Quadratic Inequalities: My answer is different from the ones provided.

Question

The problem is:

$$x^2 + 2x - 1 < 0$$

Step 1: Move the $1$ to the other side.

$$x^2 + 2x < 1$$

Step 2: Add $1$ to both sides to complete the quadratic equation.

$$x^2 + 2x + 1 < 2$$

Step 3: Factor the quadratic equation.

$$(x + 1)^2 < 2$$

Step 4:

$$(x + 1) < \pm \sqrt2$$

Step 5:

$$x_1 < 1.41 - 1 \qquad and \qquad x_2 < -1.41 - 1$$

Step 6:

$$x_1 < 0.41 \qquad and \qquad x_2 < -2.41$$

My Answer is:

$$SC: (-\infty, -2.41)$$

The possible answers are:

$$ \begin{align} &a)\quad (-\sqrt2,\ \sqrt2)\\ &b)\quad (-\sqrt2 - 1,\ -\sqrt2 + 1)\\ &c)\quad (1 - \sqrt2,\ 1 + \sqrt2)\\ &d)\quad (-\sqrt2 - 1,\ \sqrt2 - 1)\\ &e)\quad (-2 - \sqrt2,\ 3- \sqrt2)\\ \end{align} $$

Where did I go wrong, what could I do better?

Answer 1

From this step: $$(x + 1)^2 < 2$$ you get: $$|x + 1| < \sqrt{2}$$ so you get $$x + 1 < \sqrt{2}$$ $$x + 1 > -\sqrt{2}$$ which simplifies to: $$x < \sqrt{2} - 1$$ $$x > -\sqrt{2} - 1$$ so your answer is

d) $(-\sqrt2 - 1; \sqrt2 - 1)$

You can visualize the answer with this graph:

Answer 2

for $x\geq -1$ we have $$x<\sqrt{2}-1$$ and for $x<-1$ we get $$-1-\sqrt{2}<x$$

Answer 3

If the square of a number is less then $a$ then it is wrong (and meaningless) to say that that number is less that $\pm\sqrt{a}$. Instead, from $t^2<a$ it follows that $-a<t<a$. So in your case $-\sqrt2<x+1<\sqrt2$.