2. 2x(x+3)=0
2x^2+3x=0
[i don't know what to do.]
3. y^2+8y=0
4. x^2+3x=28
5.2x^2+9x-5=0
2x^2+9x=-5
[please help me by starting off the problems, or explaining for each.]
2. 2x(x+3)=0
There are 2 terms: 2x and (x+3).
If any term in a product is 0 the whole product is 0. So, what value of x would make 2x=0 and what value of x would make x+3 = 0?
Sorry, I meant to say factor not term.
If any "factor" in a product is 0 the whole product is 0.
There are 2 "factors".
x=-3 ?
Aren't you supposed to solve these equations for x?
For #2, if
2x(x+3)=0, then either x = 0 or x = -3 will satisfy equation. Those are the solutions,
For #4, note that x^2 + 3x - 28 =
(x+7)(x-4). Since this is zero, then x = -7 and x = 4 satisfy the original equation, and are solutions
The two equations that you wrote for #5 are inconsistent. If the first one is correct, then what you need to do is factor 2x^2+9x-5
Try (2x-1)(x+5)
If that equals zero, then either factor is zero, and that tells you what x is.
i still don't understand , i'm sorry .
2x(x+3)=0
I am assuming that you need to find the values of x that will be true for the above.
When you multiply several items (each of these is called a factor) together the answer (called the product) is 0 if ANY of them are 0.
Example:
multiply (a)(b)(c)
If any of them, a or b or c is 0, then multiplying them always gives 0.
The problem has 2 factors:
2x
(x+3)
If either of these is 0 then 2x(x+3) is 0.
The answers are 0 and -3.
It is always a good idea to substitute back to check.
For x=0
2x(x+3)
=2(0)(0+3)
-0(3)
=0
For x=-3
2(-3)(-3 + 3)
=(-6)(0)
=0
The line
-0(3)
should have been
=0(3)
See below:
For x=0
2x(x+3)
=2(0)(0+3)
=0(3)
=0
mkay , thanks (:
so , what about number 3 ?
Factor y^2 + 8y = 0 into
y(y+8) = 0, and get the solutions from the factors. We are trying to teach you how to do these by yourself.
To solve these equations, we will use the method of factoring or, if necessary, the quadratic formula. Let's go through each problem step by step:
2. To solve 2x(x + 3) = 0, we start by factoring out the common factor of x:
2x(x + 3) = 0
This equation will be true if either 2x = 0 or (x + 3) = 0.
Solving 2x = 0 gives x = 0.
Solving (x + 3) = 0 gives x = -3.
Thus, the solutions to the equation 2x(x + 3) = 0 are x = 0 and x = -3.
3. To solve y^2 + 8y = 0, we can factor out a common factor of y:
y(y + 8) = 0
This equation will be true if either y = 0 or (y + 8) = 0.
Solving y = 0 gives y = 0.
Solving (y + 8) = 0 gives y = -8.
Thus, the solutions to the equation y^2 + 8y = 0 are y = 0 and y = -8.
4. To solve x^2 + 3x = 28, we first arrange the equation in standard quadratic form by subtracting 28 from both sides of the equation:
x^2 + 3x - 28 = 0
Next, we can factor the quadratic equation:
(x + 7)(x - 4) = 0
This equation will be true if either (x + 7) = 0 or (x - 4) = 0.
Solving (x + 7) = 0 gives x = -7.
Solving (x - 4) = 0 gives x = 4.
Thus, the solutions to the equation x^2 + 3x = 28 are x = -7 and x = 4.
5. To solve 2x^2 + 9x - 5 = 0, we can apply the quadratic formula. The standard quadratic formula states that for any equation of the form ax^2 + bx + c = 0, the solutions are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 2, b = 9, and c = -5. Substituting these values into the quadratic formula, we can find the solutions.
x = (-9 ± √(9^2 - 4*2*(-5))) / (2*2)
x = (-9 ± √(81 + 40)) / 4
x = (-9 ± √(121)) / 4
x = (-9 ± 11) / 4
This gives us two possible solutions:
x = (-9 + 11) / 4, which simplifies to x = 2/4 = 1/2
x = (-9 - 11) / 4, which simplifies to x = -20/4 = -5
Thus, the solutions to the equation 2x^2 + 9x - 5 = 0 are x = 1/2 and x = -5.