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Question Video: Finding an Unknown in a Quadratic Equation Given That Its Roots Are Equal Mathematics • 9th Grade

Given that the roots of the equation βˆ’18π‘₯Β² + 3π‘˜π‘₯ βˆ’ 72 = 0 are equal, determine all possible values of π‘˜. For each value of π‘˜, work out the roots of the equation.

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Video Transcript

Given that the roots of the equation negative 18π‘₯ squared plus three π‘˜π‘₯ minus 72 equals zero are equal, determine all possible values of π‘˜. For each value of π‘˜, work out the roots of the equation.

Now, in fact, the equation given here is a quadratic equation. So let’s recall what we know about quadratic equations that can give us information about its roots. Suppose we have an equation of the form π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐 equals zero. For this equation to be quadratic of course, π‘Ž, the coefficient of π‘₯ squared, cannot be equal to zero. The discriminant of this equation, which we can represent using this Ξ” symbol, is given by 𝑏 squared minus four π‘Žπ‘. In other words, we square the coefficient of π‘₯. Then we subtract four times the coefficient of π‘₯ squared times the constant.

Because this discriminant is derived from the quadratic formula, where we find its square root, the discriminant itself can tell us information about the number of roots. For instance, if this value is positive, then there are two real roots to the equation. If it’s equal to zero, and assuming our values for π‘Ž, 𝑏, and 𝑐 are real, then we have one real root. If it’s negative, there are no real roots.

Now, we’re told that the roots of the equation are equal. In other words, essentially, we have one root. So let’s find the discriminant of our equation and set it equal to zero. We can see that π‘Ž, the coefficient of π‘₯ squared, is negative 18, 𝑏 is three π‘˜, and 𝑐 is negative 72. Then, the discriminant of our equation is three π‘˜ squared minus four times negative 18 times negative 72. This simplifies to nine π‘˜ squared minus 5184. And of course this is equal to zero since our roots are equal. Essentially, there’s one root.

Let’s now solve this equation for π‘˜. We begin by adding 5184 to both sides of this equation. So nine π‘˜ squared equals 5184. Then, dividing through by nine, and we find π‘˜ squared is 576. To solve this equation for π‘˜, we’ll take the square root of both sides, remembering of course to take both the positive and negative square root of 576. This means π‘˜ is equal to the positive or negative square root of 576. And that’s positive or negative 24. So those are the two possible values of π‘˜ that ensure that the roots of our equation are equal.

Now that we have those, let’s work out their corresponding roots. Given the quadratic equation, its roots are π‘₯ equals negative 𝑏 plus or minus the square root of the discriminant all over two π‘Ž. Now, of course, our discriminant is zero and the square root of zero is also zero. So the single root or the repeated root is given by negative 𝑏 over two π‘Ž. And of course there are two scenarios. We’ve said π‘˜ could be positive or negative 24. Now, if π‘˜ is 24, our value for 𝑏, which was three π‘˜, is three times 24, which is 72. And if π‘˜ is negative 24, 𝑏 is negative 72.

Since π‘₯ is negative 𝑏 over two π‘Ž, when π‘˜ is 24, π‘₯ is negative 72 over two times negative 18. And that’s equal to two. In a similar way, when 𝑏 is negative 72, π‘₯ is negative negative 72 over two times negative 18, which is negative two. So when π‘˜ is 24, we have a repeated root π‘₯ equals two. And when π‘˜ is negative 24, we have a repeated root negative two. And we can write this as shown. When π‘˜ is 24, the roots are two and two. And when π‘˜ is negative 24, the roots are negative two and negative two.

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