Question

4(4 – x2) – 3x(x – 3) = 8 + 12x – 7x2

Answer 1

Ответ:

Объяснение:

Смотри

Answer 2

премер не правельно записан

Объяснение:

Answer 3

To solve the given equation, let's simplify and solve for $x$:

$4(4 - x^2) - 3x(x - 3) = 8 + 12x - 7x^2$

First, distribute the terms:

$16 - 4x^2 - 3x^2 + 9x = 8 + 12x - 7x^2$

Combine like terms on both sides:

$-7x^2 - 3x^2 + 12x + 16 - 9x = 8$

Combine the quadratic terms:

$-10x^2 + 3x + 16 = 8$

Move all terms to one side to set the quadratic equation to zero:

$-10x^2 + 3x + 16 - 8 = 0$

Combine the constants:

$-10x^2 + 3x + 8 = 0$

Now, you can either factor the quadratic or use the quadratic formula to find the solutions. In this case, it's not factorable easily, so let's use the quadratic formula:

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

For the quadratic equation $ax^2 + bx + c = 0$, in our case:

$a = -10, \quad b = 3, \quad c = 8$

Now plug these values into the quadratic formula:

$x = \frac{-3 \pm \sqrt{3^2 - 4(-10)(8)}}{2(-10)}$

Simplify the expression under the square root:

$x = \frac{-3 \pm \sqrt{9 + 320}}{-20}$

$x = \frac{-3 \pm \sqrt{329}}{-20}$

So, the solutions are:

$x = \frac{-3 + \sqrt{329}}{-20} \quad \text{and} \quad x = \frac{-3 - \sqrt{329}}{-20}$

These are the two solutions for the given quadratic equation.