# Math Help - Polynomials help please!

Hi guys, first post here. I'm from the United Kingdom, and am currently studying the IB (doing HL Mathematics). I'm not sure how the American system of mathematics works (so I won't abbreviate the word mathematics haha) but I'm pretty sure these are Algebra questions.

If anyone could give me pointers on how to start these, I'd really appreciate it

2. ## Re: Polynomials help please!

12) $x^2-kx+(k-1)$ factors easily and you can read the roots right off. Use the quadratic formula if you can't figure out the factors.

b) is trivial once you figure out (a)

13) from the quadratic formula what must be true to ensure real roots? What does that translate to here? Is it always true?

14) just solve the inequality for k. Remember the problem says "for all x"

3. ## Re: Polynomials help please!

Originally Posted by Bude8
Hi guys, first post here. I'm from the United Kingdom, and am currently studying the IB (doing HL Mathematics). I'm not sure how the American system of mathematics works (so I won't abbreviate the word mathematics haha) but I'm pretty sure these are Algebra questions.

If anyone could give me pointers on how to start these, I'd really appreciate it
Yes, this is algebra. And in the US we say math instead of maths.

All three questions are essentially asking you to solve a quadratic equation with real coefficients. Such an equation has two distinct real solutions, one real solution, or no real solution. There are three methods to find such solutions analytically, but the most common are through factoring and through the quadratic formula.

Factoring method

If it is "obvious" how to factor a quadratic, this method is fastest:

$ax^2 + bx + c = a(x - u)(x - v) \implies au^2 + bu + c = 0 = av^2 + bv + c.$

Personally if I do not see a factoring almost immediately, I proceed to the

Formula Method:

$ax^2 + bx + c = 0 \implies x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$

I like to use this formula this way.

I first calculate $d = b^2 - 4ac.$

$d < 0 \implies no\ real\ solution.$ I can stop looking for a real solution if if d < 0.

$d = 0 \implies solution = \dfrac{-b}{2a}.$ One real solution.

$d > 0 \implies solutions = \dfrac{-b \pm \sqrt{d}}{2a}.$ Two distinct real solutions.

Personally I'd attack all three of your problems using the quadratic formula.

4. ## Re: Polynomials help please!

Originally Posted by JeffM
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Originally Posted by romsek
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Thanks for both of your help

@JeffM, I can do the maths 90% of the time, it's just that with the IB questions, sometimes I don't even know where to begin, but after a few hints or clues (which is all I need the majority of the time), I'm able to do them myself