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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.
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.
Is the square root of 68 irrational or rational?
The square roots are irrational.
Can the answer to a quadratic equation be a decimal?
Yes. You can calculate the two roots of a quadratic equation by using the quadratic formula, and because there are square roots on the quadratic formula, and if the radicand is not a perfect square, so the answer to that equation has decimal.
What are the 4 quadratic equations?
actoring, using the square roots, completing the square and the quadratic formula.
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.
The square root of 50 is it rational or irrational?
The square roots of 50 are irrational.
Is the square root of 163 rational or irrational?
The square roots of 163 are irrational.
The square root of 84 is it rational or irrational?
The square roots of 84 are irrational.
Why do you use square roots to solve quadratic equations?
Because it's part of the quadratic equation formula in finding the roots of a quadratic equation.
Are all square roots irrational?
No.
