Thelma Said:

What are the 3 ways on how to do this Algebra 1 problem? And how do you do it?

We Answered:

1) Graphing it. Make a grid on paper and graph the curve. This problem can be accurately solved by graphing. The zero on the right side, that is where y would be. In this problem, y is set to zero.
what yoy need to do is graph y=x^2 + 10x + 9 and see what x is when y =0. (hint, both answers are on the left side so start with x=0 and go negative)

2) factoring 0= (x +- a)(x+-b)

since the highest x has a power of 2 (x^2), there are two binomial factors.

( )( )

Since both +/- signs are plus, the signs in the middle are both +

( + )( + )

in order to get an x^2, two x's must be multiplied and since there is no coefficient in fron of the x^2, you know that there is no coefficent in fron of the x's in the ()'s.

(x+ )(x+ )

Here is the hard part. Look at the third number --> 9
How do we make 9?
3X3 or 9X1

The numbers in the brackets could be either 3 , 3 OR 9, 1

(x+3)(X+3)

X times X, plus 3 times X, plus 3 times X, plus 3 times 3

X^2 +3x +3x +9 = x^2 +6x +9 NO!

try the other choice

(X+1)(X+9)

X times X, plus 1 times X, plus 9 times X, plus 1 times 9
= X^2 +1X +9X +9 = x^2 + 10x +9 YES!!!!!!!
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(x+1)(X+9)=0
how do we get a product to be zero?
one (or both) number(s) must be zer0

X+1=0 X = -1
X+9=0 X=-9

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3) You know in power rangers how at the end of the battle, no matter how hard it is, the giant robot takes out its giant sword and waves it and all the monsters die? This is it..
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
~~~THE QUADRATIC EQUATION!!!!!!!!!!!!!!!!!!!!!!!!!! ~~~
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

x= -b +- sqrt( b^2 -4ac) ALL OVER 2a

a is the coefficient of x^2
b is the coefficient of x
c is the other number


x= (-10 +- sqrt(100- 4(1)(9)) ) / 2(1)

x= (-10+8)/2 = -1 AND x= (-10-8)/2 = -9

the +- means plus or minus.

Tracy Said:

Algebra 1 Problem regarding Systems of Linear Equations?

We Answered:

How old ARE each of them.

Let
b = Bob's present age
m = Maria's present age
s = Sarah's present age

b + 2 = s + m
b + 5 = 2(m + 5)
m - 2 = (1/2)(s - 2)

Remove the parentheses and fractions.
b + 2 = s + m
b + 5 = 2m + 10
2m - 4 = s - 2

Simplify.
b = s + m - 2
b = 2m + 5
s = 2m - 2

Subtract the first equation from the second.
0 = -s + m + 7
s = m + 7

Now we have two equations in s and m.
s = 2m - 2
s = m + 7

Equation the two equations and solve for m.
s = 2m - 2 = m + 7
m = 9

Solve for s.
s = m + 7 = 9 + 7 = 16

Solve for b.
b = 2m + 5 = 2*9 + 5 = 23

So the ages of the people are:
Bob = 23
Maria = 9
Sarah = 16

Bobby Said:

How do you work this Algebra 1 problem out?

We Answered:

Ok so you have x amount of guests to begin with.
So x equals half of x minus three, then you add five and subtract two, so far leaving you with x=.5x, then you subtract six and then add 16. x=.5x+10, so then you subtract .5x which equals .5x=10, then divide by .5, giving you 20=x. Since only twelve people are attending, you would subtract 12 from 20, and you get 8.

Georgia Said:

Can somebody help me and show the work with this Algebra 1 problem? thanks a lot?

We Answered:

2x+1/4(8x)= -8 (8x goes in for y because of second equation)
2x+2x= -8
4x= -8
x= -2
sub in for second equation to get y
y=8(-2)
y=-16
check both answers in first equation
2(-2) +1/4(-16) = -8
-4+(-4)= -8
-4-4=-8
yes, x=-2, y=-16

Sheila Said:

How would you go about doing this algebra 1 problem?

We Answered:

The reciprocal of x = 1/x

So your problem is x + 1/x = 13/6

Multiply by x, move to left side: x^2 -(13/6)x + 1 = 0

Using the quadratic formula, your answers are:

(-b +/- sqrt(b^2 -4ac))/2a where a = 1, b = -13/6, and c = 1

(-(-13/6) +/- sqrt((-13/6)^2 -4(1)(1)))/2 = (13/6 +/- sqrt(189/144 -4))/2
= (13/6 +/- sqrt(25/36))/2 = (13/6 +/- 5/6)/2

= 8/12 or 18/12 = 2/3 or 3/2

You could also have factored the equation as (x-2/3) (x-3/2) and gotten the same answer.

Jeffery Said:

How would you work this Math / Algebra 1 problem?

We Answered:

The hypotenuse length is the square root of (a(squared) + b)squared))

So assuming the legs are the same length (you didn't say whether they were or not)
length of both legs as 1
1 => then hypotenuse is 1(square root of 2) or square root of (1+1)
as 2 => then hypotenuse is 2(square root of 2) or square root of (4+4)
as 3 => then hypotenuse is 3(square root of 2) or square root of (9+9)
as 4=> then hypotenuse is 4(square root of 2) or square root of (16+16)
as 5=> then hypotenuse is 5(square root of 2) or square root of (25+25)

Javier Said:

How would you solve this Algebra 1 problem?

We Answered:

just pretend you have a equal sign

2x +3 = 9
2x = 6
x = 3

now put back the inequality

x>3

so to graph x=3, it is a straight vertical line at x = 3

to graph x > 3 draw a dotted vertical line at x =3

Now interpret what you have

x has to be greater than (or bigger) than 3

so lightly shade the right side of the graph

cheers! Good Luck