A point (3*sqrt(5), d + 6) is 3d units away from the origin. What is the smallest possible value of d?
+1632 To graph it out, we would see a right triangle with legs of length \(3\sqrt{5}\) and \(d + 6\). If the hypotenuse is \(3d\), then we can use the pythagorean theorem to get this equation:
\((3\sqrt{5})^2 + (d + 6)^2 = (3d)^2\)
Simplified:
\(45 + d^2 + 36 + 12d = 9d^2\)
Combine like terms and subtraction:
\(-8d^2 + 12d + 81 = 0\)
Now we have a quadratic, we can apply the quadratic formula \(d\) = \(-b {+\over} \sqrt{b^2 - 4ac}\over2a\) where a is the coefficient of \(d^2\), b is the coefficient of \(d\), and c is the constant of the equation.
Plugging in the values, we get:
\(-3 {+\over} 3\sqrt{19}\over-4\) = d
Since we need the smallest value of d, and d can't be a negative distance away from something, then we will use the subtraction operation. This simplifies to:
\(d = {3 + 3\sqrt{19}\over4}\)
