Solve by using tables. Give each answer to at most two decimal places. –2x2 – 4 = –8x (1 point) 0.59, 3.41 –3.41, –0.59 1.17, 6.83 –1.41, 1.41
congrats on ur first question
Thanks. Do you know the answer to this?
you need to rewrite the equation to make it easy... and let it equal zero \[2x^2 -8x + 4 = 0\] which will become \[2(x^2 -4x + 2 )= 0\] you need to solve \[x^2 -4x + 2 = 0\] by completing the square or the general quadratic formula.
Not to sound stupid or anything, but what is the general quadratic formula?
a method for solving a quadratic that can't be factorised.... and I'm curious about using tables to solve this...
the general quadratic formula is use for a quadratic \[ax^2 + bx + c = 0\] which is written the the general equation the you use \[\ x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Thanks.
reading your question again... I think that they want you to substitute to find the solution s e.g. x. 0.59 is -2(0.59)^2 - 4 = 8(0.59)
if it is then that will be the solution pair
Solve by using tables. Give each answer to at most two decimal places. –2x2 – 4 = –8x (1 point) (1 pt) 0.59, 3.41 (0 pts) –3.41, –0.59 (0 pts) 1.17, 6.83 (0 pts) –1.41, 1.41 1 /1 point
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