Question Video: Finding the Solution Set of Quadratic Equations Involving Proportion Mathematics • 7th Grade

Given that 𝑦/𝑧 = 𝑧/𝑛 = 𝑛/π‘š = 4/11, determine the solution set of the equation 𝑦π‘₯Β² βˆ’ 2𝑧π‘₯ + 𝑛 = 0.


Video Transcript

Given that 𝑦 over 𝑧 equals 𝑧 over 𝑛 equals 𝑛 over π‘š, which equals four elevenths, determine the solution set of the equation 𝑦π‘₯ squared minus two 𝑧π‘₯ plus 𝑛 equals zero.

Now, whilst it might not look like it, we actually have a quadratic equation here. 𝑦, negative two 𝑧, and 𝑛 are real constants. And so if we can find the values of these constants using the information about the relationship between 𝑦, 𝑧, 𝑛, and π‘š, we’ll be able to solve the quadratic equation by either factoring or using the quadratic formula.

So here is the relation we’ve been given. We’re told that 𝑦 and 𝑧 are in the same proportion to one another as 𝑧 and 𝑛 such that 𝑦 over 𝑧 equals four elevenths as does 𝑧 over 𝑛. Now, in fact, we also know that the same relationship holds for 𝑛 over π‘š. But we’re not given any information about π‘š in the rest of the question. So we’re going to ignore this.

Instead, we’ll use these two equations to find values for 𝑦, negative two 𝑧, and 𝑛. In fact, consider the first equation. Let’s make 𝑦 the subject by multiplying both sides by 𝑧. So 𝑦 equals four elevenths 𝑧. We’re going to make 𝑛 the subject in our second equation. To do so, we begin by multiplying by 𝑛 and then dividing both sides by four elevenths. When we divide by a fraction, of course, we multiply by the reciprocal of that fraction. So 𝑧 divided by four over 11 is 11 over four 𝑧.

Now, we see that we have expressions for 𝑦 and 𝑛 in terms of 𝑧. And this is useful because if we substitute them into our quadratic equation, we will get an equation purely in terms of π‘₯ and 𝑧. So let’s see what that looks like. 𝑦π‘₯ squared becomes four elevenths 𝑧π‘₯ squared. And we replace 𝑛 with 11 over four 𝑧. So our quadratic equation is four elevenths 𝑧π‘₯ squared minus two 𝑧π‘₯ plus 11 over four 𝑧 equals zero.

Now, we know that 𝑧 itself cannot be equal to zero since when we divide some number 𝑦 by 𝑧, we get a constant four elevenths. This means we can divide this entire equation by 𝑧. And so we have a quadratic equation four elevenths π‘₯ squared minus two π‘₯ plus 11 over four equals zero. To get rid of the fractions here, we can multiply through by the lowest common multiple of 11 and 44, and that’s 44. Four elevenths times 44 becomes four times four, which is 16. Negative two times 44 is negative 88. And 11 over four times 44 is the same as 11 times 11, which is 121. And this of course is all equal to zero.

We can then factor the expression on the left-hand side. And we do so by noting that both 16π‘₯ squared and 121 are square numbers. So we might deduce that each expression contains four π‘₯ and 11. In fact, if we have four π‘₯ minus 11 times four π‘₯ minus 11, we get 16π‘₯ squared and 121 as required. But four π‘₯ times negative 11 plus negative 11 times four π‘₯ gives us negative 88π‘₯. So our quadratic equation can now be written as four π‘₯ minus 11 squared equals zero.

Let’s solve for π‘₯. Since we have zero on the right-hand side, we can simply take the square root. So four π‘₯ minus 11 equals zero. Adding 11 to both sides, we get four π‘₯ equals 11. And finally, dividing through by four, and we get π‘₯ is equal to 11 over four.

And so there is simply one solution to the equation. Using set notation, we say it’s the set containing the single element 11 over four.

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