Question

(x-1)⁴+21(x-1)²-100=​0

Answer 1

Ответ:

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Answer 2

To solve the equation $(x-1)^4 + 21(x-1)^2 - 100 = 0$, you can use a substitution to simplify the equation. Let $y = (x - 1)^2$. Then, your equation becomes:

$y^2 + 21y - 100 = 0$

This is now a quadratic equation in $y$, and you can use the quadratic formula to solve for (y:

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

In this case, $a = 1$, $b = 21$, and $c = -100$. Plugging these values into the quadratic formula:

$y = \frac{-21 \pm \sqrt{21^2 - 4(1)(-100)}}{2(1)}$

Now, calculate the discriminant:

$D = 21^2 - 4(1)(-100) = 441 + 400 = 841$

So:

$y = \frac{-21 \pm \sqrt{841}}{2}$

Now, you can take the square root of 841:

$y = \frac{-21 \pm 29}{2}$

Now, calculate the two possible values of $y$:

1. $y_1 = \frac{-21 + 29}{2} = 4$
2. $y_2 = \frac{-21 - 29}{2} = -25$

Now, recall that $y = (x - 1)^2$. So, we have:

1. $4 = (x - 1)^2$
2. $-25 = (x - 1)^2$

For the first equation, take the square root of both sides:

$x - 1 = \pm 2$

This gives you two possible values for $x$:

1. $x_1 = 1 + 2 = 3$
2. $x_2 = 1 - 2 = -1$

For the second equation, you have:

$(x - 1)^2 = -25$

Since the square of a real number cannot be negative, there are no real solutions for this equation.

So, the real solutions to the original equation $(x-1)^4 + 21(x-1)^2 - 100 = 0$ are $x = 3$ and $x = -1$.