- Rearrange \(7x^2 = 9 + x\) to standard form: \(7x^2 - x - 9 = 0\).
- Identify coefficients: \(a = 7\), \(b = -1\), \(c = -9\).
- Apply the quadratic formula: \(x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(7)(-9)}}{2(7)}\).
- Simplify and match with Option B: \(\boxed{B}\).
Explanation
- Rewrite the equation in standard form
We begin by writing the given quadratic equation \(7x^2 = 9 + x\) in the standard quadratic form, which is \(ax^2 + bx + c = 0\). To do this, we subtract \(x\) and \(9\) from both sides of the equation:
\[7x^2 - x - 9 = 0.\]
From this standard form, we identify the coefficients:
- \(a = 7\)
- \(b = -1\)
- \(c = -9\)
- Apply the quadratic formula
The quadratic formula to solve \(ax^2 + bx + c = 0\) is given by:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\]
Substituting the values \(a = 7\), \(b = -1\), and \(c = -9\) into the formula, we have:
\[x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(7)(-9)}}{2(7)}.\]
Simplifying the terms, we get:
- \(-(-1) = 1\)
- \((-1)^2 = 1\)
- \(-4(7)(-9) = 4 \times 7 \times 9 = 252\)
- \(2a = 14\)
Thus, the formula becomes:
\[x = \frac{1 \pm \sqrt{1 + 252}}{14} = \frac{1 \pm \sqrt{(-1)^2 - 4(7)(-9)}}{14}.\]
- Evaluate options and determine the correct one
Comparing our result \(x = \frac{1 \pm \sqrt{(-1)^2 - 4(7)(-9)}}{2(7)}\) with the given options:
- Option A: \(x = \frac{-1 \pm \sqrt{(-1)^2 + 4(7)(9)}}{2(7)}\)
- Option B: \(x = \frac{1 \pm \sqrt{(-1)^2 - 4(7)(-9)}}{2(7)}\)
- Option C: \(x = \frac{1 \pm \sqrt{(-1)^2 + 4(7)(9)}}{2(7)}\)
- Option D: \(x = \frac{-1 \pm \sqrt{(1)^2 - 4(7)(-9)}}{2(7)}\)
We can see that Option B matches our derivation, noting that \(4(7)(-9)\) is \(-4(7)(9)\), so the subtraction of the term \(-4ac\) results in \(+4(7)(9)\). Specifically, \(b^2 - 4ac = (-1)^2 - 4(7)(-9) = 1 - 4(7)(-9) = 1 + 252\). Therefore, the expression \(1 \pm \sqrt{(-1)^2 - 4(7)(-9)}\) over \(2(7)\) is correct.
Examples
Quadratic equations are fundamental in modeling real-world physical phenomena. For example, if you throw a ball, its height over time is described by a quadratic equation. Solving for the roots tells you exactly when the ball hits the ground. Understanding how to correctly apply the quadratic formula allows you to calculate these critical moments in physics, engineering, and many other technical fields.