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What indicates that the roots of a quadratic are irrational?
In the quadratic equation, b^2 - 4ac < 0.
What does the discriminant have to be in order for the roots of a quadratic to be irrational?
The discriminant must be a positive number which is not a perfect square.
Which of the folllowing indicates that the roots of a quadratic are irrational?
The discriminant is the expression inside the square root of the quadratic formula. For a quadratic ax² + bx + c = 0, the quadratic formula is x = (-b +- Sqrt(b² - 4ac))/(2a). The expression (b² - 4ac) is the discriminant. This can tell a lot about the type of roots. First, if the discriminant is a negative number, then it will have two complex roots. Because you have a real number plus sqrt(negative) and real number minus sqrt(negative). You asked about irrational. If the discrimiant is a perfect square number {like 1, 4, 9, 16, etc.} then the quadratic will have two distinct rational roots (which are real numbers). If the discriminant is zero, then you will have a double root, which is a real rational number. So if the discrimiant is positive, but not a perfect square, then the roots will be irrational real numbers. If the discriminant is a negative number which is not the negative of a perfect square, then imaginary portion of the complex number will be irrational.
Does there exist a quadratic equation whose coefficients are irrational but both the roots are rational?
None, if the coefficients of the quadratic are in their lowest form.
Is the square root of 163 rational or irrational?
The square roots of 163 are irrational.
