Question

(x+4)^2+3(x-3)(x+4)+(x-3)^2=0 Срочнооооо помогите решшиииитььь!!))))

Answer 1

Ответ в файле

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Answer 2

To solve the equation (x+4)^2 + 3(x-3)(x+4) + (x-3)^2 = 0, we can start by expanding and simplifying the equation. Let's break it down step by step:

Step 1: Expand the equation

Expand each term in the equation using the distributive property:

(x+4)^2 + 3(x-3)(x+4) + (x-3)^2 = 0

Expanding the first term: x^2 + 8x + 16

Expanding the second term: 3(x^2 + x - 12)

Expanding the third term: x^2 - 6x + 9

Step 2: Simplify the equation

Combine like terms:

x^2 + 8x + 16 + 3x^2 + 3x - 36 + x^2 - 6x + 9 = 0

Simplifying the equation: 5x^2 + 5x - 11 = 0

Now we have a quadratic equation in the form ax^2 + bx + c = 0.

Step 3: Solve the quadratic equation

To solve the quadratic equation, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In our case, a = 5, b = 5, and c = -11.

Substituting the values into the quadratic formula:

x = (-5 ± √(5^2 - 4 * 5 * -11)) / (2 * 5)

Simplifying further:

x = (-5 ± √(25 + 220)) / 10

x = (-5 ± √245) / 10

The solutions to the equation are:

x = (-5 + √245) / 10 and x = (-5 - √245) / 10

Please note that these are the exact solutions to the equation. If you need decimal approximations, you can calculate them using a calculator.

I hope this helps! Let me know if you have any further questions.