Question

(2x^2-1)(2x^2+1)-12(x^2+7)=131

Answer 1

Ответ:

±3

Объяснение:

(2x^2-1)(2x^2+1)-12(x^2+7)=131

4x^4-1-12x^2-84=131

4x^4-1-12x^2-84-131=0

4x^4-12x^2-216=0

x^4-3x^2-54=0

Пусть x^2 = t

t^2-3t-54=0

По теореме Виета:

t1=9

t2= -6

Обратная замена:

x²=9

x=±3

x²=-6

x∈∅ , в действительных решениях.

Answer 2

To solve the equation, let's simplify the left side and set it equal to 131:

(2x^2 - 1)(2x^2 + 1) - 12(x^2 + 7) = 131

Expanding the products and simplifying, we have:

(4x^4 - 1) - 12x^2 - 84 = 131

Rearranging terms and combining like terms:

4x^4 - 12x^2 - 1 - 84 = 131

4x^4 - 12x^2 - 85 = 131

Next, let's move 131 to the left side:

4x^4 - 12x^2 - 131 - 85 = 0

4x^4 - 12x^2 - 216 = 0

Now, let's divide the equation by 4 to simplify further:

x^4 - 3x^2 - 54 = 0

At this point, we can factor the equation:

(x^2 - 9)(x^2 + 6) = 0

Now, setting each factor equal to zero:

x^2 - 9 = 0 or x^2 + 6 = 0

Solving each equation separately:

For x^2 - 9 = 0:

x^2 = 9

Taking the square root of both sides:

x = ±3

For x^2 + 6 = 0:

x^2 = -6

This equation has no real solutions since the square of a real number cannot be negative.

Therefore, the solution to the original equation (2x^2 - 1)(2x^2 + 1) - 12(x^2 + 7) = 131 is x = ±3.