MaTh HeLppPPP!!!?

Question

solve using the qudratic formula:
x*2-3x-10=0

*2=means x squared


please show me how to solve this in a step by step process...it is a practice question on a practice test....if you have any questions...'email message' me from my profile!

-----*THANK YOU*-----

Answer 1

x^2-3x-10=0
x^2-5x+2x-10=0
x(x-5)+2(x-5)=0
(x-5)(x+2)=0
x-5=0 or x=5
x+2=0 or x=-2
solnset is 5,-2

Answer 2

Quadratic formula

ax^2 + bx + c = 0 (a does not equal 0)

x = -b +- [b^2 - 4ac]^-2
-------------------------
2a

x^2 -3x -10 = 0
a= +1
b= -3
c = -10

I found that to avoid sign errors, I had to put parentheses in place of all the letters and then substitute.

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

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

Solve in order of
1. inside parentheses
2. powers and roots
3. multiplication and division
4. addition and subtraction

Tackle the b^2

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

Change the signs
Like signs = +
Opposite signs = -

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

Simplify with multiplication

x = 3 +- [9 + 40]^-2
-------------------------
2

Simplify with addition or subtraction

x = 3 +- [49]^-2
-------------------------
2

Find the square root of what is in the brackets

x = 3 +- 7
-----------
2

Now separate it out into two problems


x = 3 + 7
-----------
2

and


x = 3 - 7
-----------
2

Solve each

x = 10
-----
2

and

x = -4
--------
2

Solve

x = 5 and x = -2

To make sure these are correct, plug them back into the formula to check.

x^2 -3x -10 = 0

x = 5 and -2

For x = 5
(5)^2 -3(5) - 10 = 0

solve
25 - 15 - 10 = 0
10 - 10 = 0
0 = 0

For x = -2
(-2)^2 -3(-2) - 10 = 0
(-2)(-2) -3(-2) - 10 = 0
+4 +6 - 10 = 0
10 - 10 = 0

It checks!!!

Remember to put () around each letter BEFORE substituting, and it will save you hours of confusion.

Answer 3

There is two valid solutions to those equations because of the x squared. Sometime it is just two time the same value or no solutions.

The best way to find the first one is to do tries:
If x=1 ...2-3-10 ... no
If x=-1 ...2+3-10...no
If x=2 ...4-6-10 ... no
If x=-2 ...yes! found! EASY

Hence (x-2)(x+other)=x*2-3x-10
Look at the end of it... : -2other=-10 -->other=5

Solution: 5 and -2! DONE!

---------------------------------------

Another way to do it, complicated but always right :

First step:

x*2-3x-10=0 is similar to ax*2+bx+c=0 with:
a = 1
b = -3
c = -10

2d step:

Calculate Delta= b*2-4ac
Delta = 9-4(-10)=9+40=49

3d step:

- If Delta is positive, hence
2 solutions:
First=(-b+sqrtDelta)/2
=(3+7)/2=5
Second=(-b-sqrtDelta)/2
=(3-7)/2=-2
Done!

- If Delta is negative, more complicated! Say, no solution by now!
------------------------------------------------

GOOD LUCK!!!

Answer 4

ok sweetie here is how it goes:
x*2-3x-10=0
(x-5)(x-2)=0
so x=5, x=2
when you plug both of them into the equation you will find that x=5 is the right solution
2*2-3(2)-10=-12 (wrong)
5*2-3(5)-10=0 (right)
good luck with your math class :D

Answer 5

42

Answer 6

There are websites where you type in the question and it will show you step by step how to do it. Just search for whatever your doing, "Algebra Help" might be one

Answer 7

(2x-3)(3x+2) FOIL. First Outer inner final First: 2x times3x equals 6x^2 Outer: 2x circumstances 2 equals 4x inner: -three times 3x equals- 9x final: -three times 2 equals- 6 answer: 6x2+5x-6 stable good fortune

Answer 8

That's the way I like to do it :)
Probably you know this? (a - b)^2 = a^2 - 2ab + b^2 (^2 means square)
Than look:
x^2 -3x-10=0 /add 2,25 (it's 1,5^2)

x^2 - 2*x*1,5 + 2,25 -10 = 2,25 that's difficult. look up at the formula



(x-1,5)^2 -10 = 2,25 then:

(x-1,5)^2 = 12,25

|x-1,5| = root12,25 (=3,5)

Then we have

x-1,5 = 3,5 --> x=5 OR x-1,5=-3,5 --> x=-2


:D

Answer 9

please ask your teacher.