Question

(2х 2 + 1) 2 = 14(2х 2 + 1) – 45.

Answer 1

Ответ:

(2x²+1)²=14(2x²+1)-45. (2x²+1)=t

t²-14t+45=0

t1+t2=14

t1•t2=45

t1=9

t2= 5

2x²+1=5. 2x²+1=9

2x²=-4. 2x²=8

x²=-4÷2. x²=8÷2

x²=-2. x²=4

x=2

Відповідь: -√2, √2, 2, -2.

Answer 2

Let's solve the given equation step by step.

The equation is: $(2x^2 + 1)^2 = 14(2x^2 + 1) - 45$.

First, let's expand the left side of the equation using the square of a binomial: $(2x^2 + 1)^2 = (2x^2 + 1)(2x^2 + 1)$ $= 4x^4 + 2x^2 + 2x^2 + 1$ $= 4x^4 + 4x^2 + 1$.

Now, let's substitute this into the original equation: $4x^4 + 4x^2 + 1 = 14(2x^2 + 1) - 45$.

Next, distribute 14 to $2x^2$ and 1: $4x^4 + 4x^2 + 1 = 28x^2 + 14 - 45$.

Now, let's combine like terms: $4x^4 + 4x^2 + 1 = 28x^2 - 31$.

Move all terms to one side to set the equation equal to zero: $4x^4 + 4x^2 - 28x^2 - 1 + 31 = 0$.

Combine the x^2 terms: $4x^4 - 24x^2 + 30 = 0$.

Now, let's factor out the common terms, which is 2: $2(2x^4 - 12x^2 + 15) = 0$.

Now, we can solve for x^4 - 12x^2 + 15 = 0. Let's set $u = x^2$, then we have: $2(2u^2 - 12u + 15) = 0$.

Now we have a quadratic equation: $4u^2 - 24u + 30 = 0$.

Divide the whole equation by 2 to simplify it: $2u^2 - 12u + 15 = 0$.

Now we can use the quadratic formula to solve for u: $u = \frac{-(-12) \pm \sqrt{(-12)^2 - 4 \times 2 \times 15}}{2 \times 2}$ $u = \frac{12 \pm \sqrt{144 - 120}}{4}$ $u = \frac{12 \pm \sqrt{24}}{4}$ $u = \frac{12 \pm 2\sqrt{6}}{4}$ $u = \frac{6 \pm \sqrt{6}}{2}$.

Now, remember that we set $u = x^2$, so we have: $x^2 = \frac{6 \pm \sqrt{6}}{2}$.

Taking the square root of both sides, we get two possible solutions for x: $x = \pm \sqrt{\frac{6 \pm \sqrt{6}}{2}}$.