Welcome to Westonci.ca, your one-stop destination for finding answers to all your questions. Join our expert community now! Join our Q&A platform and connect with professionals ready to provide precise answers to your questions in various areas. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
[tex]2x^2+4x-16=0\ \ \ /:2\\\\x^2+2x-8=0\\\\\underbrace{x^2+2x\cdot1+1^2}_{(*)}-1^2-8=0\\\\(x+1)^2-1-8=0\\\\(x+1)^2-9=0\\\\(x+1)^2=9\iff x+1=-3\ or\ x+1=3\\\\x=-3-1\ or\ x=3-1\\\\x=-4\ or\ x=2\\\\\\(*)\ (a+b)^2=a^2+2ab+b^2[/tex]
[tex]2x^2+4x-16=0\ \ \ /:2\\\\x^2+2x-8=0\\\\a=1;\ b=2;\ c=-8\\\\\Delta=b^2-4ac\ if\ \Delta > 0\ then\ x_1=\frac{-b-\sqrt\Delta}{2a}\ and\ x_2=\frac{-b+\sqrt\Delta}{2a}\\\\\Delta=2^2-4\cdot1\cdot(-8)=4+32=36;\ \sqrt\Delta=\sqrt{36}=6\\\\x_1=\frac{-2-6}{2\cdot1}=\frac{-8}{2}=-4;\ x_2=\frac{-2+6}{2\cdot1}=\frac{4}{2}=2[/tex]
[tex]2x^2+4x-16=0\ \ \ /:2\\\\x^2+2x-8=0\\\\a=1;\ b=2;\ c=-8\\\\\Delta=b^2-4ac\ if\ \Delta > 0\ then\ x_1=\frac{-b-\sqrt\Delta}{2a}\ and\ x_2=\frac{-b+\sqrt\Delta}{2a}\\\\\Delta=2^2-4\cdot1\cdot(-8)=4+32=36;\ \sqrt\Delta=\sqrt{36}=6\\\\x_1=\frac{-2-6}{2\cdot1}=\frac{-8}{2}=-4;\ x_2=\frac{-2+6}{2\cdot1}=\frac{4}{2}=2[/tex]
in order to solve it, we need find the zero of the polynomial.
we find the zero of the polynomial by splitting the middle term method
2x2 -+ 4x - 16
= 2x2 + 8x - 4x -16
= 2x( x + 4)- 4(x + 4)
= (2x-)(x+4)
we find the zeroes of the factors
experimentally we find two values, 2 and -4.
Thus, values of x are 2 and -4
we find the zero of the polynomial by splitting the middle term method
2x2 -+ 4x - 16
= 2x2 + 8x - 4x -16
= 2x( x + 4)- 4(x + 4)
= (2x-)(x+4)
we find the zeroes of the factors
experimentally we find two values, 2 and -4.
Thus, values of x are 2 and -4
Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.