How would you solve this with the quadratic formula and leave the answer in 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} © Algebra House
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