Question

X²-9x-10=10 x²+3x-10 = 0

Answer 1

$1)\; \; x^2-9x-10=10\\\\x^2-9x-20=0\; \; ,\; \; D=161\\\\x_1=\frac{9-\sqrt{161}}{2}\; ,\; \; x_2=\frac{9+\sqrt{161}}{2}\\\\2)\; \; x^2+3x-10=0\; \; ,\; \; x_1=-5\; ,\; \; x_2=2\; \; (teorema\; Vieta,\; tak\; kak\\\\-5+2=-3\; ,\; \; -5\cdot 2=-10\; )$

Answer 2

To solve the quadratic equations x² - 9x - 10 = 10 and x² + 3x - 10 = 0, we need to find the values of x that satisfy each equation. Let's solve them step-by-step:

1. x² - 9x - 10 = 10

To solve this equation, we first need to rewrite it in the standard quadratic form (ax² + bx + c = 0):

x² - 9x - 20 = 0

Now, we can factorize the quadratic expression:

(x - 5)(x + 4) = 0

Now, we can apply the zero product property, which states that if the product of two factors is zero, then at least one of the factors must be zero:

Setting each factor to zero and solving for x:

x - 5 = 0 ---> x = 5 x + 4 = 0 ---> x = -4

So, the solutions for the equation x² - 9x - 10 = 10 are x = 5 and x = -4.

2. x² + 3x - 10 = 0

This equation is already in standard quadratic form, so let's proceed to factorize:

(x + 5)(x - 2) = 0

Again, apply the zero product property:

x + 5 = 0 ---> x = -5 x - 2 = 0 ---> x = 2

So, the solutions for the equation x² + 3x - 10 = 0 are x = -5 and x = 2.

In summary, the solutions for the two given quadratic equations are:

1. x = 5 and x = -4 for x² - 9x - 10 = 10
2. x = -5 and x = 2 for x² + 3x - 10 = 0