Question

2x^4+3x^2(x-2)=2(x-2)^2 Пожалуйста!!!!!!!

Answer 1

Ответ: $-1\pm\sqrt5$

Пошаговое объяснение:

$2x^4+3x^2(x-2)=2(x-2)^2\\2x^4+3x^2(x-2)-2(x-2)^2=0\\x^2=a, \ (x-2)=b,\\2a^2+3ab-2b^2=0\\1) b=0, a=0 \ \varnothing\\2) b\ne0\\2\dfrac{a^2}{b^2}+3\dfrac{a}{b}-2=0\\D=9-4\cdot2\cdot(-2)=25=5^2\\\dfrac{a}{b}=\dfrac{-3+5}{4}=\dfrac12 \Leftrightarrow b=2a\Leftrightarrow x-2=2x^2\Leftrightarrow 2x^2-x+2=0; \ D=1-4\cdot2\cdot2$

Answer 2

Sure thing! Let's solve this equation step by step.

First, let's simplify the equation:

$2x^4 + 3x^2(x - 2) = 2(x - 2)^2$

Expand the terms on both sides:

$2x^4 + 3x^3 - 6x^2 = 2(x^2 - 4x + 4)$

Distribute the 2 on the right side:

$2x^4 + 3x^3 - 6x^2 = 2x^2 - 8x + 8$

Now, let's bring all terms to one side of the equation to set it to zero:

$2x^4 + 3x^3 - 6x^2 - 2x^2 + 8x - 8 = 0$

Combine like terms:

$2x^4 + 3x^3 - 8x^2 + 8x - 8 = 0$

Unfortunately, this equation doesn't have an easily factorizable solution. You might need to use numerical methods or other advanced techniques to find the roots. If you're using a tool or software, you can input this equation to get the numerical solutions. Let me know if you need any further assistance!