Type your answer here...5x2 - 3x + 10 = 2x2x2 - 6x - 7 = 2x
x2 • (5x2 + x + 8)
(5x+2) and (x-2)
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
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)
X= (3/5 , -2)
Yes
Type your answer here...5x2 - 3x + 10 = 2x2x2 - 6x - 7 = 2x
5x2 + 8x = 7 5x2 + 8x - 7 = 0 This cannot be factorised so the solutions need to be determined using the quadratic formula The solutions are {-8 ± sqrt[82 - 4*5*(-7)]}/(2*5) = {-8 ± sqrt[64 + 140]}/10 = {-8 ± sqrt[204]}/10 = -2.22829 and 0.62829 (to 5 dp)
5x2-16x+12 = (5x-6)(x-2) when factored with the help of the quadratic equation formula
x2 • (5x2 + x + 8)
5x2-8x-4
(5x+2) and (x-2)
You put everything to the left side. For example, if you have a constant term on the right side, subtract it on both sides, so that you have an equation where the right-hand side IS zero. For example: 5x2 + 3x - 5 = 20 Subtract 20 from both sides: 5x2 + 3x - 25 = 0 Now you can apply the quadratic formula.
its 10
It is: (5x-1)(x-1) = 0 Therefore: x = 1/5 or x = 1 Use the quadratic equation formula
If that's +32, the answer is (x - 4)(5x - 8)