quadratic equations o_0?

Question

3. 10x - 6 = 5x^2

4. x + 2x^2 + 1 = -1

5. 2x^2 + x = 10

6. 2x + 1 = 2x^2


can anyone answer any of these?? my family is stumped as so am i

Answer 1

To solve, you need to rearrange so that the equations are in the form:
ax^2 + bx + c = 0
Then you can use:
x = (-b +/- SQRT(b^2-4ac))/2a

Your equations can be written as:
3. 5x^2 + (-10)x + 6 = 0, so a=5, b=-10, c=6
4. 2x^2 + 1x + 0 = 0, so a=2, b=1, c=0
5. 2x^2 + 1x + (-10) = 0, so a=2, b=1, c=10
6. 2x^2 + (-2)x + (-1) = 0, a=2, b=-2, c=-1

Then use the quadratic solution formula above to find the values of x.

Answer 2

First rearrange your equations so that all the terms are on one side and they equal 0.

The number in front of the x^2 term = "a"
The number in fron of the x term = "b"
The number with no x = "c"

Substitute "a", "b" and "c" into the quadratic formula
(I'll type it here but it would look less confusing if you see it in a math book)

-b +or- square root of (b^2-4ac) all divided by 2a

Your first one, rearranged, would be: 5x^2 -10x - 6 = 0

here "a" = 5, "b" = -10 and "c" would be -6 (make sure you include proper signs)

When you plug the numbers in the quadratic formula you should get two answers: -0.483 and +2.483

Answer 3

Use the quadratic equation. Try actually reading your text book. Quadratic equation: if you have an equation of the form ax^2+bx+c=0 then x has two values x= -b + (1/2a)sqrt(b^2-4ac) and x= -b - (1/2a)sqrt(b^2-4ac)

Answer 4

are you trying to find the solutions? if so, here's what i got. (i have an exam on this tomorrow morning.)

3. 5x^2 - 10x + 6 = 0. 10 plus or minus the square root of -20 over 10. 1 plus or minus 2i square root of 5.

4. 2x^2 + x +2 = 0. [-1 plus or minus (i)square root of 15] over 4.

5. 2x^2 + x - 10 = 0. (-1 plus or minus 9) over 4.

6. 2x^2 - 2x - 1 = 0. (1 plus or minus the square root of 3) over 2.

hope that helps. (and i hope it's right!)

Answer 5

make sure that each equation is equal to zero, and that the coefficent (the number before the squared term) is positive. then after that you can use the quadratic formula ( ((b^2)± (sqrt (b^2)-(4ac)))/2a) using this you can get the x values for your equation.

By the way : a=coefficent of x^2 term
b=coefficent of x term
c= the term with no x