Question

solve for x in the equation $3x^2 - 18x + 5 = 47$.\
\bigcirc $x = 3 \pm \sqrt{23}$\
\bigcirc $x = 3 \pm \sqrt{51}$\
\bigcirc $x = 3 \pm \sqrt{41}$\
\bigcirc $x = 3 \pm \sqrt{5}$

Explanation:

Step1: Simplify the equation

Subtract 47 from both sides: \(3x^2 - 18x + 5 - 47 = 0\)
Simplify: \(3x^2 - 18x - 42 = 0\)
Divide all terms by 3: \(x^2 - 6x - 14 = 0\)

Step2: Use the quadratic formula

The quadratic formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) for \(ax^2 + bx + c = 0\).
Here, \(a = 1\), \(b = -6\), \(c = -14\).
Calculate the discriminant: \(b^2 - 4ac = (-6)^2 - 4(1)(-14) = 36 + 56 = 92\)
Wait, wait, no, wait. Wait, in Step1, when we had \(3x^2 - 18x + 5 = 47\), subtract 47: \(3x^2 - 18x - 42 = 0\), divide by 3: \(x^2 - 6x - 14 = 0\)? Wait, no, 5 - 47 is -42, correct. But let's check again. Wait, maybe I made a mistake. Wait, original equation: \(3x^2 - 18x + 5 = 47\). Subtract 47: \(3x^2 - 18x - 42 = 0\). Divide by 3: \(x^2 - 6x - 14 = 0\). Then discriminant: \(b^2 - 4ac = (-6)^2 - 4(1)(-14) = 36 + 56 = 92\). Wait, but 92 is 4*23, so \(\sqrt{92} = \sqrt{4*23} = 2\sqrt{23}\). Then \(x = \frac{6 \pm 2\sqrt{23}}{2} = 3 \pm \sqrt{23}\). Ah, there we go. So the correct answer is \(x = 3 \pm \sqrt{23}\).

Answer:

\(x = 3 \pm \sqrt{23}\) (corresponding to the first option: \(x = 3 \pm \sqrt{23}\))