Solving Quadratic Equations?

Question

Solve each equation for x.

1. 0=-3x(2x+5)

2. 0=2x^2+32x

3. 0=18x-9x^2

Please explain too. I don't think I am supposed to figure it out using the Quadratic Formula though.

Thanks!

Answer 1

You're right for these equations you don't have to use the quadratic equation to find the roots (the numbers that will make the equation equal zero).

Just try to factor them down as much as you can and then try to figure out what numbers will make the equation equal zero.

1) 3x(2x+5)=0
You know that x=0 is a root because 3(0)(2(0)+5)=0
All that is left is to make 2x+5=0, which is easy, just solve for x.
2x=-5
x=-5/2
So for 1. x=0, x=-5/2 are the roots of the equation.

2) 2x^2+32x=0 (step one factor it down as much as you can)

2x(x+16)=0
This one is even easier to find the roots (just remember anything multiplied by zero equals zero), so the roots are:
x=0, x=-16

3)-9x^2+18x=0

Is the same as -9x(x-2)=0 (factored)

Again it is easy to see what values for x will make the equation equal zero.
x=0, x=2

Just remember that anything time zero is zero and since the equation is equalled to zero all you have to do is find what numbers you can replace x with to make the equation equal zero. Those numbers are the "roots" of the equation. Roots are very important in math and you will use them later on for calculus :)

Good luck to you!

Answer 2

Without using the quadratic, you can solve it this way.

1. Obviously, x can be zero, that works. Then solve 2x+5=0, giving you x=-5/2. So in the first one, the possible solutions are x = 0, x = -5/2

2. Factor again here, as was already done in equation 1. This gives you 2x(x+16). Similarly, x=0 is an obvious solution. And solving x+16=, then x=-16 is the second solution.

3. Factor again, giving you 9x(-x+2). Again, x can be zero, or x=2.

Answer 3

Solving Quadratic Equations?
Solve each equation for x.

1. 0=-3x(2x+5)
or x=0 or -5/2answer
2.2. 0=2x^2+32x=2x(x+16)
x=0 or-16 answer
3. 0=18x-9x^2=9x(2-x)
x=0 or2answer
I have done these by factorizationwhich is easier in these cases.But we could also use quadratic equation formula.

Answer 4

1. This one is already factored, so x is equal to the values at which EITHER factor = 0.
-3x=0 => x=0
2x+5=0 => x=-5/2

2. Factor out the greatest common factor (in this case 2x is the common factor):
0=2x(x+16)
Once again, set each factor equal to zero and solve:
2x=0 => x=0
x+16=0 => x= -16

3. Factor out the greatest common factor again (in this case, it is 9x)
0=9x(2-x)
Once again, set each factor equal to zero and solve:
9x=0 => x=0
2-x=0 => x=2

Answer 5

3x(2x+5) = 0

if u multiply two variables and obtain zero as the answer this means one of them or maybe both of them is zero.

so now we take 3x = 0
x = 0;

lets take 2x + 5 = 0
x = -5/2

same applies for question two
factorize the equation 1st
2x^2 +32x =0
2x(x+16)=0

for 2x =0
x = 0
for x + 16 =0
x = -16.

question 3 is about da same as number 2.

Answer 6

Question 1
- 6x² - 15 x = 0
6x² + 15x = 0
(3x).(2x + 5) = 0
3x = 0, (2x + 5) = 0
x = 0, x = - 5/2

Question 2
2x.(x + 16) = 0
x = 0, x = - 16

Question 3
9x² - 18x = 0
(9x).(x - 2) = 0
x = 0, x = 2

Answer 7

Do your own homework.