Question

How do you solve ${(x - 3)^2} = 5$?

Answer 1

Hint: To solve this equation, firstly, change the equation in the standard form of the quadratic equation. Then we will move on and solve the quadratic equation with the help of the formula $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$where, $\sqrt {{b^2} - 4ac}$ will be discriminate. And at last, you will find two values for the respective equation.

Complete step By Step Solution:
The given equation is${(x - 3)^2} = 5$
Convert the equation into standard quadratic form
$\Rightarrow (x - 3)(x - 3) = 5$
Multiply $(x - 3)(x - 3)$, we get,
$\Rightarrow {x^2} - 3x - 3x + 9 = 5$
Simplifying the above equation, we get,
$\Rightarrow {x^2} - 6x + 9 = 5$
Adding 5 to both sides, we get,
$\Rightarrow {x^2} - 6x + 9 - 5 = 5 - 5$
Simplifying the above equation, we get,
$\Rightarrow {x^2} - 6x + 4 = 0$
Now, solve this quadratic equation by using the formula $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
By substituting $a = 1,b = - 6$and $c = 4$ we get,
$\Rightarrow x = \dfrac{{ - ( - 6) \pm \sqrt {{{( - 6)}^2} - 4 \times 1 \times 4} }}{{2 \times 1}}$
Simplifying the above equation, we get,
$\Rightarrow x = \dfrac{{6 \pm \sqrt {36 - 16} }}{2} \\
 \Rightarrow x = \dfrac{{6 \pm \sqrt {20} }}{2} \\ $
When considering – sign, we get,
$\Rightarrow x = \dfrac{{6 - \sqrt {20} }}{2}$
We can write $\sqrt {20}$as $\sqrt {4 \times 5}$, we get,
\[\Rightarrow x = \dfrac{{6 - \sqrt {4 \times 5} }}{2}\]
We know $\sqrt 4 = 2$,we get,
$\Rightarrow x = \dfrac{{6 - 2\sqrt 5 }}{2}$
Taking 2 commons from the numerator, we get,
$\Rightarrow x = \dfrac{{2(3 - \sqrt 5 )}}{2}$
The 2 from the numerator and denominator cancel out, we get,
$\Rightarrow x = 3 - \sqrt 5$
Similarly, when considering the + sign, we get,
$\Rightarrow x = \dfrac{{6 + \sqrt {20} }}{2}$
We can write $\sqrt {20}$as $\sqrt {4 \times 5}$, we get,
\[\Rightarrow x = \dfrac{{6 + \sqrt {4 \times 5} }}{2}\]
We know $\sqrt 4 = 2$,we get,
$\Rightarrow x = \dfrac{{6 + 2\sqrt 5 }}{2}$
Taking 2 commons from the numerator, we get,
$\Rightarrow x = \dfrac{{2(3 + \sqrt 5 )}}{2}$
The 2 from the numerator and denominator cancel out, we get,
$\Rightarrow x = 3 + \sqrt 5$

The two values are $x = 3 + \sqrt 5$ or $3 - \sqrt 5$

Additional Information:
The Quadratic Equation standard form is:
$\Rightarrow a{x^2} + bx + c$$= 0$ where a, b and c are known values and cannot be 0.
Moreover, x is the unknown or variable.
The formula for the quadratic equation is: $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Here in the above equation, there is a plus/minus ($\pm$) which means that we will consider one time positive and another time negative.
\[\Rightarrow x = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}} \\
   \Rightarrow x = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}} \\ \]
In the formula above there is b2 − 4ac. It is called the Discriminant because it can "discriminate" between some types of answer:
There are two real solutions. Moreover, it states that discrimination is positive.
There is just one real solution that is that both answers are the same. Moreover, it states that discrimination is zero.
There is a pair of Complex solutions. Moreover, it states that discrimination is negative.

Note:
There is an alternative method to find the solution to the equation.
The given equation is${(x - 3)^2} = 5$
Taking square roots on both sides, we get,
$\Rightarrow {(x - 3)^{2 \times \dfrac{1}{2}}} = \pm \sqrt 5$
In the power $2 \times \dfrac{1}{2}$ cancels out and become 1
$\Rightarrow x - 3 = \pm \sqrt 5$
When considering + sign, we get,
$\Rightarrow x - 3 = \sqrt 5$
Adding 3 to both sides, we get,
$\Rightarrow x - 3 + 3 = 3 + \sqrt 5 \\
\Rightarrow x = 3 + \sqrt 5 \\ $
Similarly, when considering – sign, we get,
$\Rightarrow x - 3 = - \sqrt 5$
Adding 3 to both sides, we get,
$\Rightarrow x - 3 + 3 = 3 - \sqrt 5 \\
\Rightarrow x = 3 - \sqrt 5 \\ $
The two values are
\[x = 3 + \sqrt 5\]
 or $3 - \sqrt 5$.