Will Medal Please help!! Use Quadratic formula to solve for x: (x+2)(x-1) = 5

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Will Medal Please help!! Use Quadratic formula to solve for x: (x+2)(x-1) = 5

Mathematics
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Start off by putting into \[ax^2+bx+c=0\] form
ok, wait let me solve it first
Take your time

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Other answers:

ok, x^2 +1x -7
Good now we have the quadratic formula which is the following \[x = \frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\] where a = 1, b= 1, and c = -7
ok, wait again sorry
Answer is 2,-1
Are you sure?
yes
no no i am wrong wait
the answer is 2.15 and -3.15
i took the wrong equation
\[x= \frac{ -1 \pm \sqrt{1^2-4(1)(-7)} }{ 2(1) } \implies x = \frac{ -1 \pm \sqrt{29} }{ 2 }\] correct?
yes
That should \[x \approx -3.19,~~ \text{and}~~ x \approx 2.19\]
can i ask you an another quetion pls
Sure thing
Use Quadratic formula to solve for x: \[x + \frac{ 1 }{ x+2 } = 4\]
Same thing as earlier, try putting it into \[ax^2+bx+c=0\]
how will i do that
You can multiply through by (x+2)
??
I mean you can find a common denominator which in this case is (x+2) if you like
can we cross multiply
You can do as following \[\frac{ 1 }{ x+2 }=4-x \] now you may do as you like
\[\frac{ 1 }{ (x+2) } = (4-x)\]
ok
wait
is it = to 4x-x-5??
Hmm, not quite, \[\frac{ 1 }{ (x+2) } = (4-x) \implies 1 = (4-x)(x+2)\]
yeah then we multiply (4-x)(x+2) = 4x - x +6
an we bring it to the other side
-4x + x -6 + 1
= -4x + x -5 = 0
right??
Remember, we want our equation in \[ax^2+bx+c=0\] form, so what you wrote, does that make sense? Think of what you did in the previous problem
x^2-4x-5
= 0
\[(4-x)(x+2)=1 \implies 4x+8-x^2-2x =1 \implies 2x+8-x^2=1 \] so we can set this up as \[x^2-2x-7=0\]
ok. sorry i was offline, i had a lag
answer is = 1.9, -0.9 right??
Can you show how you got that?
ok |dw:1444558479751:dw|
|dw:1444558581896:dw|
What is a, b, and c?
b =-2, c -7
\[x = \frac{ -(-2) \pm \sqrt{(-2)^2-4(1)(-7)} }{ 2(1) } = \frac{ 2 \pm \sqrt{32} }{ 2 }\]
a = 1, b = -2, c = -7
thnxs

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