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Find all real solutions to 100x^2 + 20x + 1 = \(25\).  If you find more than one, then list the values separated by commas.

 Jan 17, 2022
 #1
avatar+360 
+2

\(120x^2+20x+1=25\)

\(120x^2+20x-24=0\)

\(x_{1,\:2}=\frac{-20\pm \sqrt{20^2-4\cdot \:120\left(-24\right)}}{2\cdot \:120}\)

\(x_{1,\:2}=\frac{-20\pm \:4\sqrt{745}}{2\cdot \:120}\)

\(x_1=\frac{-20+4\sqrt{745}}{2\cdot \:120},\:x_2=\frac{-20-4\sqrt{745}}{2\cdot \:120}\)

\(x=\frac{-5+\sqrt{745}}{60},\:x=-\frac{5+\sqrt{745}}{60}\)

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 Jan 17, 2022
 #2
avatar+36416 
+1

Use Quadratic Formula    a = 100      b = 20       c = -24   

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

 

to find x = 2/5   and  - 3/5

 Jan 17, 2022
 #3
avatar+360 
+2

hm....

you're right...

that was a much easier way thank you!

 Jan 17, 2022
 #4
avatar+2333 
+1

Mathguy, you did use the quadratic formula. You presented your work in very formal detail, and your solution is correct for the equation you presented. The roots are incorrect for the asked question because you used 120 (instead of 100) as the quadratic coefficient.

 

EP’s method is easier because you plug the coefficients into a computerized solver and it gives you the roots. There’s nothing wrong with that –the method is less prone to calculation errors and it’s very fast. However it’s a good idea to do these manually on occasion to keep the actual process active in your working memory.  Of course, check your work with a computerized solver to confirm your solution results.

 

 

GA

--. .-

GingerAle  Jan 17, 2022
 #5
avatar+360 
0

Thank you!

 Jan 22, 2022

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