Question

(x-3) ² + 4=|x-3||(x-10|-|x-6|)

Answer 1

Ответ:

Пошаговое объяснение:

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Answer 2

To solve the equation $(x - 3)^2 + 4 = |x - 3| \cdot (|x - 10| - |x - 6|)$, you can follow these steps:

1. Distribute the absolute value signs:

$(x - 3)^2 + 4 = |x - 3| \cdot |x - 10| - |x - 3| \cdot |x - 6|$

2. Consider cases for the absolute value terms:

a. When $x < 3$: In this case, both $x - 3$ and $x - 10$ are negative, and $x - 6$ is also negative. Therefore, the absolute values become negations.

$(x - 3)^2 + 4 = -(x - 3) \cdot -(x - 10) - -(x - 3) \cdot -(x - 6)$

3. Simplify the equation for the case when $x < 3$:

$(x - 3)^2 + 4 = (x - 3)(x - 10) - (x - 3)(x - 6)$

4. Expand and simplify further:

$(x - 3)^2 + 4 = x^2 - 13x + 30 - (x^2 - 9x - 18)$

5. Combine like terms:

$(x - 3)^2 + 4 = -13x + 30 - x^2 + 9x + 18$

6. Simplify:

$(x - 3)^2 + 4 = -4x + 48$

7. Now, square the left side to eliminate the square root:

$(x - 3)^2 + 4 = 16x^2 - 384x + 2304$

8. Subtract $16x^2 - 384x + 2304$ from both sides:

$(x - 3)^2 + 4 - 16x^2 + 384x - 2304 = 0$

9. Combine like terms:

$(x - 3)^2 - 16x^2 + 384x - 2300 = 0$

10. Use the quadratic formula to solve for $x$:

The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$

In this case:

• $a = 1$
• $b = 384$
• $c = (x - 3)^2 - 16x^2 - 2300$

So, the equation becomes:

$x = \frac{-384 \pm \sqrt{384^2 - 4 \cdot 1 \cdot [(x - 3)^2 - 16x^2 - 2300]}}{2 \cdot 1}$

Simplify and solve for $x$ using the quadratic formula, keeping in mind that you'll have two solutions for $x$, one for each ± sign.