Question

How do you use the quadratic formula to solve ${x^2} + 5 = 4x$?

Answer 1

Hint: Here we can solve the given quadratic equation by using quadratic formula. On doing some simplification we get the required answer.
Solving quadratic equations involves the use of the following formula
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ where $a,b$ and $c$ are taken from the quadratic equation written in its general form of $a{x^2} + bx + c = 0$ where a is the numeral that goes in front of ${x^2}$ ,$b$ is the numeral that goes in front of $x$ , and $c$ is the numeral with no variable next to it.

Complete step-by-step solution:
Given quadratic equation is ${x^2} + 5 = 4x$
Re order the given equation like this ${x^2} - 4x + 5 = 0$. Comparing this equation with the general form of the quadratic equation that is $a{x^2} + bx + c = 0$ where a is the numeral that goes in front of ${x^2}$ ,$b$ is the numeral that goes in front of $x$ , and $c$ is the numeral with no variable next to it.
 we get,
$a = 1$ ,$b = - 4$ and $c = 5$
We have the quadratic formula that is $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Substitute the values of $a,b$ and $c$ in this formula we get,
\[ \Rightarrow x = \dfrac{{ - \left( { - 4} \right) \pm \sqrt {{{\left( { - 4} \right)}^2} - 4\left( 1 \right)\left( 5 \right)} }}{{2\left( 1 \right)}}\]
On squaring the term and we get.
$ \Rightarrow x = \dfrac{{4 \pm \sqrt {16 - 20} }}{2}$
Let us subtract we get
$ \Rightarrow x = \dfrac{{4 \pm \sqrt { - 4} }}{2}$
On splitting the term and we get
$ \Rightarrow x = \dfrac{{4 \pm \sqrt {4( - 1)} }}{2}$
On rewriting we get
$ \Rightarrow x = \dfrac{{4 \pm 2\sqrt { - 1} }}{2}$
Hence, we get
$ \Rightarrow x = 2 \pm \sqrt { - 1} $
We know that the complex numbers
$i = \sqrt { - 1} $ . Using this we get the imaginary roots,
$x = 2 \pm i$

Therefore the quadratic equation has two solution that is \[x = \left\{ {2 + i,2 - i} \right\}\]

Note: Many quadratic equations cannot be solved by factoring. This is generally true when the roots, or answers, are not rational numbers. A second method of solving quadratic equations involves the use of the quadratic formula. When using the quadratic formula, you should be aware of three possibilities. These three possibilities are distinguished by a part of the formula called the discriminant. The discriminant is the value of under the radical sign, ${b^2} - 4ac$. A quadratic equation with real numbers as coefficients can have the following:
Two different real roots if the discriminant ${b^2} - 4ac$ is a positive number
One real root if the discriminant ${b^2} - 4ac$ is equal to $0$
No real root if the discriminant ${b^2} - 4ac$ is a negative number.