Question Video: Determining Whether the Roots of a Given Quadratic Equation Are Always Rational | Nagwa Question Video: Determining Whether the Roots of a Given Quadratic Equation Are Always Rational | Nagwa

# Question Video: Determining Whether the Roots of a Given Quadratic Equation Are Always Rational Mathematics • First Year of Secondary School

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If π and π are rational numbers, are the roots of the equation π₯Β² β 4πΒ³π₯ + 4πβΆ β 4πβΆ = 0 always rational?

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### Video Transcript

If π and π are rational numbers, are the roots of the equation π₯ squared minus four π cubed π₯ plus four π to the sixth power minus four π to the sixth power equals zero always rational?

First, letβs recall what it means for a number to be rational. If it is a rational number, we can write it as π over π, where π and π are integers and π is not equal to zero. Now, if π is rational, we can say that π cubed must also be rational. But why is that? Well, letβs say weβve written π as π over π. π cubed is then π over π all cubed. But we can cube a fraction by cubing the numerator and cubing the denominator. Since π and π are integers, their cubes are also integers. This means π cubed over π cubed is the quotient of two integers where the denominator is not equal to zero, so it is rational. Similarly, if we then multiply that by four, we still have a rational number.

If we multiply it by four, we get four times π cubed over π cubed. Four times an integer is still an integer, so we can deduce that four π cubed is rational. Using similar logic, we can also establish that four π to the sixth power and four π to the sixth power are both rational. Since the sum or difference of a pair of rational numbers is also a rational number, then four π to the sixth power minus four π to the sixth power is rational itself.

Now, what was the purpose in establishing whether each of these terms was rational? Well, in doing so, we can use the discriminant of the general quadratic equation, in other words, the quadratic equation of the form ππ₯ squared plus ππ₯ plus π equals zero. The quadratic formula tells me that any solutions to this equation are of the form negative π plus or minus root π squared minus four ππ over two π. The discriminant of this equation is the part inside the square root. Itβs π squared minus four ππ.

Now, letβs split the quadratic formula up a little bit. It can be written as negative π over two π plus or minus the square root of π squared minus four ππ over two π. Now, if we can establish that π and π are rational numbers, then their quotient is also rational. So if π and π are rational, negative π over two π must also be rational. Similarly, if we can also establish that the square root of the discriminant is rational, then the next part of this expression must also be rational. And for the square root of the discriminant to be rational, we must be able to establish that the discriminant itself is a square number.

So our quadratic equation has an π₯ squared coefficient of one, so π is one. The coefficient of π₯ is negative four π cubed, so thatβs π. And our constant term is four π to the sixth power minus four π to the sixth power. Now, π is rational, and weβve established that four π cubed is rational. So negative four π cubed, which is π, is also rational. This means negative π divided by two π is rational since itβs the quotient of two rational numbers. Letβs now consider the discriminant. π squared minus four ππ in this case is negative four π cubed squared minus four times one times four π to the sixth power minus four π to the sixth power. Distributing all our parentheses and this becomes 16π to the sixth power minus 16π to the sixth power plus 16π to the sixth power.

Now, in fact, these first two terms cancel to give zero. So we have a discriminant of 16π to the sixth power. If we now find the square root of the discriminant, we get four π cubed. Using the same logic as earlier, we can deduce that four π cubed is rational. This means that the second part of our quadratic formula, the square root of π squared minus four ππ over two π, must also be rational. Itβs a rational number divided by another rational number. The sum or difference of a pair of rational numbers we know is also a rational number. And that means that any values of π₯ that satisfy our equation must themselves be rational.

Now, of course, the values of π₯ that satisfy our equation are the roots of that equation. So if π and π are rational numbers, we can say yes, the roots of the equation are themselves always rational.

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