Question

How do you solve ${x^2} - 3x = - 3$ using the quadratic formula?

Answer 1

Hint: In this question, we are asked to solve the given equation using the quadratic equation formula. The quadratic equation formula is $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ .
First we have to check the existences of roots using the formula stated below and find the types of roots.
After that we have to apply the quadratic equation formula and derive the answer.

Formula used: Quadratic equation formula:
 $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
To check the roots = $\sqrt {{b^2} - 4ac} $

Complete step-by-step solution:
The given equation is ${x^2} - 3x = - 3$, we need to solve the equation using the quadratic equation formula.
First we should alter the given equation by transferring $ - 3$ to the other side.
$ \Rightarrow {x^2} - 3x = - 3$
$ \Rightarrow {x^2} - 3x + 3 = 0$
Quadratic equation formula:
 $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Before that, we have to check the roots. For that we will use this formula $\sqrt {{b^2} - 4ac} $
Clearly in${x^2} - 3x + 3 = 0$,
${\text{a = 1}}$,
${\text{b = - 3}}$,
${\text{c = 3}}$.
$ \Rightarrow \sqrt {{b^2} - 4ac} = \sqrt {9 - 12} $
$ \Rightarrow \sqrt {{b^2} - 4ac} = \sqrt { - 3} $
Therefore, the roots are not real.
Now applying ${x^2} - 3x + 3 = 0$in$\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$,
  $ \Rightarrow x = \dfrac{{3 \pm \sqrt { - {3^2} - 4(1)(3)} }}{{2(1)}}$
$ \Rightarrow x = \dfrac{{3 \pm \sqrt {9 - 12} }}{2}$
Simplifying the numerator we get,
$ \Rightarrow x = \dfrac{{3 \pm \sqrt { - 3} }}{2}$
Now we can expand the expression into two, as there is a $ \pm $ in the expression. One becomes plus and the other becomes minus.
$ \Rightarrow x = \dfrac{{3 + \sqrt { - 3} }}{2},\dfrac{{3 - \sqrt { - 3} }}{2}$
The roots are not real but they are imaginary roots.
$ \Rightarrow x = \dfrac{{3 + 3i}}{2},\dfrac{{3 - 3i}}{2}$
Now we get the value of $x$ which are equal roots.
$ \Rightarrow x = \dfrac{3}{2}(1 + i),\dfrac{3}{2}(1 - i)$
Therefore the roots of ${x^2} - 3x + 3 = 0$ are not real.

Note: Whenever the question asks you to find the roots with a quadratic equation, first you need to check the roots whether the roots exist or not and if it exists is it clear roots or equal roots. To check the roots, we have to find the value of $\sqrt {{b^2} - 4ac} $ first,
If $\sqrt {{b^2} - 4ac} $ is greater than $0$, then the roots are different and clear.
If $\sqrt {{b^2} - 4ac} $ is equal to $0$, then the roots are equal.
If $\sqrt {{b^2} - 4ac} $ is less than $0$, then the roots are not real or real roots do not exist.