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Guest
Guest
88x^2+90= 179x[/quote][/tex][/code]
Use the quadratic formula http://www.sosmath.com/algebra/quadrati ... rmula.html
G
Guest
Guest
I tried using the quadratic formula but once I get to step 1 and multiply 1/88 by 1/90 and 1/179 I get stuck. Any ideas?
You're obviously lost, alost girl;alostgirl said:
why are you given such equations if you can't work the quadratic? Bad teacher?
your equation is: 88x^2+90= 179x
SO: 88x^2 - 179x + 90 = 0 : ok?
Using the quadratic equation:
x = [179 +- sqrt(-179^2 - 4(88)(90)] / [2(88)] : do you understand that?
WHERE/HOW do you get stuff like 1/88 and 1/90 ?
Hello, alostgirl!
On this planet, we use: \(\displaystyle \,x \;=\;\L\frac{-b\,\pm\,\sqrt{b^2\,-\,4ac}}{2x}\)
Your equation is: \(\displaystyle \,88x^2\,-\,179x\,+\,90\:=\:0\)
\(\displaystyle \;\;\) which has: \(\displaystyle \,a\,=\,88,\:b\,=\,-179,\:c\,=\,90\)
so we have: \(\displaystyle \,x\;=\;\:\frac{-(-179)\,\pm\,\sqrt{(-179)^2\,-\,4(88)(90)}}{2(88)}\;\) . . . crank it out!
By the way, it factors: \(\displaystyle \
8x\,-\,9)(11x\,-\,10)\)
What Quadratic Formula are you using?
On this planet, we use: \(\displaystyle \,x \;=\;\L\frac{-b\,\pm\,\sqrt{b^2\,-\,4ac}}{2x}\)
Your equation is: \(\displaystyle \,88x^2\,-\,179x\,+\,90\:=\:0\)
\(\displaystyle \;\;\) which has: \(\displaystyle \,a\,=\,88,\:b\,=\,-179,\:c\,=\,90\)
so we have: \(\displaystyle \,x\;=\;\:\frac{-(-179)\,\pm\,\sqrt{(-179)^2\,-\,4(88)(90)}}{2(88)}\;\) . . . crank it out!
By the way, it factors: \(\displaystyle \
8x\,-\,9)(11x\,-\,10)\)Write as
\(\displaystyle \mbox{ 88x^2 - 179x + 90 = 0}\)
and compare with
\(\displaystyle \mbox{ ax^2 + bx + c = 0}\)
So we have
\(\displaystyle \mbox{ a = 88}\)
\(\displaystyle \mbox{ b = -179}\)
\(\displaystyle \mbox{ c = 90}\)
Apply the quadratic formula:
\(\displaystyle \mbox{x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}\)
\(\displaystyle \mbox{ = \frac{-(-179) \pm \sqrt{(-179)^2 - 4 \cdot 88 \cdot 90}}{2 \cdot 88}}\)
\(\displaystyle \mbox{ = \frac{ 179 \pm \sqrt{361}}{176}}\)
\(\displaystyle \mbox{ = \frac{ 179 \pm 19}{176}}\)
Therefore:
\(\displaystyle \mbox{ x = \frac{ 179 + 19}{176}} = \frac{198}{176} = \frac{9}{8{\)
\(\displaystyle \mbox{ }\)or \(\displaystyle \mbox{x = \frac{ 179 - 19}{176} = \frac{160}{176} = \frac{10}{11}}\)
That is, \(\displaystyle \mbox{x = \frac{9}{8}, \frac{10}{11}}\).
\(\displaystyle \mbox{ 88x^2 - 179x + 90 = 0}\)
and compare with
\(\displaystyle \mbox{ ax^2 + bx + c = 0}\)
So we have
\(\displaystyle \mbox{ a = 88}\)
\(\displaystyle \mbox{ b = -179}\)
\(\displaystyle \mbox{ c = 90}\)
Apply the quadratic formula:
\(\displaystyle \mbox{x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}\)
\(\displaystyle \mbox{ = \frac{-(-179) \pm \sqrt{(-179)^2 - 4 \cdot 88 \cdot 90}}{2 \cdot 88}}\)
\(\displaystyle \mbox{ = \frac{ 179 \pm \sqrt{361}}{176}}\)
\(\displaystyle \mbox{ = \frac{ 179 \pm 19}{176}}\)
Therefore:
\(\displaystyle \mbox{ x = \frac{ 179 + 19}{176}} = \frac{198}{176} = \frac{9}{8{\)
\(\displaystyle \mbox{ }\)or \(\displaystyle \mbox{x = \frac{ 179 - 19}{176} = \frac{160}{176} = \frac{10}{11}}\)
That is, \(\displaystyle \mbox{x = \frac{9}{8}, \frac{10}{11}}\).
G
Guest
Guest
thank you for your help
- Joined
- Feb 4, 2004
- Messages
- 16,582
Please post new questions as new threads, not as replies to old threads, where they tend to be overlooked.alostgirl said:
Also, since the instructions for this exercise were not included, it is difficult to ascertain whether your answer is correct. Sorry.
When you re-post this, kindly please include the instructions. Thank you.
Eliz.