Question

(8x²-3x+1)²-18(8x²-3x+1)+72=0​

Answer 1

Ответ:

(8x²-3x+1)² =t²

(8x²-3x+1)=t

маємо рівняння

t²-18t+72=0

за теоремою вієта

t1+t2=18

t1•t2=72

t1=12 t2=6

t²=12. t²=6

t=+-√12 t=+-✓6

Answer 2

To solve the given quadratic equation $(8x^2 - 3x + 1)^2 - 18(8x^2 - 3x + 1) + 72 = 0$, let's substitute $y = 8x^2 - 3x + 1$. This allows us to rewrite the equation in terms of $y$:

$$y^2 - 18y + 72 = 0$$

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

The quadratic formula is $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = -18$, and $c = 72$.

$$y = \frac{18 \pm \sqrt{(-18)^2 - 4(1)(72)}}{2 \times 1}$$

$$y = \frac{18 \pm \sqrt{324 - 288}}{2}$$

$$y = \frac{18 \pm \sqrt{36}}{2}$$

Now, we have two possible values for $y$:

1. $y_1 = \frac{18 + 6}{2} = 12$
2. $y_2 = \frac{18 - 6}{2} = 6$

Now, we'll substitute these values back to solve for $x$:

For $y_1 = 12$: $8x^2 - 3x + 1 = 12$ $8x^2 - 3x - 11 = 0$

Now, solve this quadratic equation for $x$ using the quadratic formula:

$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(8)(-11)}}{2(8)}$ $x = \frac{3 \pm \sqrt{369}}{16}$

For $y_2 = 6$: $8x^2 - 3x + 1 = 6$ $8x^2 - 3x - 5 = 0$

Now, solve this quadratic equation for $x$ using the quadratic formula:

$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(8)(-5)}}{2(8)}$ $x = \frac{3 \pm \sqrt{199}}{16}$