Question

Y = x² +10 y= 3 x²+2 x²-15x

Answer 1

Ответ:

Объяснение:

1) y = x² +10

0 = x² +10

x ∉ R

2) y= 3 x²+2 x²-15x

0 = 3 x²+2 x²-15x

3 x²+2 x²-15x = 0

5 x²-15x = 0

5x × (x-3) = 0

x × (x-3) = 0

x = 0

x-3 = 0

x = 0

x = 3

x₁ = 0, x₂ = 3

Answer 2

It appears you have two equations involving "x" and "y." Let's solve for "x" and "y" separately in each equation:

1. Equation 1: $y = x^2 + 10$
2. Equation 2: $y = 3x^2 + 2x^2 - 15x$

First, let's simplify Equation 2:

Equation 2 becomes $y = 5x^2 - 15x$.

Now, we can set Equation 1 and Equation 2 equal to each other to find the values of "x" and "y" that satisfy both equations:

$x^2 + 10 = 5x^2 - 15x$

To isolate "x," we'll move all terms to one side of the equation:

$0 = 5x^2 - x^2 - 15x - 10$

Combine like terms:

$0 = 4x^2 - 15x - 10$

Now, we have a quadratic equation in terms of "x." To solve for "x," we can use the quadratic formula:

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

In this case, $a = 4$, $b = -15$, and $c = -10$. Plug these values into the formula:

$x = \frac{-(-15) \pm \sqrt{(-15)^2 - 4(4)(-10)}}{2(4)}$

Simplify further:

$x = \frac{15 \pm \sqrt{225 + 160}}{8}$

$x = \frac{15 \pm \sqrt{385}}{8}$

Now, we have two possible solutions for "x":

1. $x = \frac{15 + \sqrt{385}}{8}$
2. $x = \frac{15 - \sqrt{385}}{8}$

These are the values of "x" that satisfy both equations. To find the corresponding values of "y," plug these values of "x" into either Equation 1 or Equation 2. Let's use Equation 1:

For the first solution:

$y = \left(\frac{15 + \sqrt{385}}{8}\right)^2 + 10$

For the second solution:

$y = \left(\frac{15 - \sqrt{385}}{8}\right)^2 + 10$

Calculate these expressions to find the values of "y" corresponding to each "x" solution.