Question Video: Solving Quadratic Equations Using the Quadratic Formula | Nagwa Question Video: Solving Quadratic Equations Using the Quadratic Formula | Nagwa

Question Video: Solving Quadratic Equations Using the Quadratic Formula Mathematics • First Year of Secondary School

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Find the solution set of the equation −5 − (5/𝑥) = 1/(𝑥²) in ℝ, giving values to one decimal place.

03:12

Video Transcript

Find the solution set of the equation negative five minus five over 𝑥 is equal to one over 𝑥 squared for all real values, giving values to one decimal place.

We will begin by trying to simplify the equation and eliminate the denominators. One way of doing this is to multiply each of our terms by 𝑥 squared. Negative five multiplied by 𝑥 squared is negative five 𝑥 squared. Negative five over 𝑥 multiplied by 𝑥 squared is negative five 𝑥 squared over 𝑥, which simplifies to negative five 𝑥. Finally, one over 𝑥 squared multiplied by 𝑥 squared is equal to one.

We will then add five 𝑥 squared and five 𝑥 to both sides of our equation. This gives us the quadratic equation zero is equal to five 𝑥 squared plus five 𝑥 plus one. Our equation is now written in the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 is equal to zero. And we can, therefore, use the quadratic formula. This states that 𝑥 is equal to negative 𝑏 plus or minus the square root of 𝑏 squared minus four 𝑎𝑐 all divided by two 𝑎. The positive and negative signs will give us two solutions.

In our equation, 𝑎 is equal to five, 𝑏 is equal to five, and 𝑐 is equal to one. These are the coefficient of 𝑥 squared, the coefficient of 𝑥, and the constant, respectively. 𝑥 is, therefore, equal to negative five plus or minus the square root of five squared minus four multiplied by five multiplied by one all divided by two multiplied by five. This simplifies to 𝑥 is equal to negative five plus or minus root five all divided by 10.

We have two solutions: 𝑥 is equal to negative five plus root five over 10 and 𝑥 is equal to negative five minus root five over 10. Typing these into the calculator, we get 𝑥 is equal to negative 0.2763 and so on and negative 0.7236 and so on. We want our answers to one decimal place. Therefore, 𝑥 is equal to negative 0.3 and negative 0.7. This can be written using set notation. The solution set of the equation negative five minus five over 𝑥 is equal to one over 𝑥 squared is negative 0.3 and negative 0.7.

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