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Types of equations in practice – Part 2

Solving other exercises about use of equations
11.4
Hello. Let us consider now the fourth exercise of the section
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introduction: types of equations. We wanted to understand where the graphs of two functions intersect. The two functions are f, which sends x to 6 times x plus 2, and g, which sends x to x squared plus 9. First of all, let us observe that these two functions are defined on all the real numbers. Therefore their domains are equal, and both are equal to the set of all real numbers. Now, which are the points of the graph of f? OK, a generic point of the graph of f has the following shape– x, f of x, where x is any real number. That is point of the type x 6 times x plus 2.
86
And the points of the graph of g have the following shape– x, g of x, which is equal to x, and x squared plus 9. What means to intersect the two graphs? I am looking for points in the plane of the following shapes– of both the following shapes. Therefore, I am looking for real numbers– x– such that 6 times x plus 2 is equal to x squared plus 9. That is, we get the equation 6 times x plus 2 equal to x squared plus 9. OK, and this equation is equivalent to what? OK, we just rearrange the terms, and we get x squared minus 6 times x, and we get plus 7 equal 0.
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OK, this is an equation of second degree. Let us compute its discriminant, delta. Then it is equal to what? To the square of the minus 6, which is 36, minus 4 times the product of the leading coefficient, which is 1, by the constant terms, which is 7.
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Then we get 36 minus 28, which is 8. Observe that the discriminant is greater than 0. What it means? It means that this equation has two different solutions. And in terms of our exercise, what means? It means that we will find two different points of intersection between the two graphs.
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Let us solve the equation. We have that the solutions of our equation are minus minus 6, which is 6, plus or minus the square root of the discriminant divided by 2 times the leading coefficient, and this is equal to what? OK, 6 divided by 2 is 3, plus or minus. And now we have the square root of 8. The square root of 8 is exactly like to write– the square root of 4 times 2. And we have to divide by 2. Now, this is equal to what? 3 plus minus, and the square root of 4 is 2. Therefore, we can simplify the square root of 4 with 2, and we remain with the square root of 2.
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OK, therefore, what we can conclude– that the two points of intersections between the two graphs are which ones? The first is the point 3 plus the square root of 2 and f of 3 plus square root of 2, which is equal to g of 3 plus the square root of 2. They are equal exactly by the computations that we have done. And the point 3 minus square root of 2, f over 3 minus square root of 2, which is equal to g of 3 minus square root of 2. OK, let us conclude computing these terms. We get 3 plus the square root of 2 here.
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And now the image of 3 plus the square root of 2 through f is 6 times 3 plus square root of 2 plus 2, which is 18 plus 2, which is 20, plus 6 times square root of 2. And here we have 3 minus square root of 2 and the image of 3 minus square root of 2 through f is 6 times 3 minus square root of 2 plus 2, which is again equal to 20 minus 6 times square root of 2. Good. These two are the points in common between the two graphs of our functions. Ciao.

The following exercise is solved in this step.

We invite you to try to solve it before watching the video.

In any case, you will find below a PDF file with the solution.

Exercise 4.

Where do the graphs of the functions (xmapsto f(x)=6x+2) and (xmapsto f(x)=x^2+9) intersect?

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Precalculus: the Mathematics of Numbers, Functions and Equations

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