Solve the equation $x^2 + 4(\sqrt{1 - x} + \sqrt{1 + x}) - 8 = 0$

Question

Solve the equation $x^2 + 4(\sqrt{1 + x} + \sqrt{1 - x}) - 8 = 0$.

Let $\sqrt{1 + x} = a$, $\sqrt{1 - x} = b$.

I tried doing this.

"$1 - x^2 = [\sqrt{(1 - x)(1 + x)}]^2 = (ab)^2$.

The original equation becomes $a + b - (ab)^2 - 7 = 0$."

But it did no good for me.

So I tried again.

"$4x^2 = [(\sqrt{1 + x})^2 - (\sqrt{1 - x})^2]^2 = (a^2 - b^2)^2 = (a - b)^2(a + b)^2$

The equation becomes $(a - b)^2(a + b)^2 + 16(a + b) - 32 = 0$

We can solve this as a quadratic equation.

$\Delta = 16^2 - 4(-32)(a - b)^2 = 128[(a - b)^2 + 2]$"

And I am done. I cannot figure this out.

So if anyone can solve this equation, I will be grateful.

Answer 1

Since our equation is not changed after replacing $x$ on $-x$, we can assume that $x\geq0$.

We'll prove that $$x^2+4\left(\sqrt{1+x}+\sqrt{1-x}\right)\leq8$$ for all $0\leq x\leq1.$

Indeed, we need to prove that $$1-\sqrt{1-x}+1-\sqrt{1+x}\geq\frac{x^2}{4}$$ or $$\frac{1}{1+\sqrt{1-x}}-\frac{1}{1+\sqrt{1+x}}\geq\frac{x}{4}$$ or $$\frac{\sqrt{1+x}-\sqrt{1-x}}{(1+\sqrt{1-x})(1+\sqrt{1+x})}\geq\frac{x}{4}$$ $$\frac{1}{(1+\sqrt{1-x})(1+\sqrt{1+x})(\sqrt{1+x}+\sqrt{1-x})}\geq\frac{1}{8}$$ or $$(1+\sqrt{1-x})(1+\sqrt{1+x})(\sqrt{1+x}+\sqrt{1-x})\leq8,$$ which is true by AM-GM and C-S: $$(1+\sqrt{1-x})(1+\sqrt{1+x})(\sqrt{1+x}+\sqrt{1-x})\leq$$ $$\leq\left(\frac{1+\sqrt{1-x}+1+\sqrt{1+x}+\sqrt{1+x}+\sqrt{1-x}}{3}\right)^3=$$ $$=\left(\frac{2+2(\sqrt{1-x}+\sqrt{1+x})}{3}\right)^3\leq\left(\frac{2+2\sqrt{2(1-x+1+x)}}{3}\right)^3=8.$$ The equality occurs for $x=0$ only, which says that $0$ is an unique root.

Answer 2

$$x^2 +4\left(\sqrt{1-x} + \sqrt{1+x}\right) - 8 =0 $$ $$x^2 - 8 = -4\left(\sqrt{1-x} + \sqrt{1+x}\right) $$

We'll raise the equation power by 2

$$x^4 -16x^2 +64 = 16( 1- x + 2\sqrt{1 - x^2} + 1-x) \Rightarrow$$

$$x^4 -16x^2 + 32 = 32\sqrt{1 - x^2} $$

again We'll raise the power:

$$x^8 -32x^6 +320x^4 -1024x^2 +1024 = 1024(1-x^2)$$

therefore

$$x^8 -32x^6 +320x^4 = 0$$

$x =0$ is a solution, and it's easy to check that it's the only solution.

Answer 3

Interestingly, we have the Taylor development

$$4(\sqrt{1-x}+\sqrt{1+x})=\color{green}{8-x^2}-\frac{5x^4}{16}+\cdots$$ and $x=0$ is a solution.

It turns out that for $x>0$

$$4(\sqrt{1-x}+\sqrt{1+x})<{8-x^2}$$

(green vs blue curves) because we have equality at $x=0$ and by differentiation

$$2\left(-\frac1{\sqrt{1-x}}+\frac1{\sqrt{1+x}}\right)<-2x$$

(black vs magenta curves) because we have equality at $x=0$ and by differentiation

$$-3\left((1-x)^{-3/2}+(1-x)^{-3/2}\right)<-2$$

(pink curve vs turquoise line).

Answer 4

Let $2y=\sqrt{1+x}+\sqrt{1-x}$

Using $2(a^2+b^2)=(a+b)^2+(a-b)^2\ge(a+b)^2\implies(a+b)^2\le2(a^2+b^2),$

$4y^2\le2\cdot2=4\implies y\le1$

$4y^2=2+2\sqrt{1-x^2}\implies1-x^2=(2y^2-1)^2\iff x^2=4y^2-4y^4$

$\implies8y=8-(4y^2-4y^4)$

$\iff0=y^4-y^2-2y+2=y^2(y^2-1)-2(y-1)=(y-1)\{y^2(y+1)-2\}$

If $y=1, x^2=4y^2(1-y^2)=?$

Else $y<1\implies y^2<1, y^2(y+1)<2$

So, the only solution is $y=1$

Answer 5

As $-1\le x\le1,$

let $\sqrt{1\pm x}=\cos2y\pm\sin2y\ge0\implies x=\sin4y$

$$0=x^2+4(\sqrt{1+x}+\sqrt{1-x})-8=\sin^24y+8\cos2y-8=0$$

$$\iff0=8(2\sin^2y)-(4\sin y\cos y\cos2y)^2=16\sin^2y(1-\cos^2y\cos^22y)$$

If $\sin y=0,\sin2y=0,\cos2y=1\implies\sqrt{1\pm x}=1\implies x=?$

Else $\sin y\ne0,\cos y\ne\pm1$ we need $\cos y\cos2y=\pm1$

If $\cos y\cos2y=-1,\cos y=-\cos2y=\pm1$ but $\cos y\ne 1$

If $\cos y\cos2y=+1,\cos y=\cos2y=\pm1$ but $\cos y\ne 1$

Answer 6

The domain $-1\le x\le 1$ suggests to use the substitution: $x=\sin \alpha$. Then: $$x^2 + 4(\sqrt{1 + x} + \sqrt{1 - x}) - 8 = 0 \Rightarrow \\ 4(\sqrt{1+\sin \alpha}+\sqrt{1-\sin \alpha})=8-\sin^2 \alpha \Rightarrow \\ 16(2+2\cos \alpha)=49+14\cos^2 \alpha +\cos^4 \alpha \Rightarrow \\ \cos^4 \alpha +14\cos^2 \alpha -32\cos \alpha +17=0 \Rightarrow \\ \cos^4 \alpha -\cos^2 \alpha +15\cos^2 \alpha - 15\cos \alpha +17\cos \alpha -17=0 \Rightarrow \\ (\cos \alpha-1)(\cos^2 \alpha (\cos \alpha +1)+15\cos \alpha +17)=0 \Rightarrow \\ 1) \cos \alpha -1=0 \Rightarrow \alpha =0 \Rightarrow \color{red}{x = 0};\\ 2) \cos^3 \alpha +\cos ^2 \alpha +15\cos \alpha +17=0 \Rightarrow \emptyset, \ \text{because} \\ \cos^3 \alpha +\cos ^2 \alpha +15\cos \alpha \ge -15.$$