Areteem: MCIII Algebra 5.8

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MCIII Algebra 5.8

Hi,

I don't understand the solution at the back of the book. Technically, the coefficient of x isn't 2(sin(pix/2)), since there's still a variable, x, in the coefficent, which is a contradivtion.

How do I solve the problem then?

Thanks

TIna Jin

The question is: Find all real roots for the equation $$x^2 - 2x\sin\dfrac{\pi x}{2} + 1 = 0.$$

The solution treats this as a quadratic equation in $x$, and uses the discriminant. There is no contradiction here. As long as $x$ is a root, the inequality involving the discriminant is true even though $x$ appears in the coefficients. Granted, this is not really a quadratic equation, but suppose $x=a$ is a real root to the original equation, then the following quadratic equation

$$x^2 - 2\left(\sin\dfrac{\pi a}{2}\right)x + 1 = 0$$

should also have a solution $x=a$. The discriminant analysis works for this quadratic equation. Essentially, there is no difference from the original method.

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I see, so does the discriminant ALWAYS work? Even when the coefficient of it has an x in it?

For example, would this work:

x^2+(F(x))x+n=0, discriminant is (F(x))^2-4n.

Thanks,

Tina

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We need to be specific in saying the following: if the equation $x^2+(F(x))x+n=0$ has a real root for $x$, then $(F(x))^2-4n\geq 0$ for the same value of $x$.

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Thank you!

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