Question Video: Finding a Root of a Quadratic Equation given the Other Root Mathematics

Given that βˆ’10 is a root of the equation 2π‘₯Β² + 13π‘₯ βˆ’ 70 = 0, what is the other root?

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Video Transcript

Given that negative 10 is a root of the equation two π‘₯ squared plus 13π‘₯ minus 70 equals zero, what is the other root?

We’re told that negative 10 is a root of our equation, which means that our quadratic must be equal to zero when π‘₯ is equal to negative 10. Essentially, it’s a solution to the equation two π‘₯ squared plus 13π‘₯ minus 70 equals zero. Now, what this actually means is that π‘₯ plus 10 must be a factor of two π‘₯ squared plus 13π‘₯ minus 70. Two π‘₯ squared plus 13π‘₯ minus 70 can therefore be written as π‘₯ plus 10 times some other binomial.

Now, let’s give that binomial a form. Let’s say it’s in the form π‘Žπ‘₯ plus 𝑏, where π‘Ž and 𝑏 are real constants. What we’re going to do is distribute the parentheses on the right-hand side of this equation and see what we get. We begin by multiplying π‘₯ by π‘Žπ‘₯. That’s π‘Žπ‘₯ squared. We then multiply the outer terms. That’s π‘₯ times 𝑏, which is 𝑏π‘₯. Next, we multiply the inner terms. That’s 10 times π‘Žπ‘₯, which is 10π‘Žπ‘₯. And finally, we multiply 10 by 𝑏, to give us 10𝑏. So we find that this is equal to two π‘₯ squared plus 13π‘₯ minus 70.

And we now use a process called comparing coefficients. We look at the coefficients of our various terms. Let’s begin by comparing our coefficients of π‘₯ squared. On the left-hand side, we have a two. And on the right-hand side, the coefficient of π‘₯ squared is π‘Ž. So when we compare coefficients of π‘₯ squared, we find π‘Ž is equal to two.

We could next compare coefficients of π‘₯. In fact though, we’re going to compare constants. We might say these are the coefficients of the π‘₯ to the power of zero terms. On the left-hand side, our constant is negative 70. And on the right-hand side, we have 10𝑏. So negative 70 equals 10𝑏. And so we’re going to divide by 10 to solve for 𝑏. 𝑏 is therefore equal to negative seven. This means that our quadratic expression can be written as π‘₯ plus 10 times two π‘₯ minus seven. We’ve replaced π‘Ž and 𝑏 with their solutions.

But we know that we’re using this to solve the equation two π‘₯ squared plus 13π‘₯ minus 70 equals zero. We already know that we have one root of negative 10. That’s found by setting π‘₯ plus 10 equal to zero. We’re now going to set two π‘₯ minus seven equal to zero and solve for π‘₯. We’ll add seven to both sides of this equation so that two π‘₯ is equal to seven. And then we’ll divide through by two. So π‘₯ is equal to seven over two or 3.5. And we could check this solution by substituting π‘₯ equals seven over two into our original equation, making sure that it is indeed equal to zero.

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