Question

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

Answer 1

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

Answer 2

`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`




`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=(-b-\sqrt\Delta)/(2a)\ and\ x_2=(-b+\sqrt\Delta)/(2a)\\\\\Delta=2^2-4\cdot1\cdot(-8)=4+32=36;\ \sqrt\Delta=√(36)=6\\\\x_1=(-2-6)/(2\cdot1)=(-8)/(2)=-4;\ x_2=(-2+6)/(2\cdot1)=(4)/(2)=2`