Question

How do you solve $ {{\left( x+2 \right)}^{2}}=16 $ ?

Answer 1

Hint: In this question, we are given an equation in terms of x and we need to find the solution to this equation. We need to find the value of x which satisfies this equation. For this, we will first take square root on both sides of the equation. After that, we will rearrange the equation. So that we can get two variables of x. We will use the property that if $ {{x}^{2}}=y\Rightarrow x=\pm \sqrt{y} $ i.e. we have to consider both signs while taking the square root.

Complete step by step answer:
Here we are given the equation as $ {{\left( x+2 \right)}^{2}}=16 $ . We need to find the value of x which satisfies the given equation. For this let us take square root on both sides of the equation we get $ {{\left( {{\left( x+2 \right)}^{2}} \right)}^{\dfrac{1}{2}}}={{\left( 16 \right)}^{\dfrac{1}{2}}} $ .
As we know that $ {{\left( {{a}^{m}} \right)}^{n}} $ can be written as $ {{a}^{mn}} $ so left side of the equation becomes $ {{\left( x+2 \right)}^{\dfrac{2}{2}}}={{\left( 16 \right)}^{\dfrac{1}{2}}}\Rightarrow {{\left( x+2 \right)}^{1}}={{\left( 16 \right)}^{\dfrac{1}{2}}}\Rightarrow x+2={{\left( 16 \right)}^{\dfrac{1}{2}}} $ .
Now let us solve the right side of the equation. As we know that $ {{\left( 4 \right)}^{2}}=16 $ and we know that $ {{\left( -4 \right)}^{2}} $ is also 16. So we can write 16 to be equal to $ {{\left( \pm 4 \right)}^{2}} $ . Hence the above equation becomes $ x+2={{\left( {{\left( \pm 4 \right)}^{2}} \right)}^{\dfrac{1}{2}}} $ .
Using the property of exponent that $ {{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}} $ we get the left side of the equation as $ x+2={{\left( \pm 4 \right)}^{\dfrac{2}{2}}}\Rightarrow x+2={{\left( \pm 4 \right)}^{1}}\Rightarrow x+2=\pm 4 $ .
This means we have two equations now, one with the right-hand side having positive 4 and the other with the negative 4. Separating the equations we get $ x+2=-4\text{ and }x+2=4 $ .
Solving them one by one we get,
Taking the first equation $ x+2=-4\Rightarrow x=-4-2\Rightarrow x=-6 $ .
Taking another equation $ x+2=4\Rightarrow x=4-2\Rightarrow x=2 $ .
So we have obtained the value of x which are 2 and -6. These values will satisfy this equation. Let us check this also.
Checking x = 2, putting x = 2 in the equation $ {{\left( x+2 \right)}^{2}}=16 $ we get,
 $ {{\left( 2+2 \right)}^{2}}=16\Rightarrow {{\left( 4 \right)}^{2}}=16 $ .
We know that $ {{\left( 4 \right)}^{2}}=4\times 4 $ . So we get $ 4\times 4=16\Rightarrow 16=16 $ .
The left-hand side is equal to the right-hand side. Therefore, x = 2 is the correct answer.
Checking x = -6, putting x = -6 in the equation $ {{\left( x+2 \right)}^{2}}=16 $ we get,
 $ {{\left( -6+2 \right)}^{2}}=16\Rightarrow {{\left( -4 \right)}^{2}}=16 $ .
We know that $ {{\left( -4 \right)}^{2}}=-4\times -4=4\times 4 $ so we get $ 4\times 4=16\Rightarrow 16=16 $ .
The left-hand side is equal to the right-hand side. Therefore, x = -6 is the correct answer.
So required values of x are 2, -6.

Note:
 Students should always try to check their answers after finding the value of x. Note that x has degree 2 so we must get two values of x. Students can also apply $ {{\left( a+b \right)}^{2}} $ to open the bracket and then solve the quadratic equation using quadratic formula or split the middle term method.