Question

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

Answer 1

x = 1, x = -1

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\\ \\√(D)=16\\ \\t_(1,2)=(-(-4)\pm 16)/(2\cdot 4)=(4\pm 16)/(8)=(5)/(2),\ -(3)/(2)`

Note that

`2^x>0\\ \\2^(-x)>0,`

then

`t=2^x+2^(-x)>0,`

so

`2^x+2^(-x)=(5)/(2)`

Solve this equation using substitution

`2^x=u\\ \\2^(-x)=(1)/(2^x)=(1)/(u)`

Thus,

`u+(1)/(u)=(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\\ \\√(D)=3\\ \\u_(1,2)=(-(-5)\pm 3)/(2\cdot 2)=(5\pm 3)/(4)=2,\ (1)/(2)`

Now,

`u=2\Rightarrow 2^x=2,\ x=1\\ \\2^x=(1)/(2)\Rightarrow 2^x=(1)/(2)=2^(-1),\ x=-1`