Question

Which equation has a single solution of x = 1.2?

a)0 = 25x2 + 60x + 36
b)0 = 25x2 – 60x + 36
c)0 = 36x2 + 60x + 25
d)0 = 36x2 – 60x + 25

Answer 1

@pitamar do you know this

Answer 2

Remember how we found the roots in the previous problems?
we had something of the kind \(f(x) = (x+a)(x+b)\) and from that we knew what values make the function result in 0.
Having a single solution means that 'a' and 'b' are equal and therefore we got something of the kind \(f(x) = (x+a)^2\).
So try and form each function in the following form: \(f(x) = (x+a)(x+b)\)
Then see which ones are capable of having a single solution, and then see if the solution 1.2 causes them to evaluate to 0

Answer 3

how would i know what is a and b

Answer 4

you have to transform an expression like 25x^2 + 60x + 36 into something of the form (x+a)(x+b) that is equal to it. You know that kind of stuff?

Answer 5

ok so how can i get it in that form

Answer 6

well. if you look at it, it turns that:
$$
(x+a)(x+b) = x(x+b) + a(x+b) = x^2 +xb + ax + ab
$$If a and b are equal we can go even further:
$$
b = a\\
x^2 +xb + ax + ab = x^2 + xa + ax + aa = x^2 + 2ax + a^2
$$
We have to try and reverse the process

Answer 7

Let's take an example.

Answer 8

this is a little confusing

Answer 9

Can you 'open' this?
$$(x+1)(x-3) = ?$$

Answer 10

yea one moment

Answer 11

how do i reopen it

Answer 12

$$
(x+1)(x-3) = x(x-3) + 1(x-3) = ...
$$

Answer 13

so i will have to distribute?

Answer 14

Ye that's what i'm asking you to do right now

Answer 15

\[x ^{2}\]

Answer 16

oh wait

Answer 17

waiting

Answer 18

its (x2 +1x)(x2-3)=(x2-3x)+(1x-3)

Answer 19

well it really is (x2-3x)+(1x-3), but I don't really understand the (x2 +1x)(x2-3) part.
Anyway, let's replace the 3 and 1 with 'a' and 'b' and you try again:
$$
(x+a)(x+b) = ?
$$

Answer 20

so it would be (x+3) (x+1)

Answer 21

No, it's a different expression, just like you did with the (x+1)(x-3) I want you to open this expression

Answer 22

so distribute (x+1)(x-3)?

Answer 23

That is what you've just distributed. Now distribute this:
$$
(x+a)(x+b) = ...
$$

Answer 24

ok (x+a)(x+b)=... x2+ab?

Answer 25

well not exactly
let's see
$$
(x+a)(x+b) = x(x+b) + a(x+b) = ...
$$

Answer 26

ok (x2+bx)+ax+ab)?

Answer 27

ye, exactly. and that's what I wrote above that you thought was confusing

Answer 28

oh ok lol thanks so whats next now

Answer 29

a and b can be any numbers. it could be 3... 5.. whatever. it will always end up in that form.
Now let's say a = b. Then we have
$$
a = b\\
(x+a)(x+b) = (x+a)(x+a) = ...?
$$

Answer 30

so it will be (x2+a)?

Answer 31

heh. again:
$$
(x+a)(x+a) = x(x+a) + a(x+a) = ?
$$

Answer 32

ok (x2+ax)+(ax+a)

Answer 33

close a*a is what?

Answer 34

will it just be aa? or just a?

Answer 35

\(a\cdot a\) (which is the same as writing \(aa\)) is basically \(a^2\)

Answer 36

oh ok

Answer 37

so its a2

Answer 38

ye, so (x+a)(x+a) ends up with what?

Answer 39

(ax)(ax)?

Answer 40

or (a2)(a2)

Answer 41

Neither is correct. go over what we did again, then try, ok?

Answer 42

ok is it (x2)(a2)

Answer 43

you tend to forget the step
$$
(x+a)(x+a) = x(x+a) + a(x+a)
$$

Answer 44

(x2+ax)+(ax+a2)

Answer 45

Correct.
Maybe this visualization will help:

Each side of this square is in length of (x+a). The area of the square (which is literally \((x+a)^2\)) is the sum of the components inside.