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Which equations could be solved using the quadratic formula?
Type your answer here...5x2 - 3x + 10 = 2x2x2 - 6x - 7 = 2x
how do i solve this 5x2 + 3x2 – x3 + 2x3 + 10x4 – 5x4?
x2 • (5x2 + x + 8)
What are the factors of this quadratic trinomial 5x2-8x-4?
(5x+2) and (x-2)
Solve the quadratic equation by completing the square x2 plus 2x plus 5 equals 0?
x2 + 2x + 5 = 0x2 + 2x + 5 - 5 = 0 - 5x2 + 2x = -5x2 + 2x + (2/2)2 = -5 + (2/2)2x2 + 2x + 12 = -5 + 1(x + 1)2 = -4sq. root of (x + 1)2 = sq. root of -4|x + 1| = 2ix + 1 = +&- 2ix + 1 - 1 = -1 +&- 2ix = -1 +&- 2i
What are the steps to solving an incomplete quadratic equation?
There are two kinds of incomplete quadratic equations:1) ax2+bx=0 (e.g. 4x2+2x=0)2) ax2+c=0 (e.g. 5x2-125=0)And the ways to solve them are as follows:Type 1) For example, 4x2+2x=0.You take out x (and a number multiplying it, if possible:)2x(2x+1)=0In order for the equation to be correct, then either the outcome of the brackets (2x+1) or the number multiplying them (2x) needs to be zero:2x+1=02x=-1x1=-0.52x=0x2=0So the two solutions are -0.5 and 0. This also works if you take out only x:x(4x+2)=04x+2=04x=-2x1=-0.5x2=0Type 2) For example, 5x2-125=0. This type of incomplete quadratic equation can be solved like a regular equation, as follows:5x2-125=05x2=125x2=25x=±5 (x1=5, x2=-5)
