Directions: Solve the following quadratic equations and mark the correct option. I. x2 + 3x - 54 = 0 II. y2 - 20y + 99 = 0
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Directions: Solve the following quadratic equations and mark the correct option. I. x2 + 3x - 54 = 0 II. y2 - 20y + 99 = 0
x > y
x < y
x ≤ y
x ≥ y
x = y or no relation can be established
I: x^2+3x-54=0 -> (x+9)(x-6)=0, roots are -9, 6. II: y^2-20y+99=0 -> (y-9)(y-11)=0, roots are 9, 11. Comparing the sets, x < y.
Using the factorization method for equation one, x squared plus three x minus fifty four equals zero gives factors of positive nine and negative six, so x equals nine or negative six. For equation two, y squared minus twenty y plus ninety nine equals zero gives factors of negative eleven and negative nine, so y equals eleven or nine. Comparing the values, both eleven and nine are greater than or equal to the maximum x value of nine, while both are strictly greater than negative six; therefore, x is less than y.