Question

Solve the following inequation: $\sqrt{{{x}^{2}}-3x-10}>\left( 8-x \right)$.

Answer 1

Hint: In order to solve this question, we should have some knowledge about the rules of inequalities, like if $a>b$, then $a-b>0$ and if $a>b$, then ${{a}^{2}}>{{b}^{2}}$. Also, while solving this question, we should remember the algebraic identity, ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ and that ${{\left( \sqrt{a} \right)}^{2}}=a$. We can solve this question by using these concepts.

Complete step-by-step answer:
In this question, we have been asked to solve an inequality, that is, $\sqrt{{{x}^{2}}-3x-10}>\left( 8-x \right)$. Now, to solve this inequality, we will first square both the sides of the given inequality. So, we will get the inequality as follows,
${{\left( \sqrt{{{x}^{2}}-3x-10} \right)}^{2}}>{{\left( 8-x \right)}^{2}}$
Now, we know that ${{\left( \sqrt{a} \right)}^{2}}=a$ and that ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$. So, we will write the term, ${{\left( \sqrt{{{x}^{2}}-3x-10} \right)}^{2}}={{x}^{2}}-3x-10$. Similarly, we will write the other term, ${{\left( 8-x \right)}^{2}}=64+{{x}^{2}}-16x$. So, we will get the above inequality as,
${{x}^{2}}-3x-10>64+{{x}^{2}}-16x$
Now, we will take all the terms on one side of the inequality. So, we will get the inequality as,
${{x}^{2}}-3x-10-{{x}^{2}}-64+16x>0$
And now, as we know only like terms show addition and subtraction properties, so we will simplify the above inequality and get,
$13x-74>0$
This can be further written as follows,
$13x>74\Rightarrow x>\dfrac{74}{13}$
Hence, we have obtained the value of $x>\dfrac{74}{13}$ and have solved the inequality given in the question.

Note: The most important thing that we need to remember while solving this question is that in the very first step we will square both the sides of the inequality, because if we square both sides of the equation by taking (8 - x) on the left side, then the inequation would become complicated and lengthy with the chances of calculation mistakes.