Question

A value of K for which the quadratic equation ${x^2} - 2x\left( {1 + 3k} \right) + 7\left( {2k + 3} \right) = 0$ has equal roots is:
(A) $1$
(B) $2$
(C) $3$
(D) $4$

Answer 1

Hint: In the given question, we are required to solve for the value of k such that the equation ${x^2} - 2x\left( {1 + 3k} \right) + 7\left( {2k + 3} \right) = 0$ has equal roots. We will first compare the given equation with the standard form of a quadratic equation $a{x^2} + bx + c = 0$. Then, we will apply the quadratic formula and find the condition for equal roots of the equation.


Complete step by step solution: 
In the given question, we are provided with the equation ${x^2} - 2x\left( {1 + 3k} \right) + 7\left( {2k + 3} \right) = 0$.
We simplify the equation by opening the brackets.
$ \Rightarrow {x^2} - 2x\left( {1 + 3k} \right) + 7\left( {2k + 3} \right) = 0$
Now, comparing the equation with standard form of a quadratic equation $a{x^2} + bx + c = 0$
Here, $a = 1$, $b = - 2\left( {1 + 3k} \right)$ and $c = 7\left( {2k + 3} \right)$.
Now, we use the quadratic formula.
$x = \dfrac{{( - b) \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
For both roots to be equal, we have,
${x_1} = {x_2}$
$ \Rightarrow \dfrac{{( - b) + \sqrt {{b^2} - 4ac} }}{{2a}} = \dfrac{{( - b) - \sqrt {{b^2} - 4ac} }}{{2a}}$]
Cancelling the common terms, we get,
$ \Rightarrow \sqrt {{b^2} - 4ac} = - \sqrt {{b^2} - 4ac} $
Using transposition method to transfer all terms to the right side of equation,
$ \Rightarrow \sqrt {{b^2} - 4ac} = 0$
Now, substituting the values of a, b and c in the expression. So, we get,
Here, $a = 1$, $b = - 2\left( {1 + 3k} \right)$ and $c = 7\left( {2k + 3} \right)$.
$ \Rightarrow \sqrt {{{\left( { - 2\left( {1 + 3k} \right)} \right)}^2} - 4\left( 1 \right)\left( {7\left( {2k + 3} \right)} \right)} = 0$
$ \Rightarrow \sqrt {4{{\left( {1 + 3k} \right)}^2} - 28\left( {2k + 3} \right)} = 0$
Squaring both sides,
$ \Rightarrow 4{\left( {1 + 3k} \right)^2} - 28\left( {2k + 3} \right) = 0$
Opening the whole square using ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$, we get,
$ \Rightarrow 4\left[ {{{\left( {3k} \right)}^{^2}} + 2\left( {3k} \right) + {1^2}} \right] - 28\left( {2k + 3} \right) = 0$
$ \Rightarrow 4\left[ {9{k^{^2}} + 6k + 1} \right] - 28\left( {2k + 3} \right) = 0$
$ \Rightarrow 36{k^{^2}} + 24k + 4 - 56k - 84 = 0$
$ \Rightarrow 36{k^{^2}} - 32k - 80 = 0$
Dividing both sides by $4$, we get,
$ \Rightarrow 9{k^{^2}} - 8k - 20 = 0$
Using splitting the middle term method, we get,
$ \Rightarrow 9{k^{^2}} - 18k + 10k - 20 = 0$
Taking $9k$ common from first two terms and $ + 10$ from last two terms, we get,
$ \Rightarrow 9k\left( {k - 2} \right) + 10\left( {k - 2} \right) = 0$
$ \Rightarrow \left( {k - 2} \right)\left( {9k + 10} \right) = 0$
Since the product of two terms is equal to zero. So, one of the two terms has to be zero. So, we get,
Either $k - 2 = 0$ or $9k + 10 = 0$
So, we get, either $k = 2$ or $k = - \dfrac{{10}}{9}$.
Hence, the values of k are: $k = 2$ or $k = - \dfrac{{10}}{9}$.
So, the correct answer is option (b) $2$.

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.