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What are the zeros of 7x^2 -- 144 = --x^4 

please give an explanation and show the work. thank you!


Sagot :

Add x^4 to the left side and factor.
x^4 + 7x^2 - 144 = 0 then factor
(x^2 + 16)(x^2 - 9) = 0 set both factors = 0
x^2 + 16 = 0
x^2 = -16 square root each side
x = + and - 4i
x^2 - 9 = 0
x^2 = 9 square root each side
x = + and - 3
[tex]7x^2-144=-x^4\\\\x^4+7x^2-144=0\\\\(x^2)^2+7x^2-144=0\\\\substitute:t=x^2\geq0\\\\t^2+7t-144=0\\\\\Delta=7^2-4\cdot1\cdot(-144)=49+576=625;\ \sqrt\Delta=\sqrt{625}=25\\\\t_1=\frac{-7-25}{2\cdot1}=\frac{-32}{2}=-16 < 0;\ t_2=\frac{-7+25}{2\cdot1}=\frac{18}{2}=9\\\\x^2=9\iff x=\pm\sqrt9\to x=-3\ \vee\ x=3[/tex]
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