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

# Question Video: Finding a Root of a Quadratic Equation given the Other Root Mathematics • First Year of Secondary School

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Given that β10 is a root of the equation 2π₯Β² + 13π₯ β 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?

Weβ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, itβ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, letβs give that binomial a form. Letβs say itβs in the form ππ₯ plus π, where π and π are real constants. What weβ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π₯. Thatβs ππ₯ squared. We then multiply the outer terms. Thatβs π₯ times π, which is ππ₯. Next, we multiply the inner terms. Thatβs 10 times ππ₯, which is 10ππ₯. And finally, we multiply 10 by π, to give us 10π. 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. Letβ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. So when we compare coefficients of π₯ squared, we find π is equal to two.

We could next compare coefficients of π₯. In fact though, weβ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 10π. So negative 70 equals 10π. And so weβre going to divide by 10 to solve for π. π is therefore equal to negative seven. This means that our quadratic expression can be written as π₯ plus 10 times two π₯ minus seven. Weβve replaced π and π with their solutions.

But we know that weβ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. Thatβs found by setting π₯ plus 10 equal to zero. Weβre now going to set two π₯ minus seven equal to zero and solve for π₯. Weβll add seven to both sides of this equation so that two π₯ is equal to seven. And then weβ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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