Quadratics using Pascal's triangle

In summary, the conversation discusses a problem involving a quadratic equation and the use of Pascal's Triangle. The problem involves simplifying an expression and using the quadratic formula, which resulted in a slightly different answer. The conversation ends with a minor gripe about the use of symbols in the problem.
  • #1
HorseRidingTic
6
0

Homework Statement


upload_2017-5-18_10-28-50.png

The problem equation is contained in the picture.

Homework Equations


Pascal's Triangle is useful is this one.

The Attempt at a Solution



The difficulty I'm having is in going between lines 2 and 3 which I've marked with a little red dot.

upload_2017-5-18_10-31-50.png


The closest I get to simplifying it is = a4 + B4 + 4aB(a2+B2) + 6a2B2 . From there I can't figure out the way in which to reduce it further.

P.S I also used the quadratic formula to solve this one (the one with the b2 - 4ac) and my answer came to 748.52 but not quite 752. Why does the quadratic formula not work here?

Thank you for your help,
Ben
 

Attachments

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  • #2
##\alpha^4 + \beta^4 + 4\alpha^3\beta + 6\alpha^2\beta^2 + 4\alpha \beta^3 = \alpha^4 + \beta^4 + 4\alpha^3\beta + \color{red}{8}\alpha^2\beta^2 + 4\alpha \beta^3 - \color{blue}{2\alpha^2 \beta^2} = \alpha^4 + \beta^4 + 4\alpha\beta(\alpha^2 + 2\alpha\beta + \beta^2) - {2\alpha^2 \beta^2}##

Now use ##a^2 + 2ab + b^2 = (a+b)^2##
 
  • Like
Likes HorseRidingTic
  • #3
Amazing! Thank you so much Buffu :)
 
  • #4
:smile:
 
  • #5
I have a minor gripe with the author of this problem.
A quadratic equation with roots ##\alpha## and ##\beta## is ##(x - \alpha)(x - \beta)##, and so ...
What is shown is not an equation, since the symbol = is not present.

Again, my gripe is with the author, not the person who started this thread.
 
  • #6
Number of people confusing between expression and equation is surprisingly high.
 

Related to Quadratics using Pascal's triangle

1. What is Pascal's triangle?

Pascal's triangle is a triangular array of numbers where each number is the sum of two numbers directly above it. It is named after the French mathematician Blaise Pascal and has many applications in mathematics, including solving quadratic equations.

2. How can Pascal's triangle be used to solve quadratics?

Pascal's triangle can be used to find the coefficients of a quadratic equation in the form ax^2 + bx + c. The first row of the triangle represents the coefficients of x^2, the second row represents the coefficients of x, and the third row represents the constant term. By using the numbers in the triangle, one can easily determine the values of a, b, and c.

3. Can Pascal's triangle be used for all types of quadratic equations?

Yes, Pascal's triangle can be used to solve all types of quadratic equations, including those with irrational or complex solutions. The triangle can also be extended to higher powers of x, allowing it to be used for polynomials of any degree.

4. What are the benefits of using Pascal's triangle to solve quadratics?

Using Pascal's triangle to solve quadratics can save time and effort compared to other methods, such as factoring or the quadratic formula. It also provides a visual representation of the coefficients, making it easier to understand the relationship between the terms in the equation.

5. Are there any limitations to using Pascal's triangle for quadratics?

While Pascal's triangle is a useful tool for solving quadratics, it can become cumbersome for equations with large coefficients. In these cases, other methods may be more efficient. Additionally, Pascal's triangle can only be used for equations in standard form, so it may not be applicable for equations in other forms, such as vertex form.

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