Question

Twice a certain whole number subtracted from 3times the square of the number leaves 133. Find the number

Answer 1

Let the whole number be denoted by \( x \). According to the problem, we have the equation based on the statement:

\[
3x^2 - 2x = 133
\]

To rearrange this, we can move all terms to one side:

\[
3x^2 - 2x - 133 = 0
\]

Now we can solve this quadratic equation using the quadratic formula, which is given by:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

In our equation, \( a = 3 \), \( b = -2 \), and \( c = -133 \). Substituting these values into the formula gives:

\[
x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-133)}}{2 \cdot 3}
\]

Calculating the discriminant:

\[
(-2)^2 - 4 \cdot 3 \cdot (-133) = 4 + 1596 = 1600
\]

Now substituting this back into the quadratic formula:

\[
x = \frac{2 \pm \sqrt{1600}}{6}
\]

Calculating \( \sqrt{1600} \):

\[
\sqrt{1600} = 40
\]

So we have:

\[
x = \frac{2 \pm 40}{6}
\]

This gives us two possible solutions:

1. \( x = \frac{42}{6} = 7 \)
2. \( x = \frac{-38}{6} = -\frac{19}{3} \) (which is not a whole number)

Thus, the only whole number solution is:

\[
\boxed{7}
\]

To verify, we can substitute \( x = 7 \) back into the original statement:

Calculating \( 3x^2 \):

\[
3(7^2) = 3 \times 49 = 147
\]

Calculating \( 2x \):

\[
2 \times 7 = 14
\]

Subtracting these gives:

\[
147 - 14 = 133
\]

Since this holds true, the solution \( x = 7 \) is correct.