4 times a number is 21 less than the square of that number. Find the positive solution

1 answer:

8 0

set up the equation 4x + 21 = x^2. getting everything on one side so you have x^2-4x-21= 0. find two numbers that multiply to -21 and add to -4, which are 3 and -7. substitute in with them as x to get x^2+3x-7x+21 = 0. factor out the x and the 7 to get x(x+3)-7(x+3) = 21. so x = -3 and 7. the positive solution would be 7.

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$\frac{2 m}{2 m+3} -\frac{2 m}{2 m-3}=1\\\\ 2 m[\frac{1}{2 m+3} -\frac{1}{2 m-3}]=1\\\\ 2 m\times \frac{[2 m-3 -2 m- 3]}{4m^2-9}=1\\\\ -6 \times 2 m=4 m^2 -9\\\\ 4 m^2 +1 2 m -9=0\\\\m=\frac{-12 \pm\sqrt{12^2-4 \times 4 \times (-9)}}{2\times 4}\\\\m=\frac{-12 \pm \sqrt {144+144}}{8}\\\\m=\frac{-12 \pm \sqrt {288}}{8}\\\\m=\frac{-12 \pm 12 \sqrt{2}}{8}\\\\m=\frac{3}{2}\times(-1 \pm \sqrt{2})$