Question

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Answer 1

Q.3

Imagine a number line. Absolute value is the distance from the origin. If two numbers have the same distance, the point on a number line will meet or be in the opposite direction.

From the assumption, we have proved that $2x^2+x-1=\pm(x^2+4x+1)$

Case 1. $2x^2+x-1=x^2+4x+1$ → $x^2-3x-2=0$

Solving the equation, you get $x=\frac{3\pm\sqrt{17} }{2}$.

Case 2. $2x^2+x-1=-(x^2+4x+1)$ → $3x^2+5x=0$

Solving the equation, you get $x=0$ or $x=-\frac{5}{3}$.

Q.4

Note that x²-x-6=(x-3)(x+2). We can see x+2 common on both sides.

Note that absolute value is positive or 0. The absolute value is x+2. Therefore x+2≥0. → x≥-2

In solving such equations with absolute value, we must be extremely careful about their signs. We must separate into two possibilities, where it becomes positive, 0 or negative.

1. If x-3 is positive or 0, we will get (x-3)(x+2)≥0.

→ |x²-x-6|=x²-x-6

Now we have one simple equation $x^2-x-6=x+2$.

→ $x^2-2x-8=0$

Solving the equation gives you $x=-2$ or $x=4$.

2. If x-3 is negative, we will get (x-3)(x+2)≤0.

→  |x²-x-6|=-(x²-x-6)

We have an equation $-x^2+x+6=x+2$.

Solving this equation, we get $x^2=4$ therefore we get x=2.