Quadratic equation?

Question

I need someone to solve this problems. If you know how to solve a quadratic equation please help me. I don't know how to solve quadratic equations but, I do know that they are not suppose to be factored. If you do not know what I'm talking about please do not post....because most likely you will just factor the problem out. Thanks!

z^2 - 2z - 15 = 0

Don't forget to show me how you got the answer. Thanks!

Answer 1

quadratic equation is of the form

a*x^2+bx+c=0

delta = b^2-4*a*c

the roots (solutions) of the equation are

x1= (-b+sqrt(delta))/ (2*a)
x2= (-b-sqrt(delta))/ (2*a)

in your case:
a=1
b=-2
c=-15

delta = (-2)^2-4*1*(-15)= 4+60=64

x1=(-(-2)+sqrt(64))/(2*1)= (2+8)/2 = 5
x2= (-(-2)-sqrt(64))/(2*1) =(2-8)/2 = -3
hope it helps

Answer 2

Husband's Answer:

The easiest way to solve these types of equations is by factoring, unless you want to spend all day plugging in numbers. But here is the answer:
z^2-2z-15=0
(z-5)(z+3)=0 or z^2-5z+3z-15=0
z-5=0 z=5
z+3=0 z=-3



Wife's Answer:

Completing the Square is another technique:
z^2-2z-15=0
z^2-2z=15 Add 15 to both sides to eliminate -15 & to
get the z's alone
z^2-2z+1=15+1 Half the z number, square the number,
& add to both sides of the equal sign
(z-1)^2=16 Convert left side into squared form & simplify
right side
z-1= +/-4 Square root both sides remembering to place
to place the positive & negative sign in front
of the 4
z-1= +4 and z-1= -4 Solve for "z"


We both came up with the same answer, but two different methods. Another method is by graphing, but unfortunately, we couldn't upload a graph. I (the wife) used to be a high school math teacher & love this stuff!!! Thanks for letting us be a part of your adventure.

Answer 3

z² - 2z - 15 = 0

The quadratic formula requires an equation in the following form.
ax² + bx + c = 0

In your example, a = 1, b = -2, and c = -15

The quadratic formula is
x (or z in this case) = {-b ±√[b² - 4ac]}/(2a)

You plug the values of a, b, and c into the formula

{-(-2) ±√[(-2)² - 4(1)(-15)]}/[(2)(1)]
= {2 ±√[4+60]}/2
= {2 ±√(64)}/2
= (2±8)/2
= 1±4
So z = 5 or z = -3

To check, plug these values back into the original equation
z² - 2z - 15 = 0
(5)² - 2(5) – 15 = 25 – 10 – 15 = 0
(-3)² - 2(-3) – 15 = 9 + 6 -15 = 0
So both answers check.

Answer 4

The way I tackle this kind of problem is this. You know that the break down of this problem will be:

(z +/- A)(z +/- B)

And you know that 15 has to be a result of A times B.
And you know that 2 has to be a result of A +/- B.

So what two numbers can do that?

15 x 1 = 15
BUT you can get 2 from 1 and 15.

3 x 5 = 15
AND you can get 2 from 5 - 3!!!

So now it's just a matter of getting the signs right.
Since the 15 is negative, we know ONE of them has to be negative.
Since the 2z is negative, then we know that the LARGE number (5) has to be negative if we are subtracting them.

(z-5)(z+3)

So those are both the factores of the equation.
We then have to make them both equal zero to get the answers.

(z-5)=0
z=5

(z+3)=0
z=-3



This is more of an explaination that me figuring it out for you. Just because I find these problems SUPER easy if I think them out in the way I explained them. Hopefully it makes them easier for you.

Answer 5

ok. i will give u a standard equation and we will comapre your given equation with the standard one and match the answers.
the standard equation is ax^2+bx+c

So
Step I:multiply 'a' with c
in our question a=1, c = -15
a * c = -15
Step II:find the factors of a*c
the ready factors are 3 and -5
so in your equation,
z^2-5z+3z-15=0
so take common in the 1st and 2nd terms
3rd and the 4th terms
u get z(z-5)+3(z-5)=0
(z+3)(z-5)=0
Step III:so the solution of the equation is
either z+3=0,z=-3
or z-5=0,z=5
hence the solution of this equation is -3 and 5

Answer 6

Hi, your first post showed you how to factor it and solve it, but if you're not to factor then use the quadratic formula. It is:
..........________
-b ± √b^2 - 4ac
------------------------ = x
........2a

where "a" is the number in front of x^2, "b" is the number in front of x and "c" is the constant.
For this problem, a = 1, b = -2 and c = -15. Substituting these:
.........________
-b ± √b^2 - 4ac
------------------------ = x
........2a
............_____________
-(-2) ± √(-2)^2 - 4(1)(-15)
------------------------------------- = x
........2(1)

Simplifying:

........______
2 ± √4 + 60
------------------ = x
........2

........___
2 ± √64
------------ = x
.....2

2 ± 8
-------- = x
...2

2 + 8......10
-------- = ----- = 5 for one answer
...2...........2

2 - 8.......-6
-------- = ----- = -3 for the other answer
...2...........2


I hope that helps, without factoring!! :-)

Answer 7

(z-5)(z+3)=0
z= 5, -3

There's no need to use the quadratic formula to find the answer because the solutions are integers. If the equation could not be simplified as shown above then you would need to use:

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

Answer 8

use the quadratic equation:

z = (-b +/- sq root (b^2 -4 ac)) / 2a

where the any equation can be shown as:
a z^2 + b z + c = 0

so a = 1
b= -2
c= -15

Answer 9

First find factors of z².
These are z and z:-
(z------).(z------)
Now require numbers that multiply to give -15 and add/subtract to give - 2. These numbers are - 5 and + 3 :-
(z - 5).(z + 3) is required answer.

Answer 10

lil' is right