Not sure about this one.
But here’s what I got:
- has 2 roots
- has 1 root
- has no root
Hi @Martin_Castaneda, your answers don’t look quite right for this one.
Remember that when you’re trying to find a value for ‘u’;
- If it’s positive there will be 2 roots
- If it’s zero there will be 1 root
- If it’s negative there will be no roots
I’ve written some additional explanations for how to solve each equation below and marked them with spoilers. You can use them to double check your work to see where you may have gone wrong.
Give the challenge another go and see how you get on. It’s absolutely ok to make mistakes when first learning all of this stuff, so don’t feel discouraged.
You may also want to use something like desmos.com/calculator to plot some of your own graphs in standard form, and then work through this method to get them into intercept form. You can then add your answer to the graph to check whether the lines match.
(1)
The first equation is y = x^2 - 5x + 10.
So you’re looking for two numbers that add up to make 5 and multiply to make 10.
Here are the steps…
Write an equation to figure out a value for ‘u’;
(5/2 + u)(5/2 - u) = 10
Expand the parentheses to get;
25/4 - u^2 = 10
Rearrange this equations to make ‘u’ the subject;
u^2 = 25/4 - 40/4
u^2 = -15/4
At this point, if the answer was positive you could take the square root to find a final answer for ‘u’ but in this case the answer is negative.
If you were to take the square root of a negative number you’d get a complex number for an answer, which is no use to us, so we can stop there and just say that this equation has no roots.
Therefore the answer for (1) is that it has no roots, so can’t be converted to intercept form.
(2)
Your answer for this one is correct.
Here are the steps:
The original equation is y = x^2 - 4x + 4
So you want two numbers that add up to 4 and multiply to 4.
(2 + u)(2 - u) = 4
Then we expand and solve for ‘u’;
4 - u^2 = 4
u^2 = 0
We could take the square root of 0 but the answer is always going to be zero. This means that there is only one root, since the vertex would be touching the x-axis.
You could write this as y = (x - 2)(x - 2) or y = (x - 2)^2.
(3)
For this one the equation was y = x^2 - 6x + 8.
So we’re looking for 2 numbers that add up to 6 and multiply to make 8.
(3 + u)(3 - u) = 8
Once again, expand this and solve for u;
9 - u^2 = 8
u^2 = 9 - 8
u^2 = 1
u = sqrt(1)
u = 1
In this case, we had a positive number, so we could take the square root to get a proper answer.
This means that the equation will have 2 roots.
Now to get the numbers for where our roots are, we just substitute in our value for ‘u’;
(3 + 1)(3 - 1) = 8
4 * 2 = 8
Therefore, y = (x - 2)(x - 4)
Not sure what went wrong. Lol.
I tried it again, and it’s:
1 no root
2 one root
3 two roots
Thanks for letting me know and for the help!
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