Question Video: Finding an Unknown in a Quadratic Equation Using the Relation between Its Coefficients and Roots Mathematics

The sum of the roots of the equation 4π‘₯Β² + π‘˜π‘₯ βˆ’ 4 = 0 is βˆ’1. Find the value of π‘˜ and the solution set of the equation.

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Video Transcript

The sum of the roots of the equation four π‘₯ squared plus π‘˜π‘₯ minus four equals zero is negative one. Find the value of π‘˜ and the solution set of the equation.

So, we’re looking, first of all, to find the value of π‘˜, which is a missing coefficient in this quadratic equation. It’s the coefficient of π‘₯. We’re going to need to use the quadratic formula to find expressions for the roots of this equation, which will be in terms of π‘˜. And we recall that, for the general quadratic equation π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐 equals zero, its roots are given by π‘₯ equals negative 𝑏 plus or minus the square root of 𝑏 squared minus four π‘Žπ‘ all over two π‘Ž.

Let’s determine the values of π‘Ž, 𝑏, and 𝑐 for this quadratic equation then, which is in the correct form with all the terms on the same side. The value of π‘Ž is the coefficient of π‘₯ squared. So π‘Ž is equal to four. The value of 𝑏 is the coefficient of π‘₯. So 𝑏 is equal to this unknown value π‘˜. And the value of 𝑐 is the constant term, so 𝑐 is equal to negative four. Substituting into the quadratic formula then, we have that π‘₯ is equal to negative π‘˜ plus or minus the square root of π‘˜ squared minus four multiplied by four multiplied by negative four all over two multiplied by four. In the denominator, two times four is eight. And within the square root, four times four times negative four is negative 64. And subtracting negative 64 is the same as adding 64. So, we have negative π‘˜ plus or minus the square root of π‘˜ squared plus 64 all over eight.

Now, we’re told this key piece of information, that the sum of the roots of this equation is negative one. Adding the expressions for our two roots then, and we have the equation on the screen. Now, as both of these fractions have a common denominator of eight, we can actually combine to a single fraction. And what we notice is that there’s actually quite a lot of simplification. We have plus the square root of π‘˜ squared plus 64, and then minus the square root of π‘˜ squared plus 64. So, these two terms will cancel each other out. We’re left with the far simpler equation negative two π‘˜ over eight is equal to negative one, which we can solve by multiplying by eight and dividing by negative two to give π‘˜ equals four.

In fact, this is illustrative of a general and really useful result. When we added our two roots together, the part under the square root canceled out due to them having different signs. So, what we were left with, the contribution from each root to the sum, was just negative 𝑏 over two π‘Ž. We therefore summed two lots of this, giving negative two 𝑏 over two π‘Ž, which simplified to negative 𝑏 over π‘Ž. In our case, we found that π‘˜ equals four. So, the value of 𝑏 in our quadratic equation is four. And so, negative 𝑏 over π‘Ž gives negative four over four, which is negative one, the correct sum for the roots. What this has shown though is that, in general, the sum of the roots of a quadratic equation is equal to negative 𝑏 over π‘Ž. And we can quote this as a general result.

Now that we know the value of π‘˜, we need to determine the values of π‘₯. We have negative four plus or minus the square root of four squared plus 64 over eight. We can simplify the surd here. The square root of 80 is equal to four root five. And then dividing through by a common factor of four gives our simplified solutions as negative one plus or minus the square root of five over two. So, we’ve completed the problem. The value of π‘˜ is four, and the solution set of the equation in surd form is negative one minus root five over two, negative one plus root five over two.

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