Question

If the roots of the equation \[\left( {2k + 3} \right){x^2} + 2\left( {k + 3} \right)x + \left( {k + 5} \right) = 0,k \in R,k \ne \dfrac{{ - 3}}{2}\] are equal then k equals:
(A) $ 1,6 $
(B) $ - 1, - 6 $
(C) $ - 1,6 $
(D) $ 1, - 6 $

Answer 1

Hint: In the given question, we are required to solve for the value of k such that the equation \[\left( {2k + 3} \right){x^2} + 2\left( {k + 3} \right)x + \left( {k + 5} \right) = 0\] has equal roots. We will first compare the given equation with the standard form of a quadratic equation $ a{x^2} + bx + c = 0 $ and then apply a quadratic formula to find the condition for equal roots of a quadratic equation.

Complete step-by-step answer:
In the given question, we are provided with the equation \[\left( {2k + 3} \right){x^2} + 2\left( {k + 3} \right)x + \left( {k + 5} \right) = 0\].
Now, comparing the equation with standard form of a quadratic equation $ a{x^2} + bx + c = 0 $
Here, $ a = \left( {2k + 3} \right) $ , $ b = 2\left( {k + 3} \right) $ and $ c = \left( {k + 5} \right) $ .
Now, using the quadratic formula, we get the roots of the equation as:
 $ x = \dfrac{{( - b) \pm \sqrt {{b^2} - 4ac} }}{{2a}} $
If both the roots of a quadratic equation are equal, then, we get,
 $ {x_1} = {x_2} $
 $ \Rightarrow \dfrac{{( - b) + \sqrt {{b^2} - 4ac} }}{{2a}} = \dfrac{{( - b) - \sqrt {{b^2} - 4ac} }}{{2a}} $ ]
Cross multiplying the terms of equation and simplifying further, we get,
 $ \Rightarrow \sqrt {{b^2} - 4ac} = - \sqrt {{b^2} - 4ac} $
Shifting all the terms to left side and dividing both sides of equation by two, we get,
 $ \Rightarrow \sqrt {{b^2} - 4ac} = 0 $
Now, we can substitute the values of a, b and c in the expression. So, we get,
 $ \Rightarrow \sqrt {{{\left( {2\left( {k + 3} \right)} \right)}^2} - 4\left( {2k + 3} \right)\left( {k + 5} \right)} = 0 $
 $ \Rightarrow \sqrt {4{{\left( {k + 3} \right)}^2} - 4\left( {2k + 3} \right)\left( {k + 5} \right)} = 0 $
Opening the brackets and expanding the whole square term using the algebraic identity $ {\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2} $ , we get,
 $ \Rightarrow \sqrt {4\left( {{k^2} + 2\left( 3 \right)k + {3^2}} \right) - 4\left( {2{k^2} + 13k + 15} \right)} = 0 $
Simplifying the expression, we get,
 $ \Rightarrow \sqrt {4\left( {{k^2} + 6k + 9} \right) - 4\left( {2{k^2} + 13k + 15} \right)} = 0 $
Factoring out $ 4 $ from the square root, we get,
 $ \Rightarrow 2\sqrt {\left( {{k^2} + 6k + 9} \right) - \left( {2{k^2} + 13k + 15} \right)} = 0 $
Dividing both sides of equation by two and squaring both sides of equation, we get,
 $ \Rightarrow {\left[ {\sqrt {{k^2} + 6k + 9 - 2{k^2} - 13k - 15} } \right]^2} = {0^2} $
Cancelling the like terms with opposite signs, we get,
 $ \Rightarrow - {k^2} - 7k - 6 = 0 $
Now, we have to solve the above quadratic equation to find the values of k. First shifting all the terms to right side of equation, we get,
 $ \Rightarrow {k^2} + 7k + 6 = 0 $
Using splitting the middle term method to solve the equation, we get,
 $ \Rightarrow {k^2} + 6k + k + 6 = 0 $
Taking common terms out of the bracket in first two and two last terms, we get,
 $ \Rightarrow k\left( {k + 6} \right) + \left( {k + 6} \right) = 0 $
Simplifying the expression further,
 $ \Rightarrow \left( {k + 1} \right)\left( {k + 6} \right) = 0 $
Now, since the product of two terms is equal to zero. So, either of the two terms should be zero. Hence, either $ \left( {k + 1} \right) = 0 $ or $ \left( {k + 6} \right) = 0 $
 $ \Rightarrow k = - 1 $ or $ k = - 6 $
Hence, the values of k are: $ - 1 $ and $ - 6 $ .
Hence, option (B) is the correct answer.
So, the correct answer is “Option B”.

Note: Quadratic equations are the polynomial equations with degree of the variable or unknown as 2. We should also know the expression of the discriminant of a quadratic equation so as to solve the question. We should take care of the calculations to be sure of the final answer.