Applying quadratic formula to inverse of a function: Calculation problem

In summary, the conversation was about trying to figure out the specific steps in a math textbook that used the quadratic formula to find an inverse of a function. The person was struggling with a step in the simplification algebra and was confused about how certain numbers were divided and how one part of the equation remained unaffected. They eventually realized their mistake and explained the correct method to reach the next step.
  • #1
self4u
2
0
Hi - I'm trying to figure out the specific steps in a math textbook. I'm trying to figure out how the textbook did its algebra here with a quadratic formula to find an inverse of a function:

View attachment 5796

Sadly there's a step I'm supposed to know by now in the simplification algebra in there 3 steps shown...and I'm missing it.

From step 2 to step 3...it looks like only part of the equation is divided by the denominator -10. I didn't think it was possible to divide by 10 with individual numbers under the square root symbol. Furthermore - step 3 shows a reduced ".2(60-h)". How did 20 go to .2? How is (60-h) completely unaffected?

Sorry for such a simple question and thank you in advance for your help!
 

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  • #2
Okay I think I've finally figured it out. I'm a dolt. But since I'm a dolt, I might as well post why I'm a dolt due to the order of operations.

Step 2A: Simply divide every single number by -10. So the method to get to Step 3 becomes:

a) (-30/-10) which becomes positive 3.

b) Then (30sq./-10) which becomes -3sq...then becoming 9.

c) Then (20/-10) which becomes -2.

d) Then (60/-10) which becomes -6. At this point we could multiple it with the (-2) above making positive 12.

e) Then (h/-10) which becomes -h/10. At this point we could multiple it with the (-2) above making -2/10h...simplified to -1/5h...or -.2h.

That gets to step 3...remembering to keep the square root above 21 - .2h.
 
  • #3
\(\displaystyle \begin{align*}\dfrac{-30\pm\sqrt{30^2+20(60-h)}}{-10}&=3\pm\dfrac{\sqrt{30^2+20(60-h)}}{-1\cdot|10|} \\
&=3\pm\dfrac{\sqrt{30^2+20(60-h)}}{-1\sqrt{10^2}} \\
&=3\mp\sqrt{\dfrac{900+20(60-h)}{100}} \\
&=3\mp\sqrt{9+\dfrac15(60-h)}=3\mp\sqrt{3^2+.2(60-h)}\end{align*}\)
 

FAQ: Applying quadratic formula to inverse of a function: Calculation problem

What is the quadratic formula?

The quadratic formula is a mathematical formula used to find the solutions (or roots) of a quadratic equation in the form of ax^2 + bx + c = 0. It is written as x = (-b ± √(b^2 - 4ac)) / 2a.

How is the quadratic formula related to the inverse of a function?

The quadratic formula can be used to find the inverse of a quadratic function by solving for x in terms of y. This means that the inverse function will have x and y switched, and the quadratic formula will be used to solve for x.

Can the quadratic formula be used for any type of function?

No, the quadratic formula can only be used for quadratic functions, which are functions with a degree of 2. It will not work for functions with higher degrees, such as cubic or quartic functions.

What is meant by "calculation problem" when applying the quadratic formula to the inverse of a function?

The term "calculation problem" refers to the difficulty in solving for the inverse of a function using the quadratic formula. This is because the formula involves complex calculations and may result in multiple solutions, making it challenging to determine the correct inverse function.

Are there any alternative methods for finding the inverse of a function instead of using the quadratic formula?

Yes, there are other methods for finding the inverse of a function, such as graphing or using algebraic manipulation. However, the quadratic formula is a commonly used method for finding the inverse of a quadratic function, especially when the function is not easily graphed.

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