Video Transcript
Find the solution set of two π¦ plus four all squared plus π¦ plus two all squared equals five in the set of real numbers.
We will begin by distributing the parentheses or expanding the brackets on the left-hand side. We can do this using the FOIL method as two π¦ plus four all squared is equal to two π¦ plus four multiplied by two π¦ plus four. Multiplying the first terms gives us four π¦ squared. The outer terms have a product of eight π¦. The inner terms also have a product of eight π¦. Finally, multiplying the last terms gives us 16. Collecting like terms, two π¦ plus four all squared is equal to four π¦ squared plus 16π¦ plus 16.
We can repeat this process when squaring π¦ plus two. This is equal to π¦ squared plus two π¦ plus two π¦ plus four. This can then be simplified to π¦ squared plus four π¦ plus four. We now have the quadratic equation four π¦ squared plus 16π¦ plus 16 plus π¦ squared plus four π¦ plus four is equal to five. Collecting like terms on the left-hand side gives us five π¦ squared plus 20π¦ plus 20. We can then subtract five from both sides of this equation so that our quadratic equation is equal to zero.
The numbers five, 20, and 15 are all divisible by five. This means that we can divide our equation by five, giving us π¦ squared plus four π¦ plus three is equal to zero. We can now solve this quadratic by factoring. As the coefficient of π¦ squared is one, the first term in each of our parentheses will be π¦. Next, we need to find two integers that have a product of three and a sum of four.
One plus three is equal to four, and one multiplied by three is equal to three. The quadratic factors into the two sets of parentheses π¦ plus one and π¦ plus three. As these have a product equal to zero, either π¦ plus one equals zero or π¦ plus three equals zero. This gives us two possible solutions: negative one and negative three. The solution set of two π¦ plus four all squared plus π¦ plus two all squared equals five are negative one and negative three.
