Question 636056
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Neither of your choices are technically accurate.  This is a fourth degree polynomial equation, therefore there are four roots, so A is incorrect because it only allows for 1 solution and B is incorrect because it doesn't allow for any.

Let *[tex \LARGE u\ =\ x^2], then

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 25u^2\ -\ 15u\ +\ 2\ =\ 0]

Factor the quadratic in *[tex \LARGE u]

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (5u\ -\ 2)(5u\ -\ 1)\ =\ 0]

(Verification of the factorization using FOIL is left as an exercise for the student)

So:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u\ =\ \frac{2}{5}]

or

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u\ =\ \frac{1}{5}]

But since *[tex \LARGE u\ =\ x^2]

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ =\ \frac{2}{5}]

or

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ =\ \frac{1}{5}]

Solve each of the above quadratic equations for its two zeros by taking the square root of both sides and considering both the positive and negative roots.

Don't forget to rationalize your denominators.

John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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