Question

If \[x\]< \[5\] then \[\sqrt{{{x}^{2}}-10x+25}+x+7\] is equal to,
(a) \[7\]
(b) \[5\]
(c) \[12\]
(d) Zero

Answer 1

Hint: Try simplifying the quadratic equation in the square root and think how the condition on \[x\] given to you can be utilised to solve this question.

Complete step by step solution:
We know the algebraic identity,
\[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\]
This same pattern we can observe in the quadratic equation, there is \[25\] which is \[{{5}^{2}}\] and there is \[-10x\] which is \[-2\times 5\times x\]. So it should immediately strike us that it is \[{{(x-5)}^{2}}\].
 Therefore, the equation now reduces to \[\sqrt{{{\left( x-5 \right)}^{2}}}+x+7\]
 And after arriving at this step, we immediately jump to the conclusion that \[\sqrt{{{\left( x-5 \right)}^{2}}}=x-5\] and hence the answer should be \[x-5+x+7=2x+2\] but this is the wrong answer.
To understand this we should know the basics of square root function and how it is defined.
\[\sqrt{{{x}^{2}}}=|x|\]
It is defined as \[x\] if \[x\] is positive and \[-x\] if \[x\] is negative. So, we collectively write it as \[|x|\].
Now let’s come to the condition which is given to us,
\[\begin{align}
  & x<5 \\
 & \Rightarrow x-5<0 \\
\end{align}\]
If \[(x-5)\] is negative, then we would follow the definition of \[\sqrt{{{x}^{2}}}\] when \[x\] is negative. Therefore, \[\sqrt{{{\left( x-5 \right)}^{2}}}=-(x-5)=5-x\]
Finally, the equation gets reduced to
\[5-x+x+7=12\] and this is our final answer.

Therefore, \[\sqrt{{{x}^{2}}-10x+25}+x+7 =12\], when $x$ < $5$. Therefore, option (C) is correct.

So, this question helps us understand the importance of the conditions given to us. If it was given that \[x>5\], then our answer would have changed and in that case our answer would be \[x-5+x+7=2x+2\].

Note:
Remember the square root of any number is always non-negative(zero or positive but not negative). You will encounter square roots in various topics of maths like pythagoras theorem, quadratic equation,length of the diagonal of a square etc. So you should be familiar with it.