Question Video: Finding the All Possible Values of a Constant That Make the Roots of a Given Quadratic Equation Not Real | Nagwa Question Video: Finding the All Possible Values of a Constant That Make the Roots of a Given Quadratic Equation Not Real | Nagwa

Question Video: Finding the All Possible Values of a Constant That Make the Roots of a Given Quadratic Equation Not Real Mathematics • First Year of Secondary School

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If the roots of the equation 24π₯Β² + 6π₯ + π = 0 are not real, find the interval which contains π.

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Video Transcript

If the roots of the equation 24π₯ squared plus six π₯ plus π is equal to zero are not real, find the interval which contains π.

So weβve been told that the roots of this quadratic equation, in which π is the constant term, are not real. We need to recall the relationship that exists between the coefficients of a quadratic equation and the type of roots that it has.

Suppose we have the general quadratic equation ππ₯ squared plus ππ₯ plus π is equal to zero. The discriminant of a quadratic equation is the quantity π squared minus four ππ. The value or, more specifically, the sign of the discriminant is what determines the type of roots that the quadratic equation will have.

If the discriminant is strictly positive, then the quadratic equation will have two real and distinct roots. If the value of the discriminant is equal to zero, then the quadratic equation has only one repeated real root. If the value of the discriminant is less than zero, then the quadratic equation has no real roots, which is the situation weβre given in this question.

So we know then that the discriminant of this quadratic must be less than zero. Letβs work out what the discriminant is equal to in terms of π. Comparing the coefficients in our quadratic with the general form, we see that π is equal to 24, π is equal to six, and π is equal to π.

Therefore, the discriminant π squared minus four ππ is equal to six squared minus four multiplied by 24 multiplied by π. This simplifies to 36 minus 96 π. Remember, the roots of this quadratic equation are not real. And so the value of the discriminant is less than zero. Therefore, we have the inequality 36 minus 96 π is less than zero.

In order to find the interval which contains π, we need to solve this inequality for π. The first step is to subtract 36 from each side. This gives negative 96 π is less than negative 36. Next, we need to divide both sides of the inequality by negative 96.

We need to be very careful here. Remember, when we divide an inequality by a negative number, we need to reverse the direction of the inequality. So the less than sign becomes a greater than sign. And we now have that π is greater than negative 36 over negative 96. The negative in the numerator and the negative in the denominator cancel out. And the fraction simplifies to three over eight, by dividing both the numerator and denominator by 12.

We have then that π is greater than three over eight. The question doesnβt ask us to give our answer as an inequality. It asks us to give the interval which contains π. If π must be greater than three over eight, then the set of possible values of π is everything from three over eight to infinity.

As the lower end of the interval is a strict inequality and the upper end is infinity, we can express this as an open interval, which is what the outward-facing square brackets indicate. π belongs in the open interval with end points three over eight and infinity.

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