for these three equations x^2-4x+3=0 x^2-7x+2=20 x^2-3x=1 solve at least one by factoring solve at least one by using the quadratic formula solve the last one with either of the two options
solve the second one by factoring and the last one by factoring also the first one you need to solve with quad eq.
First of all they all need to = 0. The first two should be factoring.. the last one is quadratic formula.
x^2 - 4x + 3 = 0 Hint (3 * 1 = 3) (3 + 1 = 4)
1. (x - 3)(x - 1)
yea hes right i copied in diff order
x^2 - 7x + 2 = 20 x^2 - 7x - 18 = 0 Factors of 18 that subtract to give you 7 What are they?
x^2 - 3x - 1 = 0 the reason you use the quadratic formula is because there are no factors of 1 that subtract to give you 3.
Confused.. what are the solutions to #1 if (x - 3)(x - 1) = 0
3 and 1????
Very good
Now on the 2nd one... the factors of 18 are 1 * 18 2 * 9 3 * 6 which pair subtracts to give you 7
2 and 9
good so you have (x - 9)(x + 2) = 0 what would be your answers for x
9 and-2
Very good. Now we need to do the quadratic formula.. do you know what that is? yes/no
i think so ax^2+bx+c
well.. you have to get it in this form first.. so we have x^2 - 3x - 1 = 0 what are a? b? c?
a=1 b=-3 c=-1
Good. the formula is x = -b +- sqrt(b^2 - 4ac) all over 2a Have you seen this the +- means a + with a - sign under it.
yes
-b I like to read as the opposite of b so the opposite of -3 is +3
x = 3 +- sqrt((-3)^2 - 4(1)(-1) ------------------------ 2(1)
x = 3 +- sqrt(9 +4) ------------- 2 do you see how I got this? the (-4)(1)(-1) is a + 4
yes
OK.. so your answer is x = 3 +- sqrt(13) ----------- 2
so 4.80 and 1.20
I don't think so..
If you are using a calculator put in (3 + sqrt(13))/2 (3 - sqrt(13))/2
3.307
-3.027
Oops.. -.3027
but those aren't the answers those are approximations since the sqrt(13) is irrational
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