1. ## Polynomial questions

I tried doing these, but I always got the wrong answer or just got stuck on one part.

1.Resolve algebraically
$x^3+4x^2-7x-10<0$
I have absolutely no clue on how to start with this one.
2. $4+squareroot(x-2)=x$
I squared both sides, but it kept getting more and more complicated and really confused me.

3.Determine the coordinates of the intersection points between the line $y=x+7$ and the circle $(x+2)^2+(y-3)^2=52$

I tried substituting the line into the circle equation, but than I didn't know where to go from there.

Thanks guys.

2. Originally Posted by Methodd
I tried doing these, but I always got the wrong answer or just got stuck on one part.

1.Resolve algebraically
$x^3+4x^2-7x-10<0$
I have absolutely no clue on how to start with this one.
Hint: factor $x^3+4x^2-7x-10 = (x+1)(...)$

2. $4+squareroot(x-2)=x$
I squared both sides, but it kept getting more and more complicated and really confused me.
Hint: rewrite $4+\sqrt{x-2}=x$ as $\sqrt{x-2}=x - 4$
Now square both sides.

3.Determine the coordinates of the intersection points between the line $y=x+7$ and the circle $(x+2)^2+(y-3)^2=52$

I tried substituting the line into the circle equation, but than I didn't know where to go from there.
Yes, your approach is correct. What did you get after substituting? Simplify, and you should have a quadratic equation.

3. Yes, your approach is correct. What did you get after substituting? Simplify, and you should have a quadratic equation.
I got $x=-13+-\sqrt-900/20$

4. Careful! I think you've made the mistake of square rooting both sides. You can't do that here because $\sqrt{x^2+a^2} \neq x + a$
Let $x = 3$ and $a = 4$ to prove this quite simply yourself.

Anyway, going back to your question:
$y = x + 7$
Substituting that into $(x+2)^2+(y-3)^2=52$ gives:

$(x+2)^2+(x+7-3)^2=52$
$(x+2)^2+(x+4)^2=52$

Do not square root here.
You have to expand the brackets.
$x^2 + 4 x + 4 + x^2 + 8x + 16 = 52$
Tidy from here and factorize - you get a nice answer.

5. Thanks guys, that really helped!