whadduptori
Use the quadratic formula to solve the equation?
4x2 – 6x – 10 = 0 is anybody able to show me work?
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Rezz5
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the quadratic formula is -b (±)Sqrt[b-4ac]/2a\[-b \pm \frac{ \sqrt{b-4ac}}{ 2a}\]
Rezz5
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where
ax^2 +bx +c
just substitute and you will get your answer.
jim_thompson5910
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\[\Large x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
\[\Large x = \frac{-(-6)\pm\sqrt{(-6)^2-4(4)(-10)}}{2(4)}\]
\[\Large x = \frac{6\pm\sqrt{36-(-160)}}{8}\]
I'll let you finish.
whadduptori
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I have all that work! I just get confused with the last part? like would i subtract 36-(-160)
Rezz5
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36--160?
well double negative is positive,
so?
36+160
jim_thompson5910
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36-(-160) is the same as 36+160
whadduptori
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so it would become 196?
jim_thompson5910
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it would
whadduptori
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|dw:1352933217966:dw|
whadduptori
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so would that be the answer?
jim_thompson5910
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you can keep going and there's a +- between the 6 and square root
jim_thompson5910
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you should have
\[\Large x = \frac{6\pm\sqrt{196}}{8}\]
but you can keep going
Rezz5
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(6*14)/8
=(3*14)/4
=(3*7)/2
=21/2
whadduptori
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yes i have that!
Rezz5
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okay, what is wrong?
whadduptori
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wait where did you get the 6*14/8
jim_thompson5910
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the square root of 196 is 14, so...
\[\Large x = \frac{6\pm\sqrt{196}}{8}\]
becomes
\[\Large x = \frac{6\pm14}{8}\]
Rezz5
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above is very clear and correct,
so first you do the plus sign
which is
(6+14)/8
this is ONE OF YOUR x's
the you do
(6-14)/8
this is your other x value
so your x values are:
5/2
and
-1
whadduptori
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THank you so much! I didnt know how to find that! but you just made it clear for me! thank you both for your help! (:
Rezz5
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your most welcome,