Solve: $\sqrt{x + 3 + \sqrt{x + 14}} + \sqrt{x+3 - \sqrt{x + 14}} =4 $

The domain gives $x\geq-3$ and $(x+3)^2\geq x+14$.

Now, our equation it's $$\left(\sqrt{x + 3 + \sqrt{x + 14}} + \sqrt{x+3 - \sqrt{x + 14}}\right)^2 =16 $$ or $$2x+6+2\sqrt{(x+3)^2-x-14}=16$$ or $$\sqrt{x^2+5x-5}=5-x.$$ Now, we get also $5-x\geq0$, for which we obtain $$x^2+5x-5=x^2-10x+25$$ or $$x=2,$$ which is valid.


after squaring we get $$x+3+\sqrt{x+14}+x+3-\sqrt{x+14}+2\sqrt{(x+3)^2-(x+4)}=16$$ simplifying we obtain $$\sqrt{(x+3)^2-(x+14)}=5-x$$ squaring again $$(x+3)^2-(x+14)=25+x^2-10x$$ Can you finish?

Tags:

Radicals

Algebra Precalculus

Nested Radicals

Related

Show that there exist $n$ such that $r$ divides $\binom{p^n}{q^n}$ will a nonempty countable and compact subset of a metric space always contain an isolated point? (1) Sum of two factorials in two ways; (2) Value of $a^{2010}+a^{2010}+1$ given $a^4+a^3+a^2+a+1=0$. Proving that $\frac{ab}{c^3}+\frac{bc}{a^3}+\frac{ca}{b^3}> \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ Does there exist an injection $f:\mathbb{R}\to P(\mathbb{N})$ that satisfies these conditions? Cantor Set (Hausdorff) $f:\mathbb{R}\rightarrow\mathbb{R}$ is not one-to-one if $[f(x)]^2-4\cdot f(x^5)+3=0,\forall x\in\mathbb{R}$? How to use Stokes theorem in the presence of holes? Let $T$ be any subset of $\{1,2,3,...,100\}$ with $69$ elements. Prove that one can find four distinct integers such that $a+b+c=d$. Prove that $\sum_{l=0}^{n} \binom{n}{l}^2 (x+y)^{2l} (x-y)^{2(n-l)} = \sum_{l=0}^{n} \binom{2l}{l} \binom{2(n-l)}{n-l} x^{2l}y^{2(n-l)}$ Is there a *simple* example showing that uncorrelated random variables need not be independent? System with a Lyapunov function over $\mathbb{R}^n$ but not globally asymptotically stable

Recent Posts