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if AB=2x^2+3x-5,BC=x^2+5x-4,and AC=x^2,solve for x and find AC
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is B the mid-point of A and C ?
No |dw:1348798939629:dw|
ok, so u have AB=BC so, 2x^2+3x-5 = x^2+5x -4
subtraxt x^2 from both sides, what u get ?
1x^2+3x-5=5x-4
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subtract 5x and add 4 on both sides
1x^2-2X-1=0
now can u solve this quadratic ?
Yes thx'
Sorry No
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is the question correct ? verify it
i got x=1+√2 x=1-√2
b^2-4ac is incorrect....
so what are you tring to say did i do it wrong...
x^2-2x-1 = 0 did u use quadratic formula ?
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\[x=-b+-\sqrt{b^2-4ac}\]
that formula?
so what is b^2-4ac there ?
oh wait,
x=1+√2 x=1-√2 is correct
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how do i plug that in to find out what AC is
(1+√2 )^2 = 1+2+2√2 = 3+2√2
or (1-√2 )^2 = 1+2-2√2 = 3-2√2
ok?
ok
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