Parabolas and lines...

Question

Sooo... I have just received this question. 'Draw the graph of $y = x^2 + 3x - 2$'. Now, I can do this just fine. Then it says 'draw a line on the graph to solve the following equations. $x^2 + 3x + 5 = 0$.' I have drawn the line $y = -9$, through assumptions of $x$-intercept and turning point shift. Then it says '$x^2 + 4x - 4 = 0$'. I have no idea how to do this- is the line streight or curved; should I draw a line where $x = 4.8$ and $0.9$? There is a final part: '$x^2 - x - 1 = 0$' Please reply quickly- the homework is for tomorrow! Thanks in advance.

Answer 1

You were able to solve the first question by figuring that

$$x^2 + 3x + 5 = (x^2 + 3x - 2) + 7$$

And so the equation is equivalent to

$$x^2 + 3x + 5 = -7$$

so that the solutions are the points where your parabola cuts the line $y=-7$ (and not $y=-9$, but I assume that was just a typo or an arithmetical error).

Now observe that

$$x^2+4x-4=(x^2+3x+5)+x-9$$

so that the equation is equivalent to

$$x^2+3x+5=-x+9$$

and use similar logic as in the first question.