Question

which equation has a solution that is greater than 6?
x³ = 216
x³ = 100
x² = 12
x² = 40

Explanation:

Step1: Solve \(x^3 = 216\)

Take the cube root of both sides: \(x=\sqrt[3]{216}=6\) (since \(6^3 = 216\)).

Step2: Solve \(x^3 = 100\)

Take the cube root: \(x=\sqrt[3]{100}\approx4.64\) (since \(4^3 = 64\), \(5^3 = 125\), so \(\sqrt[3]{100}\) is between 4 and 5, less than 6).

Step3: Solve \(x^2 = 12\)

Take the square root: \(x=\sqrt{12}\approx3.46\) (since \(3^2 = 9\), \(4^2 = 16\), so \(\sqrt{12}\) is between 3 and 4, less than 6).

Step4: Solve \(x^2 = 40\)

Take the square root: \(x=\sqrt{40}\approx6.32\) (since \(6^2 = 36\), \(7^2 = 49\), so \(\sqrt{40}\) is between 6 and 7, greater than 6).

Answer:

The equation \(x^2 = 40\) has a solution greater than 6.