Question

2^(10x^2-8x-23)+2^(5x^2-4x-12)-3=0

Answer 1

2^(10x^2-8x-23)+2^(5x^2-4x-12)-3=0

2^(10x^2-8x-24+1)+2^(5x^2-4x-12)-3=0

2*2^(10x^2-8x-24)+2^(5x^2-4x-12)-3=0

2^(5x^2-4x-12) = a > 0

2a^2 + a - 3 = 0

D = 1 - 4*(-3)*2 = 25

a12 = (-1 +- 5)/4 = 1     - 3/2

a1 = -3/2 < 0 нет

a2 = 1

2^(5x^2-4x-12) = 1

2^(5x^2-4x-12)=2^0

5x^2-4x-12 = 0

D = 16 + 4*5*12 = 256 = 16^2

x12 = (4 +- 16)/10 = 2    -6/5

ответ х = {-6/5, 2}

Answer 2

To solve the equation $2^{10x^2-8x-23} + 2^{5x^2-4x-12} - 3 = 0$, you can use a substitution method. Let's define $y = 2^{5x^2-4x-12}$. Then the equation becomes:

$y^2 + y - 3 = 0.$

Now solve this quadratic equation for $y$ using the quadratic formula:

$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$

where $a = 1$, $b = 1$, and $c = -3$.

$y = \frac{-(1) \pm \sqrt{(1)^2 - 4(1)(-3)}}{2(1)}$

$y = \frac{-1 \pm \sqrt{1 + 12}}{2}$

$y = \frac{-1 \pm \sqrt{13}}{2}.$

Now, since $y = 2^{5x^2-4x-12}$, you can solve for $x$ using the equation $2^{5x^2-4x-12} = \frac{-1 \pm \sqrt{13}}{2}$ by taking the logarithm of both sides.

$5x^2-4x-12 = \log_2 \left(\frac{-1 \pm \sqrt{13}}{2}\right).$

Solve for $x$ by using the quadratic formula on this equation.