Answered

Select the approximate values of x that are solutions to f(x) = 0, where
f(x) = -9x2 + 4x + 9.

Sagot :

[tex]f(x)=-9x^2+4x+9\\\\a=-9;\ b=4;\ c=9\\\\\Delta=b^2-4ac\to\Delta=4^2-4(-9)(9)=16+324=340\\\\x=\frac{-b\pm\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt{340}=\sqrt{4\times85}=2\sqrt{85}\\\\x=\frac{-4\pm2\sqrt{85}}{2(-9)}=\frac{2\pm\sqrt{85}}{9}\\\\x_1\approx1.25\ and\ x_2\approx0.8[/tex]
Kate200468id
[tex]f(x) = -9x^2 + 4x + 9\ \ \ and\ \ \ f(x)=0\\\\-9x^2 + 4x + 9=0\ \ \ \Leftrightarrow\ \ \ 9x^2 - 4x + \frac{4}{9} -\frac{4}{9}- 9=0\\\\(3x-\frac{2}{3})^2=9\frac{4}{9}\ \ \ \Leftrightarrow\ \ \ (3x-\frac{2}{3})^2= \frac{85}{9} \\\\3x-\frac{2}{3}= \frac{ \sqrt{85} }{3} \ \ \ or\ \ \ \ \ \ 3x-\frac{2}{3}= -\frac{ \sqrt{85} }{3}\\\\3x= \frac{2+ \sqrt{85} }{3} \ \ \ \ \ \ or\ \ \ \ \ \ 3x= \frac{2- \sqrt{85} }{3}\\\\ x= \frac{2+ \sqrt{85} }{9} \ \ \ \ \ \ \ \ or\ \ \ \ \ \ \ x= \frac{2- \sqrt{85} }{9}\\\\[/tex]

[tex]x\approx1.247\ \ \ \ \ \ \ \ or\ \ \ \ \ \ \ x\approx-0.802[/tex]

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