 # 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

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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! 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! What is maths for video games course about?

Hi Kaushal,

I’ve already replied in your other thread. 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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