'I get that when you factorise it, it becomes: (a-b)(a+b)+(a-b)
but HOW does it change into (a-b)(a+b+1)'
Take out a common factor of (a-b). That's all there is to it.
Since the first part is (a-b)(a+b), it turns into 1(a+b) when you divide that part by (a-b) for the factorising, which turns out as simply a+b.
The second part is simply 1, since you are dividing (a-b) by itself for the factorising.
So when you take out the factor of (a-b), you get:
(a-b)(a+b[from the first part]+1[from the second part])
The logical step from (a-b)(a+b)+(a-b) IS (a-b)(a+b+1), that's pretty much the simplest way to put it, lol.
Pretty confusing...
(a-b)*(a+b)+(a-b)
(a-b)*(a+b)+(a-b)*1 //multiplying by 1 doesn't change it at all
(a-b)*((a+b)+1) //(a-b) is a mutual multiplier
(a-b)*(a+b+1) // remove unnecessary brackets
Okay, I really suck at Quadratic Equations and I think its time I asked for help. There is only one question I need help in:
1) For the quadratic equation, x^2 - 4x + c = 0 where the equation has only one solution... Find the value of c.
How do I do that? Full working out please! The answer is 4 but I dont know why!!!!!
I think the c should be removed.
Since:
X^2 = x * x
X^2 - 4x = 0
X^2 = 4X
4 = X
Quoted:
2) I also suck at factorising. I can factorise using bridge method, cross method, etc....but the really hard ones...... like this one is hard!!!
What are the steps required to solve this???
TRIAL AND ERROR? OR AN ACTUAL METHOD???
Factorise: 16x^2 - y^2 - 6y - 9
The answer is: (4x - 3 -y) (4x + 3 + y)
BUT I DONT KNOW WHY :( I used the cross method and stuff, but It still does work
Factorise: a^2 - b^2 + a - b
The answer is (a-b) (a+b+1)
I get that when you factorise it, it becomes: (a-b)(a+b)+(a-b)
but HOW does it change into (a-b)(a+b+1)
16x^2 - y^2 - 6y - 9
16*x*x - y*y - 6*y -9
1. x is double
2. y is double
3. 6 and 9 shows that there is a double 3, since 3+3 = 6 and 3*3 = 9
4. Since x is double, it can be assumed that the factor multiplying 16 is also double, which is 4
5. X's are not in all parts of the formula and Y's are not in all parts of the formula, which means, there are minusses involved.
So what do we now know:
1. (X)(X)
2. (x + y)(x + y)
3. (x + y + 3)(x + y + 3)
4. (4x + y + 3)(4x + y + 3)
5. (4x - y - 3)(4x + y + 3)
'I get that when you factorise it, it becomes: (a-b)(a+b)+(a-b)
but HOW does it change into (a-b)(a+b+1)'
Take out a common factor of (a-b). That's all there is to it.
Since the first part is (a-b)(a+b), it turns into 1(a+b) when you divide that part by (a-b) for the factorising, which turns out as simply a+b.
The second part is simply 1, since you are dividing (a-b) by itself for the factorising.
So when you take out the factor of (a-b), you get:
(a-b)(a+b[from the first part]+1[from the second part])
The logical step from (a-b)(a+b)+(a-b) IS (a-b)(a+b+1), that's pretty much the simplest way to put it, lol.
Pretty confusing...
(a-b)*(a+b)+(a-b)
(a-b)*(a+b)+(a-b)*1 //multiplying by 1 doesn't change it at all
(a-b)*((a+b)+1) //(a-b) is a mutual multiplier
(a-b)*(a+b+1) // remove unnecessary brackets
Or just think harder on "where the equation has only one solution..."
Factorise: a^2 - b^2 + a - b
The answer is (a-b) (a+b+1)
I get that when you factorise it, it becomes: (a-b)(a+b)+(a-b)
but HOW does it change into (a-b)(a+b+1)
'I get that when you factorise it, it becomes: (a-b)(a+b)+(a-b)
but HOW does it change into (a-b)(a+b+1)'
Take out a common factor of (a-b). That's all there is to it.
Since the first part is (a-b)(a+b), it turns into 1(a+b) when you divide that part by (a-b) for the factorising, which turns out as simply a+b.
The second part is simply 1, since you are dividing (a-b) by itself for the factorising.
So when you take out the factor of (a-b), you get:
(a-b)(a+b[from the first part]+1[from the second part])
The logical step from (a-b)(a+b)+(a-b) IS (a-b)(a+b+1), that's pretty much the simplest way to put it, lol.
You only know x = -2 since you know the answer, that cheating :o
I think the c should be removed.
Since:
X^2 = x * x
X^2 - 4x = 0
X^2 = 4X
4 = X
Too bad you can't remove C since it's unknown value... also we are looking for C which is 4 and X is -2... but nice try XD
Besides I doubt that he will ever reply :O
Okay, I really suck at Quadratic Equations and I think its time I asked for help. There is only one question I need help in:
1) For the quadratic equation, x^2 - 4x + c = 0 where the equation has only one solution... Find the value of c.
How do I do that? Full working out please! The answer is 4 but I dont know why!!!!!
I think the c should be removed.
Since:
X^2 = x * x
X^2 - 4x = 0
X^2 = 4X
4 = X
What are the steps required to solve this???
TRIAL AND ERROR? OR AN ACTUAL METHOD???
Factorise: 16x^2 - y^2 - 6y - 9
The answer is: (4x - 3 -y) (4x + 3 + y)
BUT I DONT KNOW WHY :( I used the cross method and stuff, but It still does work
Factorise: a^2 - b^2 + a - b
The answer is (a-b) (a+b+1)
I get that when you factorise it, it becomes: (a-b)(a+b)+(a-b)
but HOW does it change into (a-b)(a+b+1)
16x^2 - y^2 - 6y - 9
16*x*x - y*y - 6*y -9
1. x is double
2. y is double
3. 6 and 9 shows that there is a double 3, since 3+3 = 6 and 3*3 = 9
4. Since x is double, it can be assumed that the factor multiplying 16 is also double, which is 4
5. X's are not in all parts of the formula and Y's are not in all parts of the formula, which means, there are minusses involved.
So what do we now know:
1. (X)(X)
2. (x + y)(x + y)
3. (x + y + 3)(x + y + 3)
4. (4x + y + 3)(4x + y + 3)
5. (4x - y - 3)(4x + y + 3)
Check:
(4x - y - 3)(4x + y + 3) = 4x*4x + 4x*y + 4x*3 + -y*4x + -y*y + -y*3 + -3*4x + -3*y + -3*3 =
16x^2 + 4xy + 12x - 4xy - y^2 - 3y - 12x - 3y - 9 =
16x^2 - y^2 - 6y - 9
Problem solved :D
1) For the quadratic equation, x^2 - 4x + c = 0 where the equation has only one solution... Find the value of c.
How do I do that? Full working out please! The answer is 4 but I dont know why!!!!!
2) I also suck at factorising. I can factorise using bridge method, cross method, etc....but the really hard ones...... like this one is hard!!!
What are the steps required to solve this???
TRIAL AND ERROR? OR AN ACTUAL METHOD???
Factorise: 16x^2 - y^2 - 6y - 9
The answer is: (4x - 3 -y) (4x + 3 + y)
BUT I DONT KNOW WHY :( I used the cross method and stuff, but It still does work
Factorise: a^2 - b^2 + a - b
The answer is (a-b) (a+b+1)
I get that when you factorise it, it becomes: (a-b)(a+b)+(a-b)
but HOW does it change into (a-b)(a+b+1)