Question Video: Solving Quadratic Equations by Factoring | Nagwa Question Video: Solving Quadratic Equations by Factoring | Nagwa

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Question Video: Solving Quadratic Equations by Factoring Mathematics • First Year of Secondary School

Find the solution set of (𝑥 + 9)² = (𝑥 + 9) in ℝ.

02:30

Video Transcript

Find the solution set of 𝑥 plus nine all squared is equal to 𝑥 plus nine in the set of real numbers.

There are many ways of approaching this problem. We might notice by inspection that if we substitute 𝑥 equals negative nine into both sides of the equation, we get an answer of zero. Whilst this will be one solution to our equation, we need to ensure we have all the real solutions. We can do this by firstly expanding our parentheses. 𝑥 plus nine all squared is equal to 𝑥 plus nine multiplied by 𝑥 plus nine. We can distribute the parentheses here by using the FOIL method. This gives us 𝑥 squared plus nine 𝑥 plus nine 𝑥 plus 81.

Collecting like terms, this simplifies to 𝑥 squared plus 18𝑥 plus 81. We know this is equal to 𝑥 plus nine. Subtracting 𝑥 and subtracting nine from both sides gives us 𝑥 squared plus 17𝑥 plus 72 is equal to zero. This quadratic can be solved by factoring. As the coefficient of 𝑥 squared is equal to one, the first term in each of our parentheses will be 𝑥. We then need to find two integers that have a product of 72 and a sum of 17. Nine times eight is equal to 72, and nine plus eight is equal to 17. 𝑥 squared plus 17𝑥 plus 72 is equal to 𝑥 plus nine multiplied by 𝑥 plus eight.

As the product of our parentheses equals zero, either 𝑥 plus nine equals zero or 𝑥 plus eight equals zero. This gives us two solutions negative nine and negative eight. When 𝑥 is equal to negative eight, both sides of the equation equal one. And when 𝑥 is equal to negative nine, both sides of our equation equal zero. The solution set of 𝑥 plus nine all squared equals 𝑥 plus nine is negative eight and negative nine.

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