Find (ƒ – g)(x) where ƒ(x) = –x2, g(x) = –(x + 4)2.

Mathematics

1 answer:

Alchen [17]4 years ago

3 0

(f - g)(x) = f(x) - g(x) = -x^2 - - (x + 4)^2 = -x^2 - -(x^2 + 8x + 16)

= -x^2 + x^2 + 8x + 16

= 8x + 16

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2logx=3-2log(x+3) solve for x

AVprozaik [17]

Answer:

$\large\boxed{x=\dfrac{-3+\sqrt{40+10\sqrt{10}}}{2}}$

Step-by-step explanation:

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$x^2+3x=10^{1\frac{1}{2}}\\\\x^2+3x=10^{1+\frac{1}{2}}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\x^2+3x=10\cdot10^\frac{1}{2}\qquad\text{use}\ \sqrt[n]{a}=a^\frac{1}{n}\\\\x^2+3x=10\sqrt{10}\qquad\text{subtract}\ 10\sqrt{10}\ \text{from both sides}\\\\x^2+3x-10\sqrt{10}=0\\\\\text{Use the quadratic formula}\\\\ax^2+bx+c=0\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\a=1,\ b=3,\ c=-10\sqrt{10}\\\\b^2-4ac=3^2-4(1)(-10\sqrt{10})=9+40\sqrt{10}\\\\x=\dfrac{-3\pm\sqrt{40+10\sqrt{10}}}{2(1)}=\dfrac{-3\pm\sqrt{40+10\sqrt{10}}}{2}\\\\x=\dfrac{-3-\sqrt{10+10\sqrt{10}}}{2}\notin D$