Question

(x²-7)²-4(x²-7)+4=0​

Answer 1

Ответ:

(x²-7)²-4(x²-7)-45=0

замена (х²-7)=t

t²-4t-45=0

D=16+180=196; √D=14

t1=(4+14)/2=18/2=9

t2=(4-14)/2=-10/2=-5

х²-7=9

х²=16

х=±4

и х²-7=-5

х²=2

х=±√2

Объяснение:

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Answer 2

To solve the equation $(x^2 - 7)^2 - 4(x^2 - 7) + 4 = 0$, you can use substitution to simplify it. Let's make a substitution to simplify the equation:

Let $y = x^2 - 7$. Now, we can rewrite the equation in terms of $y$:

$y^2 - 4y + 4 = 0$

This is now a quadratic equation in $y$. To solve for $y$, you can use the quadratic formula:

$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this case, $a = 1$, $b = -4$, and $c = 4$. Plugging these values into the quadratic formula:

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

Simplify further:

$y = \frac{4 \pm \sqrt{16 - 16}}{2}$

$y = \frac{4 \pm \sqrt{0}}{2}$

Since the discriminant ($b^2 - 4ac$) is zero, there is only one real root for $y$:

$y = \frac{4}{2}$

$y = 2$

Now that we have found $y$, we need to go back to the original variable $x$ and solve for $x$:

$x^2 - 7 = 2$

Add 7 to both sides:

$x^2 = 2 + 7$

$x^2 = 9$

Now, take the square root of both sides, considering both positive and negative square roots:

$x = \pm 3$

So, the solutions to the equation $(x^2 - 7)^2 - 4(x^2 - 7) + 4 = 0$ are $x = 3$ and $x = -3$.