how do i solve for x for these?

Question

x^2 + 4x - 5 = 0

3x^2 - 2x - 3 = 2

Answer 1

Factorization.

(x + 5)(x - 1)

and

(3x - 5)(x +1)

Answer 2

x^2 + 4x - 5 = 0

I'm sorry to say that the factors just jumped out at me.
But here's kind of how the thought process went....
In elementary examples, if you have an x² term with the coefficient 1 (i.e., (1)(x²)) you can generally presume your factors will look like
(x + something)(x + something)

Looking at the 5, the factors that produce 5 are 1 and 5... (1)(5)=5. Again we're talking elementary examples.

Another clue that the numeric coefficients are probably integers is that the middle term is an integer.

So now we're looking at (x+5)(x-1) or (x-5)(x+1) I know the signs have to be different because it's a -5. if they were the same the product would be positive.
Now, I look at the middle term of the original equation. +4x

I know that the middle term has to be 5x-x (the first set of factors), or -5x+x
5x-x=4x So (x+5)(x-1) are the factors.
If (x+5)(x-1)=0, either x+5=0, or x-1=0
if x+5=0, then x=-5 I added -5 to both sides
Similarly
if x-1=0, then x=1 I just added 1 to both sides.
===================
The second one is a little trickier.
First you have to get 0 on one side of the equation.
I'd do that by adding -2 to both sides
That gives
3x² - 2x - 3 = 2
3x² - 2x – 5 = 0
Now… The factors don’t jump out at me… but the thought process is the same.
Because the factors of 3 are 1 and 3, I’m pretty certain the factors will look like
(3x + something)(x + something)
The factors for 5 are most likely to be 1 and 5. And because it’s -5, I know the signs are not the same. So we have
(3x + 5)(x – 1), or (3x – 5)(x + 1), or (3x+1)(x – 5), or (3x – 1)(x + 5)
Right away I eliminate the latter two sets of factors because the result is going to be -14x or 14x
(-1)(3x)+(5)(x)= 2x, not -2x, so this isn’t it
(3x)(1) + (-5)(x)= -2x. That’s the one we’re looking for.
So the factors are
(3x – 5)(x + 1)
Complete this one the same way as the last.
Set each factor to zero, and solve for x.

Answer 3

1) X^2 + 4x - 5 = 0

Factor it

(x + 5) (x-1) = 0

To make the equation true, either (x+5) is 0 or
(x - 1) is zero

x + 5 = 0
x = -5

or
x - 1 = 0
x = 1

So x = -5 or x = 1
To check:
if x = -5
(-5)^2 + 4(-5) - 5 = 0
25 - 20 - 5 = 0
0 = 0

if x = 1
1^2 + 4(1) - 5 = 0
1 + 4 - 5 = 0
0 = 0

2) Use the quadratic formula

First make the equation in the form
ax^2+ bx + x = 0
3x^2 - 2x - 5 = 0

a = 3
b = -2
c = -5

Quadratic formula is

x = -b +/- sqrt b^2 - 4ac divided by 2a

x = -(-2) +/- sqrt (-2)^2 - 4(3)(-5) / 2(3)

x = 2 +/- sqrt 4 + 60 / 6

x = 2 +/- sqrt 64 / 6

x = 2 +/- 8 / 6

x = 2 + 8/ 6 = 10/6 = 5/3 or 1 2/3

x = 2 - 8/6 = -6/6 = -1

x = 1 2/3 and -1

Answer 4

use the following formula
if ax^2 + bx + c =0 then x= (-b ± √(b^2 - 4a*c))/2a
(i) for x^2 + 4x - 5 = 0
a = 1 ; b = 4 ; c = -5
so x = (-4 ± √(4^2 - 4*1*(-5)))/2*1
= (-4 ± √(16 + 20))/2
= (-4 ± √(36))/2
= (-4± 6)/2
= (-4 + 6)/2 , (-4 -6)/2
= 1,-5
(ii) for 3x^2 - 2x - 3 = 2
a = 3 ; b = -2 ; c = -5
so x = (2 ± √(4 - 4*3*(-5)))/2*3
= (2 ± √(4 + 60))/6
= (2 ± √64)/6
= (2± 8)/6
= (2 + 8)/6 , (2 - 8)/6
= 5/3 , -1

Answer 5

The extreme fee(s) is even as the inequality is merely truly real. So set it to equals, and remedy: x^2 + x = 30 x^2 + x - 30 = 0 x = [-a million +/- sqrt( a million + 4*(-30) ) ] / 2 x = [-a million +/- sqrt( 121 ) ] / 2 x = [-a million +/- 11 ] / 2 x = -12/2, or x = 10/2 x = -6 or 5 somewhat more effective questioning shows that if x < -6, or x > 5, then the unique inequality is real. verify the severe values: (-6)^2 - 6 = 36 - 6 = 30 (5)^2 + 5 = 25 + 5 = 30

Answer 6

1) Factor the expression such as:

(x-1)(x+5)=0

This goes to reason that x-1=0 or x+5=0
Thus, x=1 or x=-5

Number two follows the same way, just subtract 2 from both sides to have a 0 on the right.

3x^2-2x-5=0

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

Thus x=5/3 or -1

Answer 7

You could solve these equations simply by middle term factorisation:
x^2 + 4x - 5 = 0
or, x^2 +5x - x -5= 0
or, x(x+5)-(x+5)=0
or,(x-1)(x+5)=0

Therefore, x-1=0,
x=1

or, x+5=0
x= -5

The second equation can be solved the same way.

The general algorithm is-

A) See the equation in the form of ax^2 + bx + c
B)Think of two numbers whose product is (a X c), and whose sum is b.
C)Split the middle term into those two no.s as coefficients of x.
D)Now you'll be able to express the polynomial as the product of two expressions, which is equal to zero.
E)you'll get two values of x.

For equations where the middle term can not be split as easily, you can use the formula:
http://en.wikipedia.org/wiki/Quadratic_equation#Quadratic_formula

Answer 8

factor
(x+5)(x-1)=0
so one of the terms in brackets must be 0
either x+5=0 or x-1=0 so x=-5 or 1

Second one...
set = 0 by subtracting 2 from both sides
3x^2-2x-5=0
Factor..
(3x-5)(x+1)=0 ... check by FOIL if you don't buy it!

same as above

either ....
3x-5=0 >>> add 5 to both sides
3x=5<<<< div. both sides by 3
x=5/3

or....
x+1=0
x=-1

so x=5/3 or -1

Answer 9

these are quadratic equations, and i think the easiest way is to use the formula

find out about it here: http://en.wikipedia.org/wiki/Quadratic_formula#Quadratic_formula

and so the 1st answer is
[-4 plus/minus sqrt (4^2 - 4*1*(-5))] divide by 2*1
this equals to
1 or -5

for the second answer, you have to make one side zero so it becomes 3x^2 - 2x - 5 = 0
and put it in the formula the answer is 1 and -1/3

Answer 10

Question 1
(x + 5).(x - 1) = 0
x = - 5, x = 1

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