Sarah7143id
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Solve for x : 2x^2+4x-16=0

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[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]
Tadvisohil886id
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

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