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I have recently started to self-study Spivak's Calculus book, and to be honest, my math is kind of rusty so I was just wanting to confirm that I have done this question correctly.

The question: Suppose that $b^2 -4c < 0$. Show that there are no numbers x satisfying $x^2+bx+c=0$.

My solution:

I used completing the square to arrive at the formula $(x+\frac{b}2)^2-\frac{b^2-4c}4 = 0$. Breaking it down I can conclude that $(x+\frac{b}2)^2 \ge 0$ (any number squared is greater than or equal to 0) and that $\frac{b^2-4c}4 < 0$ (as specified by the question). Therefore, any number greater than or equal to 0 subtracted by a negative number needs to be greater than 0.

Sorry if this question seems kind of pointless, it's just taking me quite a while to slog through these questions and I want to make sure that I am doing it correctly.

Daniel B
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  • You have done it correctly. – Kavi Rama Murthy Aug 20 '20 at 06:20
  • Your reasoning seems to be correct. Worst case scenario, $(x+\frac{b}{2})^2=0$ and $-\frac{b^2-4c}{4}>0$, and cannot equal zero. Also, $b^2-4ac$ is the discriminant of a quadratic polynomial, and is only defined, or equivalently, a quadratic polynomial has real roots, if the discriminant is greater than or equal to zero. so this polynomial will never have a real root, and therefore always be greater than zero. This makes more sense if you plot a graph of a quadratic polynomial that never crosses the x-axis. – C Squared Aug 20 '20 at 06:24
  • It is also interesting to note that $x^2+bx+c$ will be either always greater than zero or always less than zero – DatBoi Aug 20 '20 at 07:08

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