Inequality : $\sqrt {x} - 6 - \sqrt{10} -x \geqslant1$

Question

I have solved it by squaring both sides and got inequality $x \geqslant 17/2$ but after that, the solution part have concluded on the equation

$4x^2 + 289 - 68x \geqslant4(10 - x)$

How this equation is formed.

Answer 1

You should have done (true on the interval $[8,10]$, where the left hand side of the inequality is positive at least) : $$ \sqrt{x-6} - \sqrt{10-x} \geq 1 \iff x - 6 + (10 - x) - 2 \sqrt{(x-6)(10-x)} \geq 1 \\ \iff \frac 32 \geq \sqrt{(x-6)(10-x)} \iff 9 \geq 4(x-6)(10-x) \\ \iff 9 \geq 4(-x^2 - 60 + 16 x) \iff 4x^2 - 64x + 249 \geq 0 \\ \iff x^2 - 16x + 62.25 \geq 0 \iff (x-8)^2 \geq 1.75 \iff x \notin (8 - \sqrt{1.75},8 + \sqrt{1.75}) $$

So, the answer is $8 + \sqrt{1.75} \leq x \leq 10$. Therefore, your answer is wrong, and you should use my working to verify where it is wrong. The other answer seems to point out your mistake.

Suppose we don't square immediately, but transpose , say $\sqrt{10-x}$ to the other side before squaring, then we get: $$ \sqrt{x-6} \geq 1 + \sqrt{10-x} \iff x-6 \geq 1 + (10-x) + 2 \sqrt{10-x} \\ \iff x - 8.5 \geq \sqrt{10-x} \iff x^2 - 17x + 72.25 \geq 10 - x \\ \iff 4x^2 - 68x + 289 \geq 4(10 - x) $$

which is the equation given in the solution, and which will also lead to the right answer. The problem with squaring both sides immediately is that it is not true that $a \geq b \iff a^2 \geq b^2$, but it is true if $a,b$ are both positive.

Answer 2

You've fallen into the old trap: we don't have $(a+b)^2 = a^2 + b^2$!! As my old math teacher would say, every time you make this mistake, a kitten dies.

To clarify, you got your answer by squaring both sides and erroneously ended up with: $$ (x-6)-(10-x)\geq 1 $$ After rearranging, you get $x\geq 17/2$. But this is incorrect; squaring both sides will in fact give you: $$ (x-6)+(10-x)-2\sqrt{x-6}\sqrt{10-x} \geq 1 $$ You will need to then rearrange the equation and square again to get your solution.