Simple Trick!

Algebra Level 4

Let $a$ be a real number and let $(4a+3)^4+(4a-1)^4=322$ . If the positive solution for $a$ can be expressed as $\dfrac{n\sqrt{k}-x}{y}$ , with $n$ , $k$ , $x$ and $y$ all co-prime positive integers, find $(n+k)^{y-x}$

The answer is 216.

2 solutions

Clive Chen
Apr 2, 2016

Instead of expanding first, we can substitute, and make $b=4a+1$ . Our new equation is $(b+2)^4 + (b-2)^4=322$ . The odd powers of $b$ cancel out, and we are left with $2b^4 + 48b^2+32=322$ , so $b^4 + 24b^2+16=161$ . We can complete the square, and say $(b^2+12)^2-144+16=161$ , simplifying to $(b^2+12)^2=289$ . We now get $b^2+12=\pm 17$ . However, we can only take the positive root as $b$ is real, and we get $b=\sqrt{5}$ , so if we plug $b=4a+1$ back in, we get $a=\frac{\sqrt{5}-1}{4}$ $n=1, k=5, x=1, y=4$ , so $(n+k)^{y-x}$ = $6^3= \boxed{216}$

What do you mean about the odd power of b cancel out. Why it can ?

Daniel Sugihantoro - 5 years, 2 months ago

When we expand, we choose one term from each parentheses and multiply the terms together. If we pick an odd number of b's, we also pick and odd number of 2's and -2's. For each odd power of b, when we expand, the coefficients will be additive inverses, so we can cancel them out.

Clive Chen - 5 years, 2 months ago

Did you mean to write $\boxed{216}$ ?

Ralph James - 5 years, 2 months ago

Yes thanks.

Clive Chen - 5 years, 2 months ago

William Isoroku
Apr 6, 2016

Let $x=4a+1$

Simple Trick!

The answer is 216.

2 solutions

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