Sarah7143
Answered

Welcome to Westonci.ca, your ultimate destination for finding answers to a wide range of questions from experts. Discover comprehensive answers to your questions from knowledgeable professionals on our user-friendly platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

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

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]
tadvisohil886
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

Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.