Question

How do you solve ${{\left( 3x-1 \right)}^{2}}=12$ using the square root property?

Answer 1

Hint: We are given an equation as ${{\left( 3x-1 \right)}^{2}}=12$ , we have to find a solution using a square root property.
To do so we first learn what type of equation we have, we will find how many solution we can have for given equation after that we will find property as square root which are useful to find the solution, we will use that is ${{a}^{2}}=b$ then a is given as $a=\pm \sqrt{b}$ , using all this we will solve our problem, we will also use algebraic tool so simplify.

Complete step by step solution:
We are given a equation as ${{\left( 3x-1 \right)}^{2}}=12$ , we can see that it has a square over the variable term so our equation is a quadratic equation, quadratic equation has ‘2’ solution at most as its highest power is 2, so for ${{\left( 3x-1 \right)}^{2}}=12$ , we can find two solution –
Now, we will look for the property of square root.
It is when we find the root as any value. We can have a possible two square root for any value that if we have ${{a}^{2}}$ , then it can have two roots and its root will be opposite of one another.
So, let if ${{a}^{2}}=b$ then $a=\pm \sqrt{b}$ .
So, we will use $a=\pm \sqrt{b}$ is ${{a}^{2}}=b$ to solve our problem.
Now as we have that ${{\left( 3x-1 \right)}^{2}}=12$ .
So, we take $a={{\left( 3x-1 \right)}^{2}}$ and $b=12$ .
So, we have ${{a}^{2}}=b$ , so it is the same as $a=\pm \sqrt{b}$ .
So, we get –
$\Rightarrow \left( 3x-1 \right)=\pm \sqrt{12}$ .
Now as $12=2\times 2\times 3$ .
So, $\Rightarrow 3x-1=\pm \sqrt{12}$
Becomes, $3x-1=\pm 2\sqrt{3}$ (as 2 making a pair so it will came out)
Now, we add ‘1’ on both sides, we get –
$\Rightarrow 3x-1+1=1\pm 2\sqrt{3}$
By simplifying, we get –
$\Rightarrow 3x=1\pm 2\sqrt{3}$
Now, we divide both sides by ‘3’, we will get –
$\Rightarrow \dfrac{3x}{3}=\dfrac{1\pm 2\sqrt{3}}{3}$ .
So,
$\Rightarrow x=\dfrac{1\pm 2\sqrt{3}}{3}$ .
By separation terms, we get solution as –
$x=\dfrac{1+2\sqrt{3}}{3}$ and $x=\dfrac{1-2\sqrt{3}}{3}$ .

Note: Remember this, we have a quadratic equation, so it must have that it has 2 solutions. Also if we solve that ${{a}^{2}}=b\Rightarrow a=\sqrt{b}$ then, we are doing an incomplete solution, we need to consider all possible values.
So, we must take $a=\pm \sqrt{b}$ , because we should know that square of 2 is 4 and also square of -2 is 4. Hence the square root of 4 may be 2 or -2 two possibilities are there.