Question

I used Cauchy-Schwarz inequality but it took too much time to finish. Is there any way faster?​

Answer 1

Answer:

x = 1, x = -1

Step-by-step explanation:

Given $4(2^{2x}+2^{-2x})-4(2^x+2^{-x})-7=0$

Use substitution

$t=2^x+2^{-x}$

Square previous equality:

$t^2=(2^x+2^{-x})^2\\ \\t^2=(2^{x})^2+2\cdot 2^x\cdot 2^{-x}+(2^{-x})^2\\ \\t^2=2^{2x}+2\cdot 2^{x-x}+2^{-2x}\\ \\t^2=2^{2x}+2\cdot 2^0+2^{-2x}\\ \\t^2=2^{2x}+2\cdot 1+2^{-2x}\\ \\t^2=2^{2x}+2+2^{-2x}\\ \\2^{2x}+2^{-2x}=t^2-2$

Substitute it into the equation:

$4(t^2-2)-4t-7=0\\ \\4t^2-8-4t-7=0\\ \\4t^2-4t-15=0$

Solve this quadratic equation:

$D=(-4)^2-4\cdot 4\cdot (-15)=16+16\cdot 15=16(1+15)=16\cdot 16\\ \\\sqrt{D}=16\\ \\t_{1,2}=\dfrac{-(-4)\pm 16}{2\cdot 4}=\dfrac{4\pm 16}{8}=\dfrac{5}{2},\ -\dfrac{3}{2}$

Note that

$2^x>0\\ \\2^{-x}>0,$

then

$t=2^x+2^{-x}>0,$

so

$2^x+2^{-x}=\dfrac{5}{2}$

Solve this equation using substitution

$2^x=u\\ \\2^{-x}=\dfrac{1}{2^x}=\dfrac{1}{u}$

Thus,

$u+\dfrac{1}{u}=\dfrac{5}{2}\\ \\2u^2+2=5u\ \ [\text{Multiplied by 2u}]\\ \\2u^2-5u+2=0\\ \\D=(-5)^2-4\cdot 2\cdot 2=25-16=9\\ \\\sqrt{D}=3\\ \\u_{1,2}=\dfrac{-(-5)\pm 3}{2\cdot 2}=\dfrac{5\pm 3}{4}=2,\ \dfrac{1}{2}$

Now,

$u=2\Rightarrow 2^x=2,\ x=1\\ \\2^x=\dfrac{1}{2}\Rightarrow 2^x=\dfrac{1}{2}=2^{-1},\ x=-1$