algebra test 1

write in terms of i and simplify: neg number under a lil square root house)

just find the square root and throw an i on there. if it isn't a perf square, simplify the radical (radicand?) by finding a perf square that multiplies to it. for examp, for -48 it would be 4 squared (16) times 3. put the perf square outside the house, then the i, then the number u multiplied the perf square by inside the house.

perform this operation and write in a+bi: 1-2i/3+4i

multiply by denominator's opposite (ex: 3-4i). foil. multiply the i squared terms by -1 because i squared is negative 1. combine like terms. simplify.

perform this operation and write in a+bi: (1+2i)2

put it in a2 + 2ab + b2 format (example: 1 squared + 2(1)(2i) + 2i squared). then just combine ya like terms

simplify the expression and write the result in the form a+bi: square root of 1 - square root of 2 over square root of 3.

simplify each number in the root house by finding their square root, and rmr the square root of any neg number will have an i thrown on there. delete any numbers that cancel out (ex: in 4-4iroot3/-2, delete the 4-4 so it becomes -2+iroot3)

solve using square root property: 3xsquared - 9 = 0

start by bringing the constant numbers (2) over to the other side and change its sign (ex: 3xsquared = 9). divide both sides by the coefficient of the xsquared (ex: xsquared = 3). then take the square root of both sides (ex: x = 3 in square root house). the answer is always plus or minus the square root of the constant so the answer would be x = +/- 3 in square root house. if the constant is negative, then throw an i on the end.

solve using the zero product property when the prob looks like THIS: 1x squared = 2x.

TWO ANSWERS. move the 2x over so the whole thing equals zero (dnt forget to change the sign!). u should have smth like 1xsquared - 2x = o. divide by the gcf (in this case, just x) so u just have x-2=0. ur two answers would be x=0 or x-2=0 which equals 2, so the two answers are x=0 and x=2. my understanding of this one is kinda shaky lol.

determine if a relation is a function using ordered pairs

an x can only have one y value, but a y can have multiple x values. so, if the pairs were (2,3) and (2,4) then it would not be a function because 2, the x value, cannot have more than one y value. if the pairs were (2,3) and (3,3) then it would be a function because each x only has one y value.

solve using quadratic equation: 1x - 2 = -3xsquared

bring all the terms over so it equals zero. if needed, reorder the numbers on the left side of the equal sign so the first term is the xsquared, the second is the plain x, and the third is the constant. then just do the quadratic formula. if it seems like u can't simplify it anymore once ur in the form that's like 1+/- 2 in the square root house/3, just leave it like that.

multiply complex numbers and write in terms of a+bi (this type of prob: (1-2i)(3+4i))

foil the numbers. multiply the i squared number by -1 to cancel out the i so it's just a norma number w/ no i (ex: 4i squared times neg 1 just becomes 4). combine like terms. u dun it!

solve by completing the square: xsquared - 2x - 3 = 0

start by moving the constant over to the other side of the equal sign. then, take the second term on the left side of the equation (ex: 2x), divide it by 2 (ex: 2/2 = 1), and then squaring the product u get (1 squared = 1). add that number to both sides (equation becomes xsquared + 2x + 1 = 3 + 1). combine the like terms on the right side of the equal sign. factor the left side - since it's a perfect square, the factored version is just x+ whatever half the second term is twice (ex: in this equation, half of 2 was 1, so the factored version is (x+1)(x+1)). then turn that into (x+1)sqaured = right side of equation (ex: (x+1) squared = -2). get the square root of both sides and make the right side +/- (ex: x+1 squared is just x+1, 4 becomes +/-2.) then split the prob into two and solve for x+1 = 2 and x+1 = -2.

finding the center and radius of a square: (x + 1) 2 + ysquared = 12

the center is the second term in the x parentheses (ex: 1) and the second term in the y parentheses (here it would be 0 because it's just the y squared alone). change the signs, so the center is -1,0. the radius is just the square root of the product (ex: square root of 12 is 2root3.)

finding the equation of a circle from its radius and center: given a circle with center (2,2) and radius 3, find the standard equation of the circle and graph it

the equation of a circle is (x - h)2 + (y - k)2 = r2. h is the first number in the circle's coordinates (in this case, 2) so plug that in. the k is the second number in the in the circle's coordinates (in this case, also 2) so plug that in. the r2 is the radius squared, so just square the radius to find that (in this case, it's 9 because 3squared is 9). therefore the equation is (x-2)squared + (y-2)squared = 9. to graph, just plot a point at the given center and count the radius's number of units out from the center on each side to form a circle.

determine if the relation is a function when it looks like this: x2+y2=1.

this equation would be a circle, so no, it's not a function because a vertical line would pass through more than once.

solve using the zero property when the prob look like THIS: 1t(t+2) = 3+ - 4

this type of prob has TWO ANSWERS. start by distributing the 1t to the numbers in the parentheses. then move the numbers on the other side of the equal sign over (dnt forget to change their signs) so the whole thing equals zero. u should have 4 terms: a t squared, two regular ts, and a normal number with no variable at the end. combine the two regular ts in the middle so u have three terms. then use the quadratic equation to find ya two answers.

finding the equation of a circle from given endpoints: the endpoints of a diameter of a circle are (2,1) and (10, −5). write an equation of the circle in standard form

to find the center of the circle, use the MIDPOINT FORMULA. this is x1+x2/2, y1+y2/2. plugging in the numbers from the ex it would be 2+10/2, 1+-5/2. then just add and divide to get the two points of the center. here it would be 12/2=6 and -4/2 = -2. so the center is (6,-2). then to get the radius use the DISTANCE FORMULA. this is (x1-x2)2 + (y1-y2)2. plug in the center and one of the points. square the sum and that's ur radius. then plug everything u got into the circle formula (x-h)2+(y-k)2 = r2 if one the points in ur center is negative then it's going to be plus that number instead of subtracting.

find x- and y- intercepts of absolute value equations: |x|-1

to find the x intercept, u solve the equation in the little absolute value bars. if it were x-2 or somethin, ud solve it and it would be (2,0)/(-2,0). if it's just x alone in there, it's always (1,0) and (-1,0). the y intercept is the thing being thrown on at the end - in this case, -1, so it would be (0,-1)

finding the equation of a circle given the center and a point on the circle: given a circle where the center is (4,8) and another point on the circle is (-4,2), write the standard equation

use the distance formula: (x1-x2)2 + (y1-y2)2. plug in the xs from the center and the point and the ys from the center and the point. in this case it would be (4-(-4) which is 8 and (8-2) which is 6. square both of these values. in this case it would be 64 and 36. add these together. this number is ur radius. then plug the center and the radius into the circle formula (x-h)2+(y-k)2+r2. sample answer would be (x-4)2+(y-8)2=100.

determine if a relation is a function using a graph

vertical line test - if a vertical line would pass through the relation more than once on the graph, it's not a function. if it would pass through only once, then it is a function.

find x- and y-intercepts of this type of function: f(x) = xsquared − 63

y is just the thing hanging out at the end. in this case it's -63, so the y-intercept is (0, -63). to find x, sub in 0 for f(x) because f(x) is the same as y. the equation can then b solved using the square root property.