Question

Can someone please solve and illustrate how to solve the following quadratic equations: y^2-4y-8; x^2+3x-10=0; and y^2-y-6? Separately of course, with illustrative steps for each. Please and thanks a bunch!

Answer 1

Do you mean:
y^2-4y-8=0
x^2+3x-10=0
y^2-y-6 =0

If so = what methods do you know for solving?
Factorisation, Complete the square and Quadratic Formula are most used...

Answer 2

factoring, were supposed to solve by factoring

Answer 3

and yes, those are the problems that I am referring to.

Answer 4

OK - first of all - don't be confused by the use of y in these equations

This is just a shorthand for any unknown, and hence you can substitute x if it makes you things look more familiar.

Answer 5

not all quadratrics can be solved by factorising (not factoring)

do you know how to find out whether it can?

Answer 6

A response using Mathematica v9 Home Edition is attached.

Answer 7

I got your attachment and still am very confused

Answer 8

MrNood no I do not

Answer 9

the attachment doesn't help you to solve for yourself anyway.

Do you understand the standard form of the quadratic?

ax^2+bx+c = 0

(note - as I said - you can exchange y for x if oyu like...)

Answer 10

just in case I wasn't clear, these equations are each separate examples given to me by my classmates to practice. They are not one big problem, nor are they linked together in anyway other than how they are supposed to be solved.

Answer 11

I understand.

so - to factorise you need

(x+m)(x+n)=0

OK?

Answer 12

k

Answer 13

when you expand that it becomes

x^2 +mx +nx + mn

= x^2 +x(m+n) +mn

OK? - please don't say OK if oyu don't follow.....

Answer 14

yes I believe I follow so far

Answer 15

so we are looking foe the 2 numbers m and n that give the equation we need to factorise.

Take ift from me for now - the first one doesn't factorise

SO

the second one
x^2 +3x-10 = (x+m)(x+n)

you cne see from the expansion that
m+n = 3
mn=-10

can you find 2 numbers that work for these equations? (It's mainly trial and error..)

Answer 16

numbers that would work for what? to use for the variables?

Answer 17

From the expansion above, compared to your equation you know that

m+n = 3
mn=-10

Find m and n so these are true

Answer 18

5 + -2 for the first and 5 times -2 for the second

Answer 19

Good

so

(x+5)(x-2) = x^2 +3x -10

OK?

Answer 20

ok

Answer 21

so going back to the original equation:

x^2 +3x -10=0

we can change it to
(x+5)(x-2) =0

OK?

Answer 22

ok

Answer 23

so the 2 answers are

x+5=0
x-2 = 0

solve for the 2 separate values of x....

Answer 24

-5 for the first one and 2 for the second one

Answer 25

That is correct.

I have to go - but

For the third question:

y^2-y-6 =0

(y+m)(y+n)=y^2-y-6

SO
m+n = -1
mn = -6

Do the same as above for the solution...

Answer 26

You're welcome, don't mention it

Answer 27

ok I am trying to write this out and explain it in a post for my class, and have no idea how to show my work of working out the problem. I understand what was said, and I understand the solution, but how to write it all out and illustrate the steps to solve?

Answer 28

First - NOT ALL QUADRATICS CAN BE FACTORISED.

the general form is
ax^2 + bx +c =0

We will deal for now with the situation where a = 1:

If it can be factorised then by definition

x^2 + bx +c= (x+m)(x+n)=0

so if you expand the brackets
ax^2+bx+c = x^2+ x(m+n) +mn

so you can see that
m+n= b
mn = c

SO we need to find 2 numbers m & n such tha tey multiply to make c, and add to make b.

The best way to do this is to look at the factors of c, and find a pair that add up to b

When you have found m and n you can write the equation

(x+m)(x+n)=0

so either (x+m) = 0 or (x+n) =0

These two solutions for x are the solution for the original quadratic