Question

If $9{x^2} + kx + 1 = 0$ has equal roots, find the value of $k$.

Answer 1

Hint: In this problem we have given a quadratic equation. Also given that the roots of the given quadratic equation are the same. And we are asked to the value of $k$. We know that if the roots of the quadratic equations are the same then the discriminant value is equal to zero. By using this clue we are going to solve this problem.

Formula used: Discriminant $\left( D \right) = {b^2} - 4ac$

Complete step-by-step solution:
The quadratic equation is $9{x^2} + kx + 1 = 0$.
Let us compare the given quadratic equation with $a{x^2} + bx + c = 0$.
$ \Rightarrow a = 9,b = k,c = 1$.
We know that, Discriminant $\left( D \right) = {b^2} - 4ac$
Given that the roots are the same, so discriminant is equal to zero.
$ \Rightarrow {b^2} - 4ac = 0 - - - - - \left( 1 \right)$.
Now, substitute the values in equation $\left( 1 \right)$ then we get,
$ \Rightarrow {k^2} - 4 \times 9 \times 1 = 0$
$ \Rightarrow {k^2} - 36 = 0$
We can write the above equation as $\left( {k + 6} \right)\left( {k - 6} \right) = 0 - - - - - \left( 2 \right)$
If the product of two terms is equal to zero then one of the terms must be equal to zero or both the terms might be zero. So equate both the terms of equation $\left( 2 \right)$to zero.
$ \Rightarrow \left( {k - 6} \right) = 0$ and $\left( {k + 6} \right) = 0$
$\therefore $ $k = 6$ and $k = - 6$



Note: A quadratic equation is an equation of the second degree, meaning it contains at least one term that is squared. The standard form is $a{x^2} + bx + c = 0$ where $a,b$ and $c$ being constants, or numerical coefficients, and $x$ is an unknown variable. As well as being a formula that yields the zeros of any parabola, the quadratic formula can also be used to identify the axis of symmetry of the parabola, and the number of real zeros the quadratic equation contains. Also quadratic equations are actually used in everyday life, as when calculating areas, determining a product's profit or formulating the speed of an object.