Answer:
x = 1, x = -1
Step-by-step explanation:
Given [tex]4(2^{2x}+2^{-2x})-4(2^x+2^{-x})-7=0[/tex]
Use substitution
[tex]t=2^x+2^{-x}[/tex]
Square previous equality:
[tex]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[/tex]
Substitute it into the equation:
[tex]4(t^2-2)-4t-7=0\\ \\4t^2-8-4t-7=0\\ \\4t^2-4t-15=0[/tex]
Solve this quadratic equation:
[tex]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}[/tex]
Note that
[tex]2^x>0\\ \\2^{-x}>0,[/tex]
then
[tex]t=2^x+2^{-x}>0,[/tex]
so
[tex]2^x+2^{-x}=\dfrac{5}{2}[/tex]
Solve this equation using substitution
[tex]2^x=u\\ \\2^{-x}=\dfrac{1}{2^x}=\dfrac{1}{u}[/tex]
Thus,
[tex]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}[/tex]
Now,
[tex]u=2\Rightarrow 2^x=2,\ x=1\\ \\2^x=\dfrac{1}{2}\Rightarrow 2^x=\dfrac{1}{2}=2^{-1},\ x=-1[/tex]