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Christina said that when $a, b,$ and $c$ are rational numbers and $a$ and $c$ have opposite signs, the quadratic equation $a x^{2}+b x+c=0$ must have real roots. Do you agree with Christina? Explain why or why not.

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Algebra

Chapter 5

QUADRATIC FUNCTIONS AND COMPLEX NUMBERS

Section 3

The Discriminant

Equations and Inequalities

Quadratic Functions

Complex Numbers

Polynomials

Oregon State University

McMaster University

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in this problem. The person claims that if the coefficients on the quadratic equation ah and see have ah opposing signs, then the solutions toothy quadratic equation will be too real roots. So in order to prove this, let's set a equal to a equal to minus C so instead of C will use minus a. So the the part of the quarter equation that determines whether or not it'll have aerial route is the section of B squared minus for a C, which is called the determinant. If it's greater than zero, we have two real roots. If it's equal to zero, we have just one repeated route, and if it's less than zero, it'll have to imaginary roots. So if we just continue downward Ah, we have B squared minus four, a times minus a and that will be equal to B squared plus for a squared thes square. Terms cannot go Weston zero, because anything squared will become positive and the numbers are being added together, so there's nothing that it could possibly go under. The hero of AIDS to bigger B is too big, so that means that this expression will always be, ah greater than zero if any values plugged in for A and B. So if the signs are opposing, there will be two really roots for the solution of the polynomial.

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