Answer:
[tex]x\approx -7.405 \text{ or } x \approx 0.405[/tex]
Step-by-step explanation:
1. Set up the equation
Let x = the number. Then
7x = seven times the number and
x² = the square of the number
x² + 7x = seven times the number added to the square of the number
x² + 7x = 3
Subtract 3 from each side
x² + 7x - 3 = 0
2. Solve the quadratic equation
Use the quadratic formula: a = 1; b = 7; c = -3.
[tex]x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\=\dfrac{-7\pm\sqrt{(7)^2-4\times1\times(-3)}}{2\times1}\\\\=\dfrac{-7\pm\sqrt{49 + 12}}{2}\\\\=\dfrac{-7\pm\sqrt{61}}{2}\\\\x = \dfrac{-7 - \sqrt{61}}{2} \text{ or } x = \dfrac{-7 + \sqrt{61}}{2}\\\\\mathbf{x\approx -7.405} \text{ or } \mathbf{x \approx 0.405}[/tex]
The graph of your quadratic crosses the x-axis at (-7.405,0) and (0.405,0).
