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One positive integer is 4 less than twice another. The sum of their squares is 544. Find the integers.

User Owen Zhao
by
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1 Answer

2 votes

Taking x and y as the integers.

Let's say that x is 4 less than twice y. This is:


x=2y-4

The sum of the squares of x and y is 544, this is:


x^2+y^2=544

Use the first equation and replace this expression for the value of x in the second equation:


\begin{gathered} (2y-4)^2+y^2=544 \\ 4y^2-16y+16+y^2=544 \\ 5y^2-16y+16-544=0 \\ 5y^2-16y-528=0 \end{gathered}

Solve the quadratic equation


\begin{gathered} y=\frac{16\pm\sqrt[]{16^2-4(5\cdot-528)}}{2\cdot5} \\ y=(16\pm104)/(10) \\ y1=(16+104)/(10)=(120)/(10)=12 \\ y2=(16-104)/(10)=-(44)/(5) \end{gathered}

y can take 2 values, but, as we know, x and y must be positive integers, the value of y that meets this condition is 12. It means one of the integer is 12.

Use this value and the first equation to find the second integer:


\begin{gathered} x=2y-4 \\ x=2\cdot12-4 \\ x=24-4 \\ x=20 \end{gathered}

The integers are 20 and 12.

User Wcyn
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