How would you solve this with the quadratic formula and leave the answer in
simplest radical form?
y² + 6y - 1 = 0
simplest radical form?
y² + 6y - 1 = 0
Quadratic formula
-b ±√b² - 4ac
------------------ = x
2a
y² + 6y - 1 = 0
a = 1, b = 6, c = -1
-6 ± √6² - 4(1)(-1)
---------------------- = y {substituted into the quadratic formula}
2(1)
-6 ± √36 + 4
--------------- = y {squared the 6 and multiplied on the end}
2
-6 ± √40
--------- = y {added 36 and 4}
2
-6 ± √4√10
------------- = y {broke 40 into 4 and 10}
2
-6 ± 2√10
--------- = y {evaluated square root of 4 to be 2}
2
2(-3 ± √10)
--------------- {factored 2 out of top}
2
y = -3 ± √10 {cancelled the 2's}
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