how do i do this? x^2(x^2-7)+12
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Dec 11, 2005 #2 Hello, yasaminG! I assume that it's supposed to be an <u>equation</u>. how do i do this? x2(x2 − 7) + 12\displaystyle \;\;x^2(x^2\,-\,7)\,+\,12x2(x2−7)+12 Click to expand... We have: x4 − 7x2 +12 = 0\displaystyle \;x^4\,-\,7x^2\,+12\;=\;0x4−7x2+12=0 Factor: (x2 − 4)(x2 − 3) = 0\displaystyle \;(x^2\,-\,4)(x^2\,-\,3)\;=\;0(x2−4)(x2−3)=0 Then: x2 −4 = 0 ⇒ x2 = 4 ⇒ x = ±2\displaystyle \;x^2\,-4\:=\:0\;\;\Rightarrow\;\;x^2\,=\,4\;\;\Rightarrow\;\;x\,=\,\pm2x2−4=0⇒x2=4⇒x=±2 And: x2 − 3 = 0 ⇒ x2 = 3 ⇒ x = ±3\displaystyle \;x^2\,-\,3\:=\:0\;\;\Rightarrow\;\;x^2\,=\,3\;\;\Rightarrow\;\;x\,=\,\pm\sqrt{3}x2−3=0⇒x2=3⇒x=±3
Hello, yasaminG! I assume that it's supposed to be an <u>equation</u>. how do i do this? x2(x2 − 7) + 12\displaystyle \;\;x^2(x^2\,-\,7)\,+\,12x2(x2−7)+12 Click to expand... We have: x4 − 7x2 +12 = 0\displaystyle \;x^4\,-\,7x^2\,+12\;=\;0x4−7x2+12=0 Factor: (x2 − 4)(x2 − 3) = 0\displaystyle \;(x^2\,-\,4)(x^2\,-\,3)\;=\;0(x2−4)(x2−3)=0 Then: x2 −4 = 0 ⇒ x2 = 4 ⇒ x = ±2\displaystyle \;x^2\,-4\:=\:0\;\;\Rightarrow\;\;x^2\,=\,4\;\;\Rightarrow\;\;x\,=\,\pm2x2−4=0⇒x2=4⇒x=±2 And: x2 − 3 = 0 ⇒ x2 = 3 ⇒ x = ±3\displaystyle \;x^2\,-\,3\:=\:0\;\;\Rightarrow\;\;x^2\,=\,3\;\;\Rightarrow\;\;x\,=\,\pm\sqrt{3}x2−3=0⇒x2=3⇒x=±3
