Question

How do you find the discriminant of ${x^2} - 12x + 4 = 0?$

Answer 1

Hint: In this question, we are going to find the discriminant of the given equation.
To find the discriminant of the given equation, the given equation must be of the quadratic form.
Compare the quadratic form to the given equation and substitute the values in the quadratic formula.
We can get the required result.

Formula used: The quadratic formula can be written as
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Here ${b^2} - 4ac$ is the formula for discriminant.
There are three possible outcomes for the discriminant.
${b^2} - 4ac < 0$, the quadratic equation has no real solutions.
${b^2} - 4ac = 0,$ The quadratic equation has only one solution or two real and equal solutions.
${b^2} - 4ac > 0,$ The quadratic equation has two real and distinct solutions.

Complete step-by-step solution:
In this question, we are going to find the discriminant of the given equation.
First write the given equation in the quadratic form and mark it as $\left( 1 \right)$
${x^2} - 12x + 4 = 0....\left( 1 \right)$
Here $a = 1,\,b = - 12,\,c = 4$
It has a discriminant $\Delta $ given by the formula:
$\Delta = {b^2} - 4ac$
$ \Rightarrow \Delta = {\left( { - 12} \right)^2} - 4\left( 1 \right)\left( 4 \right)$
On simplify the term and we get,
$ \Rightarrow \Delta = 144 - 16$
On subtracting the term and we get
$ \Rightarrow \Delta = 128$
On squaring the term and we get
$ \Rightarrow \Delta = 2{\left( 8 \right)^2}$
Since $\Delta > 0,$ this quadratic equation has two distinct real roots, but since $\Delta $ is not a perfect square, those roots are irrational.
We can find the roots using the quadratic formula.
Substitute those values in the quadratic formula
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
On putting the values and we get,
$ \Rightarrow x = \dfrac{{ - \left( { - 12} \right) \pm \sqrt {{{\left( { - 12} \right)}^2} - 4\left( 1 \right)\left( 4 \right)} }}{{2\left( 1 \right)}}$
On simplify the term and we get,
$ \Rightarrow x = \dfrac{{12 \pm \sqrt \Delta }}{2}$
Then, we get
$ \Rightarrow x = \dfrac{{12 \pm \sqrt {2{{\left( 8 \right)}^2}} }}{2}$
On squaring the term and we get,
$ \Rightarrow x = \dfrac{{12 \pm 8\sqrt 2 }}{2}$
Let us split the term and we get
$ \Rightarrow x = \dfrac{{12 + 8\sqrt 2 }}{2},x = \dfrac{{12 - 8\sqrt 2 }}{2}$
On simplify the term and we get
$ \Rightarrow x = \dfrac{{4\left( {3 + 2\sqrt 2 } \right)}}{2},x = \dfrac{{4\left( {3 - 2\sqrt 2 } \right)}}{2}$
Let us divide the term and we get
$ \Rightarrow x = 2\left( {3 + 2\sqrt 2 } \right),x = 2\left( {3 - 2\sqrt 2 } \right)$
Hence, we get
$ \Rightarrow x = 6 \pm 4\sqrt 2 $

Therefore the value of x is $6 \pm 4\sqrt 2 $.

Note: There are various methods to solve the quadratic equation. Some of them are as follows: factoring, using the square roots, completing the square and the quadratic formula.
The discriminant can be positive, zero, or negative, and this determines how many solutions are there to the given quadratic equation. The discriminant determines the nature of the roots of a quadratic equation. The word nature refers to the types of numbers the roots can be namely real, rational, irrational or imaginary.