Question

How do you solve ${x^2} + 10x = 4$ by completing the square?

Answer 1

Hint: Here we have a polynomial, it is desirable to express in the form of ${a^2} + {b^2}$ , to do this, we can artificially introduce a constant which allows us to factor a perfect square out of the expression. Notice that by simultaneously adding and subtracting, we have not changed the value of the expression.

Complete step by step answer:
An expression like ${x^2} + 10x = 4$ may be thought of as an “incomplete” square. To make it complete, we would need to add ${n^2}$ to the expression. We can figure out what to use for $n$ by realizing that the $10x$ in the middle need to be $2nx$. So we want $n = 5$ and ${n^2} = 25$.
Of course we can just add a number to an expression without changing the value of the expression. So if we want to keep the same value we will have to make up for adding $25$. We do this by also subtracting $25$. That does not change the value of ${x^2} + 10x$ , but it does change the way it is written.
We write ${x^2} + 10x + 25 - 25 = 4$. If we grouping this way,
$ \Rightarrow \left( {{x^2} + 10x + 25} \right) - 25 = 4$
Then we have a perfect square minus $25$
And we write,
$ \Rightarrow {\left( {x + 5} \right)^2} - 25 = 4$
Solving an equation by completing square,
$ \Rightarrow {\left( {x + 5} \right)^2} - 25 = 4$
Adding$25$in both sides of the equation, it becomes
$ \Rightarrow {\left( {x + 5} \right)^2} - 25 + 25 = 4 + 25$
On addition,
$ \Rightarrow {\left( {x + 5} \right)^2} + 0 = 29$
Adding zero,
$ \Rightarrow {\left( {x + 5} \right)^2} = 29$
Taking square on both sides of the above equation, we get,
$ \Rightarrow \sqrt {{{\left( {x + 5} \right)}^2}} = \sqrt {29} $
$ \Rightarrow \left( {x + 5} \right) = \pm \sqrt {29} $
And the last equation above is satisfied exactly when
$x + 5 = \sqrt {29} $ or $x + 5 = - \sqrt {29} $
So, $x = - 5 + \sqrt {29} $ or $x = - \left( {5 + \sqrt {29} } \right)$

Then the solution of the given equation is $\left\{ { - 5 + \sqrt {29} , - \left( {5 + \sqrt {29} } \right)} \right\}$.

Note: we can also find the solution of the given quadratic equation by using the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ . By this formula we can easily get the solution involving with simple steps.
Given equation is ${x^2} + 10x = 4$, reordering the term ${x^2} + 10x - 4 = 0$,
Consider$a = 1$ , $b = 10$ and $c = - 4$, substitute this value into the formula,
$x = \dfrac{{ - 10 \pm \sqrt {100 + 16} }}{2}$
On simplified,
$ \Rightarrow x = \dfrac{{ - 10 \pm \sqrt {116} }}{2}$
We just factorized root term,
$ \Rightarrow x = \dfrac{{ - 10 \pm \sqrt {4 \times 29} }}{2}$
Simplified the root term
$ \Rightarrow x = \dfrac{{ - 10 \pm 2\sqrt {29} }}{2}$
Taking common $'2'$
$ x = \dfrac{{2( - 5 \pm \sqrt {29} )}}{2}$
We cancel the common term, we get
$ \Rightarrow x = - 5 \pm \sqrt {29} $
Finally we get the solution of the equation.