Asked by Anna-Marie
wow. i can't believe i forget this from last year.
5x^2 - 11x - 36 = 0
theres some kind of box trick or somethin to factor that but i forget!
5x^2 - 11x - 36 = 0
theres some kind of box trick or somethin to factor that but i forget!
Answers
Answered by
Damon
Three possible tricks
1. quadratic equation
2. complete the square
3. factor it
1. quadratic equation
2. complete the square
3. factor it
Answered by
Damon
factoring
(5x+9)(x-4) = 0
x = -9/5
or
x = 4
(5x+9)(x-4) = 0
x = -9/5
or
x = 4
Answered by
mk-tintin
i'm french so i don't know how do u solve this equation in usa
5x^2 - 11x - 36 = 0
11²-4*5*(-36)=121+720=841
sqrt(841)=29
solutions are (11-29)/(2*5) and (11+29)/(2*5)
5x^2 - 11x - 36 = 0
11²-4*5*(-36)=121+720=841
sqrt(841)=29
solutions are (11-29)/(2*5) and (11+29)/(2*5)
Answered by
Anna-Marie
umm... i don't remember how to do any of it. like to factor and get the ( ) ( ) can you tell me how to do that
Answered by
Anna-Marie
ooh.. damon how did you get the (5x+9)(x-4) = 0 that's what i need to know
Answered by
Damon
complete the square
5 x^2 - 11 x = 36
x^2 - 11/5 x = 36/5
x^2 -11/5 x + 121/100 = 36/5 + 121/100
(x-11/10)^2 = 841/100
x-11/10 = +/- 29/10
x = (11+29)/10 = 4
or
x = (11-29)/10 = -18/10 = -9/5
5 x^2 - 11 x = 36
x^2 - 11/5 x = 36/5
x^2 -11/5 x + 121/100 = 36/5 + 121/100
(x-11/10)^2 = 841/100
x-11/10 = +/- 29/10
x = (11+29)/10 = 4
or
x = (11-29)/10 = -18/10 = -9/5
Answered by
Damon
Oh, ok, I guessed'
Like so
5 is only 5 * 1
so there has to be something like
(5x + ;;;) (x - :::)
now the Plus and minus may change and 36 has factors
2*2*3*3
which could be 12 and 3
or 6 and six
or 9 and 4
turns out 9 and 4 work
Like so
5 is only 5 * 1
so there has to be something like
(5x + ;;;) (x - :::)
now the Plus and minus may change and 36 has factors
2*2*3*3
which could be 12 and 3
or 6 and six
or 9 and 4
turns out 9 and 4 work
Answered by
Damon
Tin tin is using some form of quadratic equation
x = -b/2 +/- (1/2) sqrt{ b^2 - 4 a c}
if in form
a x^2 + b X + c = 0
x = -b/2 +/- (1/2) sqrt{ b^2 - 4 a c}
if in form
a x^2 + b X + c = 0
Answered by
Damon
x = -b/2a +/- (1/2a) sqrt{ b^2 - 4 a c}
if in form
a x^2 + b X + c = 0
if in form
a x^2 + b X + c = 0
Answered by
Damon
By the way, the quadratic equation really is completing the square
a x^2 + b x + c = 0
x^2 + (b/a) x = -(c/a)
x^2 + (b/a) x + (b/2a)^2 = -(c/a) +(b/2a)^2
(x + b/2a)^2 = [-c/a + (b/2a)^2 ]
so
x+ b/2a = +/- sqrt (b^2/4a^2 - 4 a c/4a^2)
or
x = -b/2a +/- (1/2a)sqrt(b^2-4 a c)
a x^2 + b x + c = 0
x^2 + (b/a) x = -(c/a)
x^2 + (b/a) x + (b/2a)^2 = -(c/a) +(b/2a)^2
(x + b/2a)^2 = [-c/a + (b/2a)^2 ]
so
x+ b/2a = +/- sqrt (b^2/4a^2 - 4 a c/4a^2)
or
x = -b/2a +/- (1/2a)sqrt(b^2-4 a c)
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