# Quadratics using Pascal's triangle

1. May 18, 2017

### HorseRidingTic

1. The problem statement, all variables and given/known data

The problem equation is contained in the picture.

2. Relevant equations
Pascal's Triangle is useful is this one.

3. 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.

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

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2. May 18, 2017

### Buffu

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

Now use $a^2 + 2ab + b^2 = (a+b)^2$

3. May 18, 2017

### HorseRidingTic

Amazing! Thank you so much Buffu :)

4. May 18, 2017

### Buffu

5. May 18, 2017

### Staff: Mentor

I have a minor gripe with the author of this problem.
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. May 18, 2017

### Buffu

Number of people confusing between expression and equation is surprisingly high.

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