {\alpha - \beta } \right)^2} = 4{m^2} + 12m + 9 = {\left( {2m + 3} \right)^2}\]
Now take square root on both sides we have,
\[ \Rightarrow \left( {\alpha - \beta } \right) = \sqrt {{{\left( {2m + 3} \right)}^2}} = \pm \left( {2m + 3} \right)\]
Now when, \[\left( {\alpha - \beta } \right) = \left( {2m + 3} \right)\]............................... (4)
Now add equation (2) and (4) we have,
 $ \Rightarrow \left( {\alpha + \beta + \alpha - \beta } \right) = 3 + \left( {2m + 3} \right) $
 $ \Rightarrow 2\alpha = 2m + 6 $
 $ \Rightarrow \alpha = m + 3 $
Now from equation (2) we have,
 $ \Rightarrow m + 3 + \beta = 3 $
 $ \Rightarrow \beta = - m $
So the roots of the quadratic equation is (m + 3, -m)
Now when, \[\left( {\alpha - \beta } \right) = - \left( {2m + 3} \right)\].................. (5)
Now add equation (2) and (5) we have,
 $ \Rightarrow \left( {\alpha + \beta + \alpha - \beta } \right) = 3 - \left( {2m + 3} \right) $
 $ \Rightarrow 2\alpha = - 2m $
 $ \Rightarrow \alpha = - m $
Now from equation (2) we have,
 $ \Rightarrow - m + \beta = 3 $
 $ \Rightarrow \beta = m + 3 $
So the roots of the quadratic equation is (-m, m + 3)
So this is the required answer.
Hence option (B) is the correct answer.

Note: There can be another method to solve this problem that is using Dharacharya formula, for any general quadratic equation of the form $ a{x^2} + bx + c = 0 $ the roots can be given as $ \bar x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $ . This gives two roots that is one by considering the positive sign and other by considering the negative sign. Substitution of direct value in this will help finding the roots.