Math - Less than Two Roots - Challenge

In this lecture we looked at quadratic equations that have less than two roots and found a quick way of checking this using the method we learnt in the previous lecture.

For your challenge, take the following 3 equations and work out how many roots they have;

  1. y = x^2 - 5x + 10
  2. y = x^2 - 4x + 4
  3. y = x^2 - 6x + 8

Post your answer below and remember to use spoiler tags.

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  1. y = x² - 5x + 10
    (2.5 + u)(2.5 - u) = 10
    6.25 - u² = 10
    u² = 6.25 - 10
    u = √-3.75
    This is an imaginary number, so the curve of the equation has no root.

  2. y = x² - 4x + 4
    (2 + u)(2 - u) = 4
    4 - u² = 4
    u² = 4 - 4
    u = 0
    That means that the curve of this equation has 1 root, of (2,0).

  3. y = x² - 6x + 8
    (3 + u)(3 - u) = 8
    9 - u² = 8
    u² = 9 - 8
    u = 1
    That means that the curve of this equation has 2 roots of (2,0) and (4,0).

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  1. u^2 = 25/4 - 40/4 It’s going to be negative so no roots
  2. u^2 = 0 so one root at (2,0)
  3. y = (x-2)(x-4), roots at (2,0) and (4,0)
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  1. I got to u = the square root of -15/4 = Zero roots
  2. u = 0 = One root
  3. u = 1 = 2 roots
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More and more challenging! :slight_smile:

  1. y = x^2 - 5x + 10
    (-5/2 - u)(-5/2 + u) = 10
    25/4 - u^2 = 10
    u^2 = 25/4 - 10 = (25 - 40)/4 = - 15/4
    u = sqrt(- 15/4)
    => Zero roots

  2. y = x^2 - 4x + 4
    (-2 - u)(-2 + u) = 4
    4 - u^2 = 4
    u^2 = 0
    u = 0
    => One root (x = 0)

  3. y = x^2 - 6x + 8
    (-3 - u)(-3 + u) = 8
    9 - u^2 = 8
    u^2 = 9 - 8 = 1
    u = sqrt(1) = 1
    => Two roots (x = -4, x = -2)

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@Carl_Gustafsson, you’re close but you’ve made a similar mistake each time which has lead to the wrong locations for your roots.

When you half b you also want to flip the signs.
So for the first one it’s (5/2 + u)(5/2 - u) = 10, the second is (2 + u)(2 - u) = 4 and lastly (3 + u)(3 - u) = 8.

So once you’ve got an answer for u you can plug that back in and get the correct results of:

  1. No roots
  2. One root at (2,0)
  3. Two roots at (2,0) and (4,0)
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Oh. Thanks!
How silly of me - I missed that I should have changed the sign. If I had been really thorough I would of course have converted back and seen whether it returned the same equation as the initial one or not.

Don’t worry, it happens to the best of us! :smiley:

What is maths for video games course about?

Hi Kaushal,

I’ve already replied in your other thread. :slight_smile:

1). y=x^2−5x+10
((5/2)+u)((5/2)−u)=10
(25/4)−u^2=10
−u^2=(40/4)−(25/4)
u^2=−(15/4)
No root

2). y=x^2−4x+4
(2+u)(2−u)=4
4−u^2=4
u^2=0
(2+0)(2-0)=4
y=(x−2)^2
1 intercept @ (2, 0)

3.) y=x^2−6x+8
(3+u)(3−u)=8
9−u^2=8
u^2=1
u=1
(3+1)(3−1)=8
(4)∗(2)=8
y=(x−2)(x−4)
2 intercepts @ (2, 0), (4, 0)

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Equation 1: no root
Equation 2: 1 root
Equation 3: 2 roots[

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