CPhill

Assuming compound interest and one compounding per year, we have

500( 1 +. 042)³   ≈ £565.68

I believe you want this....  (243)^(1/4).....you can put it straight into the on-site calculator in this form and get your answer....

Nice graphic, Melody......how did you do that ??? (Was it a straight copy ??)

I believe what you have written is known as a "Googol"....We couldn't write this number out, even if we wanted to!!!  (There isn't enough time in our lives to do so, nor physical space in the Universe to do so !!!......)

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Edit......"Googolplex"....thanks, Melody

Let x be the number of motorcycles and y be the number of cars...each car has 4 wheels and each motorcycle has 2...so we have

4x + 2y = 310    we have more variables than equations, so there will be multiple integer solutions

These square root equations can be a little "sticky"..... here we go......

√(2x + 4) - √(x + 1)   =    √( 3x - 19)      first, square both sides....this gives

2x + 4 - 2√[(2x + 4)(x + 1)] + x + 1  = 3x - 19     simplify 

3x + 5 - 2√[(2x + 4)(x + 1)] = 3x - 19    leave the root on the left and move everything else to the right side by subtrating 3x, 5 from both sides....this gives us

- 2√[(2x + 4)(x + 1)] = -24      divide both sides by -2

√[(2x + 4)(x + 1)]  = 12           square both sides

(2x + 4)(x + 1) = 144               simplify

2x^2 + 6x + 4 = 144               subtract 144 from both sides

2x^2 + 6x - 140   = 0              divide through by 2

x^2 + 3x - 70   = 0                  factor

(x + 10) (x - 7) = 0               setting each factor to 0, we have posible solutions of x = -10 and x = 7

Notice that -10 makes all three original roots negative and we can't take a square root of a negative number (this is known as an "extraneous" solution)

Check x = 7 in the original problem....

√(2(7) + 4) - √(7 + 1)   =    √( 3(7) - 19)

√(18) - √(8)   =    √( 2)

3√(2) -  2√(2)   =    √( 2)

√( 2) = √( 2)

Yep...that works right out !!!

Be sure to ask if you have questions....these problems are a little difficult.....

6x^3+38x^2+18x+2=0

x^3+2x-5=x+4    rearrange this as

x^3+2x-5-x-4  = 0

x^3+ x - 9  = 0

Using the on-site solver, we have.....

$${{\mathtt{x}}}^{{\mathtt{3}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{9}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\left({\frac{{\sqrt{{\mathtt{2\,191}}}}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{9}}}{{\mathtt{2}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\times\,}}\left({\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right){\mathtt{\,-\,}}{\frac{\left({\frac{{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\left({\mathtt{3}}{\mathtt{\,\times\,}}{\left({\frac{{\sqrt{{\mathtt{2\,191}}}}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{9}}}{{\mathtt{2}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}\right)}}\\
{\mathtt{x}} = {\left({\frac{{\sqrt{{\mathtt{2\,191}}}}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{9}}}{{\mathtt{2}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\times\,}}\left({\frac{{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right){\mathtt{\,-\,}}{\frac{\left({\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\left({\mathtt{3}}{\mathtt{\,\times\,}}{\left({\frac{{\sqrt{{\mathtt{2\,191}}}}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{9}}}{{\mathtt{2}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}\right)}}\\
{\mathtt{x}} = {\left({\frac{{\sqrt{{\mathtt{2\,191}}}}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{9}}}{{\mathtt{2}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{\left({\mathtt{3}}{\mathtt{\,\times\,}}{\left({\frac{{\sqrt{{\mathtt{2\,191}}}}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{9}}}{{\mathtt{2}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}\right)}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\mathtt{0.960\: \!087\: \!560\: \!673\: \!589\: \!7}}{\mathtt{\,-\,}}{\mathtt{1.940\: \!439\: \!221\: \!537\: \!443\: \!7}}{i}\\
{\mathtt{x}} = {\mathtt{\,-\,}}{\mathtt{0.960\: \!087\: \!560\: \!673\: \!589\: \!7}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1.940\: \!439\: \!221\: \!537\: \!443\: \!7}}{i}\\
{\mathtt{x}} = {\mathtt{1.920\: \!175\: \!121\: \!347\: \!179\: \!5}}\\
\end{array} \right\}$$

Thus, there is only one real solution to this.

111,111 = 3*7*11*13*37