Solve. The sum of the squares of two numbers is 20 . Their product is 8. Find the two numbers.

Name: Problem 44, Chapter 13: Beginning and Intermediate Algebra
Uploaded: 2020-04-09T16:21:27-08:00
Duration: 5 min 14 s
Channel: Julie Silva
Description: Solve. The sum of the squares of two numbers is 20 . Their product is 8. Find the two numbers.

Solve. The sum of the squares of two numbers is 20 . Their...

Key Concepts

Algebraic Identities

Algebraic identities, such as (a + b)² = a² + 2ab, play a crucial role in transforming and simplifying equations. These identities allow one to express combinations of terms, like sums of squares, in alternative forms that can reveal relationships between variables which are not initially obvious.

Quadratic Equations and Vieta's Formulas

When two numbers are related by their sum and product, they can be seen as the roots of a quadratic equation. Vieta's formulas relate the sum and product of the roots to the coefficients of the quadratic equation, providing a systematic approach to reconstructing and solving for the unknowns in problems involving symmetric sums.

Transcript

00:01 In this problem, we're told the sum of the squares of two numbers is 20, and that their product is 8.

00:06 We want to find the two numbers.

00:08 Well, because we don't know the numbers, i'm going to call x the first number, and i'm going to get y be equal to the second number.

00:20 So now we want to write a system of equations.

00:22 Well, the first sentence says that the sum of the squares of the two numbers is 20, meaning when i do x squared plus y squared, that's the sum of their squares, it should equal to 20.

00:34 Now, the second thing that we were told is that their product is 8.

00:38 Well, remember, product means to multiply.

00:41 So that means x times y is equal to 8.

00:46 So now what we have to do is we have to solve the system of equations.

00:50 Well, we can do this using the substitution method.

00:54 So to do that, what i'm going to do is look at that second equation.

01:00 And what i'm going to do is get y by itself.

01:03 So to do that, i'm going to divide x on both.

01:06 Sets.

01:08 So what we have is y is equal to 8 divided by x.

01:13 So now that i know what y is, all i need to do is substitute 8 over x in place of y squared in that first equation.

01:21 So i'm going to get x squared plus the quantity of 8 divided by x squared squared is equal to 20.

01:31 All right.

01:32 Well, first we have to square 8 over x.

01:34 Well, 8 squared is 64, and x squared is just x squared.

01:39 So we have x squared plus 64 over x squared is equal to 20.

01:43 Now, i'm not going to want to have that fraction, so i'm going to multiply both sides of our equation by x squared.

01:53 Well, x squared times x squared is x to the fourth.

01:57 When we multiply x squared by 64 over x squared, the x squared are going to cancel.

02:01 So we just have plus 64, and then it's equal to, well, x squared times 20 is going to equal to 20x squared.

02:12 So now i'm going to set this equal to zero.

02:15 So in other words, i'm going to subtract 20 x squared from both sides of my equation.

02:22 So what i'm left with is x to the 4 minus 20x squared plus 64, and that's equal to zero.

02:31 So now i can go ahead and factor this.

02:34 So the only difference when we factor when we have that x to the fourth term at the beginning is that both of our binomials will start with x squares.

02:44 So now i have to simply think of what multiplies to 64, but we'll add to negative 20.

02:50 Well, that's going to be negative 16 and negative 4...

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