Question

1) 4x⁴ - 5x² + 1 = 02) 9x⁴ - 9x² + 2 = 03) (x² + x - 3)² - 12 (x² + x - 3) + 27 = 0​

Answer 1

$\displaystyle\bf\\1)\\\\4x^{4} -5x^{2} +1=0\\\\x^{2} =m \ \ , \ \ m\geq 0\\\\4m^{2} -5m+1=0\\\\D=(-5)^{2} -4\cdot4\cdot 1=25-16=9=3^{2} \\\\\\m_{1} =\frac{5+3}{8} =1\\\\m_{2}=\frac{5-3}{8} =\frac{1}{4} =0,25 \\\\x^{2} =1\\\\x_{1} =-1 \ \ ; \ \ x_{2} =1\\\\\\x^{2} =0,25\\\\x_{3}=-0,5 \ \ ; \ \ x_{4} =0,5\\\\Otvet: \ -1 \ ; \ 1 \ ; \ -0,5 \ ; \ 0,5$

$\displaystyle\bf\\2)\\\\9x^{4} -9x^{2} +2=0\\\\x^{2} =m \ \ , \ \ m\geq 0\\\\9m^{2} -9m+2=0\\\\D=(-9)^{2} -4\cdot9\cdot 2=81-72=9=3^{2} \\\\\\m_{1} =\frac{9+3}{18} =\frac{2}{3} \\\\m_{2}=\frac{9-3}{18} =\frac{1}{3} \\\\\\x^{2} =\frac{2}{3} \\\\x_{1} =-\sqrt{\frac{2}{3} } =-\frac{\sqrt{6} }{3} \\\\x_{2} =\sqrt{\frac{2}{3} } =\frac{\sqrt{6} }{3} \\\\\\x^{2} =\frac{1}{3} \\\\x_{3} =-\sqrt{\frac{1}{3} } =-\frac{\sqrt{3} }{3} \\\\x_{4} =\sqrt{\frac{1}{3} } =\frac{\sqrt{3} }{3}$

$\displaystyle\bf\\Otvet: \ -\frac{\sqrt{6} }{3} \ ; \ \frac{\sqrt{6} }{3} \ ; \ -\frac{\sqrt{3} }{3} \ ; \ \frac{\sqrt{3} }{3} \\\\\\3)\\\\(x^{2} +x-3)^{2} -12(x^{2} +x-3)+27=0\\\\x^{2} +x-3=m\\\\m^{2}-12m+27=0\\\\Teorema \ Vieta:\\\\m_{1}+ m_{2} =12\\\\m_{1}\cdot m_{2} =27\\\\m_{1}=3\\\\ m_{2} =9\\\\\\x^{2} +x-3=3\\\\x^{2} +x-6=0\\\\Teorema \ Vieta:\\x_{1}+ x_{2} =-1\\\\x_{1}\cdot x_{2} =-6\\\\x_{1} =-3\\\\x_{2} =2\\\\\\x^{2} +x-3=9\\\\x^{2} +x-12=0$

$\displaystyle\bf\\Teorema \ Vieta:\\\\x_{1}+ x_{2} =-1\\\\x_{1}\cdot x_{2} =-12\\\\x_{1} =-4\\\\x_{2} =3\\\\\\Otvet: \ -3 \ ; \ 2 \ ; \ -4 \ ; \ 3$

Answer 2

Let's solve each of these equations one by one:

1. 4x⁴ - 5x² + 1 = 0

This is a quadratic equation in terms of x². You can make a substitution to simplify it:

Let y = x².

Now, the equation becomes:

4y² - 5y + 1 = 0

You can factor this equation:

(4y - 1)(y - 1) = 0

Now, solve for y:

4y - 1 = 0 => 4y = 1 => y = 1/4

or

y - 1 = 0 => y = 1

Now, substitute back for y:

For y = 1/4: x² = 1/4

Taking the square root of both sides:

x = ±√(1/4) x = ±1/2

For y = 1: x² = 1

Taking the square root of both sides:

x = ±1

So, the solutions for the equation 4x⁴ - 5x² + 1 = 0 are x = ±1/2 and x = ±1.

2. 9x⁴ - 9x² + 2 = 0

This is also a quadratic equation in terms of x². Let y = x²:

9y² - 9y + 2 = 0

You can factor this equation:

(3y - 2)(3y - 1) = 0

Now, solve for y:

3y - 2 = 0 => 3y = 2 => y = 2/3

or

3y - 1 = 0 => 3y = 1 => y = 1/3

Now, substitute back for y:

For y = 2/3: x² = 2/3

Taking the square root of both sides:

x = ±√(2/3)

For y = 1/3: x² = 1/3

Taking the square root of both sides:

x = ±√(1/3)

So, the solutions for the equation 9x⁴ - 9x² + 2 = 0 are x = ±√(2/3) and x = ±√(1/3).

3. (x² + x - 3)² - 12(x² + x - 3) + 27 = 0

Let's make a substitution to simplify this equation. Let y = x² + x - 3:

Now, the equation becomes:

y² - 12y + 27 = 0

You can factor this equation:

(y - 9)(y - 3) = 0

Now, solve for y:

y - 9 = 0 => y = 9

or

y - 3 = 0 => y = 3

Now, substitute back for y:

For y = 9: x² + x - 3 = 9

Rearrange it:

x² + x - 3 - 9 = 0

x² + x - 12 = 0

Now, factor this quadratic equation:

(x + 4)(x - 3) = 0

Solve for x:

x + 4 = 0 => x = -4

or

x - 3 = 0 => x = 3

For y = 3: x² + x - 3 = 3

Rearrange it:

x² + x - 3 - 3 = 0

x² + x - 6 = 0

Now, factor this quadratic equation:

(x + 3)(x - 2) = 0

Solve for x:

x + 3 = 0 => x = -3

or

x - 2 = 0 => x = 2

So, the solutions for the equation (x² + x - 3)² - 12(x² + x - 3) + 27 = 0 are x = -4, 3, -3, and 2.