Question

For what value of k will ${x^2} - (3k - 1)x + 2{k^2} + 2k = 11$ have equal roots?
1. 9, -5
2. -9, 5
3. 9, 5
4. -9, -5

Answer 1

Hint: We have to find the value of k from the equation. It is given in the question that the quadratic equation has equal roots which means ${b^2} - 4ac$ is equal to zero. So, we will put the values in this and simplify to find the value of k

Complete step-by-step solution:
Given: ${x^2} - (3k - 1)x + 2{k^2} + 2k = 11$
The quadratic equation has equal roots. So,
${b^2} - 4ac = 0$
So, ${b^2} = 4ac$
In this, a is equal to $1$, b is equal to $ - (3k - 1)$ and c is equal to $2{k^2} + 2k - 11$.
Substituting the values in the equation.
${( - (3k - 1))^2} = 4 \times 1 \times (2{k^2} + 2k - 11)$
$\Rightarrow 9{k^2} + 1 - 6k = 8{k^2} + 8k - 44$
$\Rightarrow 9{k^2} - 8{k^2} - 6k - 8k + 1 + 44 = 0$
$\Rightarrow {k^2} - 14k + 45 = 0$
We will split -14k as -9k and -5k in the above equation.
${k^2} - 9k - 5k + 45 = 0$
Now, we will take k common from the first two terms and -5 from the last two terms.
$k\left( {k - 9} \right) - 5\left( {k - 9} \right) = 0$
$\Rightarrow \left( {k - 9} \right)\left( {k - 5} \right) = 0$
First, we will take the first term to find a value of k.
$k - 9 = 0$
$\Rightarrow k = 9$
One value of k is 9.
Now, we will take the second term to find another value of k.
$k - 5 = 0$
$\Rightarrow k = 5$
Other value of k is 5.
So, option (3) is the correct answer.

Note: A quadratic equation is an algebraic expression of the second degree in x. The standard form of a quadratic equation is $a{x^2} + bx + c = 0$, where a, b are the coefficients, x is the variable, and c is the constant term. The roots of a quadratic equation are the two values of x, which are obtained by solving the quadratic equation. The value ${b^2} - 4ac$ is called the discriminant of a quadratic equation. For D = 0,the roots are real and equal.