+267 Hey so I have another question
Patricia is trying to solve the following equation by completing the square: $$25x^2+20x-10 = 0.$$She successfully rewrites the above equation in the following form: $$(ax + b)^2 = c,$$where $a,$ $b,$ and $c$ are integers and $a > 0.$ What is the value of $a + b + c$?
Thanks for your help
+9479 25x2 + 20x - 10 = 0 First, divide through by 25 .
\(\frac{25}{25}\)x2 + \(\frac{20}{25}\)x - \(\frac{10}{25}\) = \(\frac{0}{25}\)
x2 + \(\frac45\)x - \(\frac25\) = 0 Add \(\frac25\) to both sides of the equation.
x2 + \(\frac45\)x = \(\frac25\) Now add (\(\frac4{10}\))2 to both sides of the equation.
x2 + \(\frac45\)x + (\(\frac4{10}\))2 = \(\frac25\) + (\(\frac4{10}\))2 Factor the left side and simplify the right side.
(x + \(\frac4{10}\))2 = \(\frac{40}{100}\)+\(\frac{16}{100}\)
(x + \(\frac4{10}\))2 = \(\frac{56}{100}\)
Now, it said a, b, and c have to be integers...let's multiply both sides by 102
102 * (x + \(\frac4{10}\))2 = 102 * \(\frac{56}{100}\) And a2b2 = (ab)2 .
( 10 * (x + \(\frac4{10}\)) )2 = 100 * \(\frac{56}{100}\) Distribute the 10 .
(10x + 4)2 = 56
So.... a + b + c = 10 + 4 + 56 = 70
+9479 25x2 + 20x - 10 = 0 First, divide through by 25 .
\(\frac{25}{25}\)x2 + \(\frac{20}{25}\)x - \(\frac{10}{25}\) = \(\frac{0}{25}\)
x2 + \(\frac45\)x - \(\frac25\) = 0 Add \(\frac25\) to both sides of the equation.
x2 + \(\frac45\)x = \(\frac25\) Now add (\(\frac4{10}\))2 to both sides of the equation.
x2 + \(\frac45\)x + (\(\frac4{10}\))2 = \(\frac25\) + (\(\frac4{10}\))2 Factor the left side and simplify the right side.
(x + \(\frac4{10}\))2 = \(\frac{40}{100}\)+\(\frac{16}{100}\)
(x + \(\frac4{10}\))2 = \(\frac{56}{100}\)
Now, it said a, b, and c have to be integers...let's multiply both sides by 102
102 * (x + \(\frac4{10}\))2 = 102 * \(\frac{56}{100}\) And a2b2 = (ab)2 .
( 10 * (x + \(\frac4{10}\)) )2 = 100 * \(\frac{56}{100}\) Distribute the 10 .
(10x + 4)2 = 56
So.... a + b + c = 10 + 4 + 56 = 70
