Moreover, what "process" does it represent? What is the name of the "division" of mathematics this kind of problem belongs to?
For example, 2 + 3 would be called a simple addition problem.
Thanks!
For example, 2 + 3 would be called a simple addition problem.
Thanks!
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This type of problem is an equation with one variable, and it comes up in algebra.
x + 1/x = 7
Multiply through by x:
x^2 + 1 = 7x
Subtract 7x from each side:
x^2 - 7x + 1 = 0
Now this is a quadratic equation, which you can solve by any means. Factoring, quadratic formula, completing the square, etc. I will complete the square:
x^2 - 7x = -1
x^2 - 7x + 49/4 = 45/4
(x - 7/2)^2 = 45/4
x - 7/2 = +/- sqrt(45/4)
x - 7/2 = +/- 3sqrt(5)/2
x = (7 + 3sqrt(5)) / 2 and x = (7 - 3sqrt(5)) / 2
x + 1/x = 7
Multiply through by x:
x^2 + 1 = 7x
Subtract 7x from each side:
x^2 - 7x + 1 = 0
Now this is a quadratic equation, which you can solve by any means. Factoring, quadratic formula, completing the square, etc. I will complete the square:
x^2 - 7x = -1
x^2 - 7x + 49/4 = 45/4
(x - 7/2)^2 = 45/4
x - 7/2 = +/- sqrt(45/4)
x - 7/2 = +/- 3sqrt(5)/2
x = (7 + 3sqrt(5)) / 2 and x = (7 - 3sqrt(5)) / 2
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This is a beginning Algebra problem. An equation of degree one (the variable has a power of one) is typically called a linear equation. An equation of degree two is a quadratic equation. your equation is a quadratic equation and could also be written as x² + 1 = 7x or x² - 7x + 1 = 0. These equations are studied in second year algebra and in geometry.
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The process is "estimate and check." You know that one answer is between x = 6.8 and x = 6.9 and a second answer is between x = 0.1 and x = 0.2 because 6.85 + 1/6.85 is approximately equal to 7 and so is 0.15 + 1/0.15.
It is also possible to solve the problem by finding the zero's of the quadratic equation x^2 - 7*x + 1 = 0
(Algebra)
It is also possible to solve the problem by finding the zero's of the quadratic equation x^2 - 7*x + 1 = 0
(Algebra)
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you need to get everything under a common denominator.for this you have a 1 and an x factor as denominators so the LCD is x. Multiply x and 7 by x/x
(x^2+1-7)/x =0
(x^2-6)/x=0
you only need the top,the denominator means nothing until you check your answers.
x^2=6
x=+- sqrt 6.Both work and make this equal 0.
(x^2+1-7)/x =0
(x^2-6)/x=0
you only need the top,the denominator means nothing until you check your answers.
x^2=6
x=+- sqrt 6.Both work and make this equal 0.
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multiply both sides by x to get rid of denominator
x^2 +1 =7x
x^2 -7x +1 =0
solve with quadratic formula
unsure of the name.. this is getting rid of denominator by multiplying
x^2 +1 =7x
x^2 -7x +1 =0
solve with quadratic formula
unsure of the name.. this is getting rid of denominator by multiplying
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Algebra. Solving an equation in one variable.
answer is x = 7/2 ± 3√(5)/2
answer is x = 7/2 ± 3√(5)/2
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simple algebra problem
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infinite possibilities
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x+1/x=7
x+1=7x
6x=1
x=1/6
x+1=7x
6x=1
x=1/6