KG1234
solve for x: (4x3)^2=15



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NickR
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so (4x3)^2=(4x3)(4x3) right

KG1234
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yes

NickR
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so take that and foil it

Hero
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(4x3)(4x3) = 4x(4x  3)  3(4x  3) = 16x^2  12x  12x +9 = 16x^2  24x + 9

KG1234
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16x^224x+24=0

NickR
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but its not = to 0 its =15

Hero
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(4x3)(4x3)
= 4x(4x  3)  3(4x  3)
= 16x^2  12x  12x +9
= 16x^2  24x + 9

NickR
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my bad see you already did that

NickR
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so now is there a common number that they are all divisible by?

KG1234
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wait im at 16x^224x+9=15 right?

NickR
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ya then add the 15 to get 16x^224x+24=0

KG1234
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okay yeah i did that

NickR
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now find a common denominator

KG1234
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8?

NickR
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yup

KG1234
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so does that then turn into 2x^23x=3=0?

KG1234
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2x^23x+3 sorry

Hero
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16x^224x+9=15
16x^2  24x + 24 = 0
8(2x^2  3x + 3) = 0
2x^2  3x + 3 = 0
x^2  3/2 + 3/2 = 0
x^2  3/2x = 3/2
x^2 3/2x + 9/16 = 3/2 + 9/16
(x  3/4)^2 = 15/16
Solution is imaginary

NickR
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yes

KG1234
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then what do i do?

KG1234
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the answer is 3 + or  the square root of 15 divided by4 but idk how to get it

NickR
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i would do the quadratic formula

KG1234
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b + or  the square root of b^2 4ac divided by 2a?

Hero
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\[\frac{b \pm \sqrt{b^2  4ac}}{2a}\]

NickR
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(b+sqrt(b^24ac))/2a

NickR
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yup

KG1234
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i got it thanks to both of you!

NickR
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not a problem

Hero
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You got what? Mind sharing what you got?

KG1234
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3 + or  the square root of 15 divided by 4

KG1234
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3+ or  i square root 15 divided by 4 sorry i forgot the "i"

NickR
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yup