Question

X^2-4|x+1|+5x+4=0 как это решать?!

Answer 1

Приблизь и посмотри.Думаю поймешь.

Answer 2

To solve the equation $X^2-4|x+1|+5x+4=0$, we need to consider two cases: one when $x+1$ is non-negative ($x+1\geq0$) and another when $x+1$ is negative ($x+1<0$).

Let's proceed step by step:

Case 1: $x+1\geq0$ ($x\geq-1$): In this case, $|x+1|=x+1$, so the equation becomes: $X^2-4(x+1)+5x+4=0$

Now, simplify the equation: $X^2-4x-4+5x+4=0$ $X^2+x=0$

Now, set the equation equal to zero and solve for $x$: $X^2+x=0$ $x^2+x=0$ $x(x+1)=0$

So, the solutions for $x$ in this case are $x=0$ and $x=-1$.

Case 2: $x+1<0$ ($x<-1$): In this case, $|x+1|=-(x+1)$, so the equation becomes: $X^2-4(-(x+1))+5x+4=0$

Now, simplify the equation: $X^2+4x+4+5x+4=0$ $X^2+9x+8=0$

Now, set the equation equal to zero and solve for $x$: $X^2+9x+8=0$

This is a quadratic equation. You can use the quadratic formula to find the solutions for $x$: $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

For this equation, $a=1$, $b=9$, and $c=8$.

$x = \frac{-9 \pm \sqrt{9^2-4(1)(8)}}{2(1)}$ $x = \frac{-9 \pm \sqrt{81-32}}{2}$ $x = \frac{-9 \pm \sqrt{49}}{2}$ $x = \frac{-9 \pm 7}{2}$

So, the solutions for $x$ in this case are $x = -8$ and $x = -1.5$.

Overall, the solutions for the equation $X^2-4|x+1|+5x+4=0$ are $x=0$, $x=-1$, $x=-8$, and $x=-1.5$. Make sure to check these solutions in the original equation to ensure they are valid.