$|x|=a\iff x=a\ or\ x=-a\\==============================\\|x+3|=12\iff x+3=12\ or\ x+3=-12\ \ \ \ |subtract\ 3\ from\ both\ sides\\\\\boxed{x=9\ or\ x=-15}$

5 0

3 years ago

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the product of two consecutive integers n and n+1 is 42. what is the positive integer that satisfies the situation?

MrMuchimi

$n(n+1)=42 \\
n^2+n=42 \\
n^2+n-42=0 \\ \\
a=1 \\ b=1 \\ c=-42 \\ b^2-4ac=1^2-4 \times 1 \times (-42)=1+168=169 \\ \\
n=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-1 \pm \sqrt{169}}{2 \times 1}=\frac{-1 \pm 13}{2} \\
n=\frac{-1 -13}{2} \ \lor \ n=\frac{-1+13}{2} \\
n=\frac{-14}{2} \ \lor \ n=\frac{12}{2} \\
n=-7 \ \lor \ n=6$

The positive integer is 6.
The consecutive integers are 6 and 7.

8 0

3 years ago