Help with Quadratic Formula?

Question

x^2 + 3x – 10 = 0

x^2 = 2x – 15

4x^2 + 8x = 3

Now I am asking for more than just the answers, I need help with the entire concept

Answer 1

A clear step by step process at each Question in order.

http://i38.photobucket.com/albums/e102/_ill_Nino_/Quad1-1.jpg

http://i38.photobucket.com/albums/e102/_ill_Nino_/Quad2.jpg

http://i38.photobucket.com/albums/e102/_ill_Nino_/Quad3.jpg

just with link 3 it says Question 2, but it is Question 3 no doubt, just didnt change it when doing the formulas themselves.

And hope u can read it easier than the others, cause it took not as long to make them as to understand it =D

Enjoy

Answer 2

Let's look at the first equation. You probably can already find the answer by factoring:

x^2 + 3x - 10 = 0 into ( x + 5) ( x - 2) = 0,
so x = 2 or x = -5.

The trouble is, each time you have a problem you would have to figure out how to factor the equation again, which with more complicated problems can be a nuisance. With the quadratic equation, you don't need to come up with the factorization, you just apply the formula, which is

[ -b plus or minus sqrt( b^2 - 4ac) ] / 2a

So what are a, b and c? The coefficients from your equation when you get all the terms on one side, and 0 on the other side.

Here we already have that situation, so we're set to go.

a is the coefficient of the x ^ 2 term. In this case we have one x^2, so a = 1.

b is the coefficient of the x term. Here we have 3x, so it is 3.

c is the number without an x. Here is is -10.

So now we just put those values into the formula to find our answer:

[ -b plus or minus sqrt( b^2 - 4ac) ] / 2a

[ -(3) plus or minus sqrt( 3^2 - 4(1)(-10) ] / 2(1)
=[ -3 plus or minus sqrt( 9 + 40) ] / 2
=[ -3 plus or minus sqrt (49) ] / 2
=[ -3 plus or minus 7 ] / 2
= 4 / 2 or (-10) / 2
= 2 or -5
Or x = 2 or x = -5.

This is the same solution we got before, so our answer checks out.

For the others, remember to get all the terms on one side and the 0 on the other before you apply the formula.

Answer 3

1. x^2 + 3x - 10 = 0

1st: identify the 3 coefficients for 3 variables > a = 1; b = 3 and c = - 10

2nd: replace the coefficients with the variables in the Quadtratic Formula which is..

x = [- b +/- V`(b^2 - 4ac)] / 2a

x = [- 3 +/- V`(3^2 - 4(1)(-10))] / 2(1)

x = [- 3 +/- V`(9 - 4(1)(-10))] / 2

x = [- 3 +/- V`(9 - 4(-10))] / 2

x = [- 3 +/- V`(9 + 40)] / 2

x = [- 3 +/- V`(49)] / 2

x = [- 3 +/- 7 ] / 2

x = [- 3 +/- 7 ] / 2

3rd: you have 2 solutions; one has a positive sign & the other has a negative sign.

a. x = [- 3 + 7 ] / 2
x = 4 / 2
x = 2

b. x = [- 3 - 7 ] / 2
x = - 10 / 2
x = - 5

2. x^2 = 2x - 15

1st: set the equation to zero by moving the variables on the right, to the opposite side (left). subtract 2x from both sides.

x^2 - 2x = 2x - 2x - 15
x^2 - 2x = - 15

*Add 15 with both sides.

x^2 - 2x + 15 = - 15 + 15
x^2 - 2x + 15 = 0

2nd: identify the 3 coefficients > a = 1; b = -2 and c = 15

3rd: replace the coefficients with the variables in the Quadtratic Formula & solve....

3. 4x^2 + 8x = 3

*Follow the same format as number 2.

Answer 4

You always want to get all of your terms on one side of the equal sign first -- so you'd change the second equation from

x^2 = 2x - 15

to

x^2 - 2x + 15 = 0

by subtracting 2x from both sides of the equal sign and by adding 15 to both sides.

Once you've done that, you know what a, b, and c are -- they're the coefficients for the x^2 term, the x term, and the constant -- for the first equation above, it's 1, 3, and -10, respectively.

Then you just plug them into the quadratic equation.

x = (-b +/- SQRT(b^2 - 4ac)) / 2a

x = (-(3) +/- SQRT((3)^2 -4(1)(-10)) / 2(1)

x = (-3 +/- SQRT(9 + 40))/2

x = (-3 +/- SQRT(49))/2

x = (-3 +/- 7)/2

x = (-3 +7)/2 AND (-3 - 7)/2

x = 4/2 AND -10/2

x = 2 AND -5

Then check your work by plugging the answers back into the original equation.

x^2 + 3x - 10 = 0

2^2 +3(2) - 10 = 0 and (-5)^2 +3(-5) - 10 = 0
4 + 6 -10 = 0 and 25 -15 -10 = 0
yes and yes!

Doug

Answer 5

1st make them all equal zero.
You need to factor them out so now they should look like this:

1) x^2 + 3x - 10 = 0

2) x^2 - 2x + 15 = 0

3) 4x^2 + 8x - 3 = 0

For the 1st one, you need to get factors of 10 (10 x 1, 5 x 2)
You need to end up with 3x in the middle so 5 x 2 looks like the best way to go.

(x - 2)(x + 5) = 0

You multiply each part by each part in the other set of brackets (x x x, x x 5, -2 x x, -2 x 5)= 0
x x x = x^2
x x 5 = 5x
-2 x x = -2x
-2 x 5 = -10

add them all together and you get x^2 + 3x - 10 = 0

I suggest you have a look here for more info and help with the other 2.

http://www.purplemath.com/modules/solvquad.htm

Answer 6

ax^2 + bx + c = 0

x = [-b +- sqrt(b^2 -4 a c)] / 2a

Just fill in a, b and c

x^2 + 3x – 10 = 0

a = 1, b = 3, c = -10

b^2 - 4 a c = 9 + 40 = 49, sqrt(b^2 - 4 a c) = +- 7

so,

x = (-3 +- 7) / 2 = -5, +2


x^2 = 2x – 15
x^2 - 2x +15 = 0
a = 1, b = -2 c = 15

...

Answer 7

The second and third equations can be re-written:

x^2 - 2x + 15 = 0

4x^2 + 8x - 3 = 0


Sometimes it is useful to know what values of the variable x will make the equation correct. The quadratic formula can be useful for that.

Answer 8

x² + 3x - 10 = 0

let

a = 1

b = 3

c = - 10

Quadratic formula

x = - b ± √b² - 4ac / 2a

x = - 3 ± √(- 3)² - 4(1)(- 10) / 2(1)

x = - 3 ± √9 - (- 40) / 2

x = - 3 ± √9 + 40 / 2

x = - 3 ± √49 / 2

x = - 3 ± 7 / 2

- - - - - -

Solving for +

x = - 3 + 7 / 2

x = 4 / 2

x = 2

- - - - - - -

Solving for -

x = - 3 - 7 / 2

x = - 10 / 2

x = - 5

- - - - - - -s-

Answer 9

(1) x=-5, x=2
(2) No Real Solution
(3) No Real Solution
Please give me best answer thanks!

Answer 10

I know i`ve heard of this...but i cant seem to put my finger on it...hmmm