Question

Xв квадрате-3x-4деленное на x+1=0

Answer 1

$\frac{x^{2}-3x-4 }{x+1}=0\\\\\left \{ {{x^{2}-3x-4=0 } \atop {x+1\neq 0 }} \right.\\\\\left \{ {{x^{2}-3x-4=0 } \atop {x\neq -1 }} \right.\\\\x^{2} -3x-4=0\\\\D=(-3)^{2}-4*(-4)=9+16=25=5^{2}\\\\x_{1}=\frac{3+5}{2}=4\\\\x_{2}=\frac{3-5}{2}=-1$

Ответ : 4

Answer 2

To solve the equation $\frac{x^2 - 3x - 4}{x + 1} = 0$, we first need to find the values of $x$ that make the expression equal to zero.

To do that, we'll start by factoring the numerator:

$x^2 - 3x - 4 = (x - 4)(x + 1)$

Now, the equation becomes:

$\frac{(x - 4)(x + 1)}{x + 1} = 0$

Now, notice that we have a common factor of $x + 1$ in both the numerator and the denominator. We can cancel out this common factor:

$x - 4 = 0$

Now, isolate $x$:

$x = 4$

So, the solution to the equation is $x = 4$. However, there's one thing we need to be cautious about. The original equation had a denominator, and we should check if $x = -1$ makes the denominator zero. If it does, then $x = -1$ is not a valid solution.

Let's check:

$x + 1 = -1 + 1 = 0$

Since $x + 1$ is equal to zero when $x = -1$, we can't use $x = -1$ as it would make the original expression undefined (division by zero is not allowed).

Therefore, the only valid solution to the equation is $x = 4$.