Question

How do you solve \[3-x=\sqrt{{{x}^{2}}+15}\]?

Answer 1

Hint: This question is from the topic of algebra. In solving this question, we will first square the both sides of the equation. After that, we will break the square term of the left side of the equation. After that, we will take all the terms of \[{{x}^{2}}\] and \[x\] to the left side of the equation and all the constant terms to the right side of the equation. After solving the further equation, we will get the value of x.

Complete step by step answer:
Let us solve this question.
In this question, we have asked to solve the equation \[3-x=\sqrt{{{x}^{2}}+15}\]. That means we have to find the value of x from the given equation.
The equation which we have to solve is
\[3-x=\sqrt{{{x}^{2}}+15}\]
On squaring both sides of equation, we can write the above equation as
\[{{\left( 3-x \right)}^{2}}={{\left( \sqrt{{{x}^{2}}+15} \right)}^{2}}\]
As we know that \[\left( \sqrt{a} \right)\] can also be written as \[a\], so we can write the above equation as
\[\Rightarrow {{\left( 3-x \right)}^{2}}={{x}^{2}}+15\]
As we know that \[{{\left( a-b \right)}^{2}}\] can also be written as \[{{a}^{2}}+{{b}^{2}}-2\times a\times b\], so we can write the above equation as
\[\Rightarrow {{3}^{2}}+{{x}^{2}}-2\times 3\times x={{x}^{2}}+15\]
The above equation can also be written as
\[\Rightarrow 9+{{x}^{2}}-6x={{x}^{2}}+15\]
After putting all values of \[x\] and \[{{x}^{2}}\] to the left side of the equation and all the constant terms to the right side of equation, the above equation can also be written as
\[\Rightarrow {{x}^{2}}-{{x}^{2}}-6x=15-9\]
The above equation can also be written as
\[\Rightarrow -6x=6\]
After dividing -6 to the both side of equation, we can write the above equation as
\[\Rightarrow \dfrac{-6x}{-6}=\dfrac{6}{-6}\]
The above equation can also be written as
\[\Rightarrow x=-1\]
Hence, we have solved the equation \[3-x=\sqrt{{{x}^{2}}+15}\] and got the value of x as -1.

Note:
We should have better knowledge in the topic of algebra to solve this type of question easily. We should remember the following formula:
\[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2\times a\times b\]
If we want to check if our answer is correct or not, then we can cross-check our answer.
Let us cross-check our answer.
After putting the value of x as -1 in the equation \[3-x=\sqrt{{{x}^{2}}+15}\], we can write
\[3-\left( -1 \right)=\sqrt{{{\left( -1 \right)}^{2}}+15}\]
The above equation can also be written as
\[\Rightarrow 3+1=\sqrt{1+15}\]
The above equation can also be written as
\[\Rightarrow 4=\sqrt{16}\]
As we know that the square root of 16 is 4, so we can write
\[\Rightarrow 4=4\]
Hence, both sides of the equation are equal. So, we can say that our answer is correct.