Quadratic equation?

Question

I need someone to solve these 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!

1. x^2 - 4x - 5 = 0

2. z^2 - 2z - 15 = 0

3. y^2 = 2y + 3

4. r^2 = 5 - 4r

Once again, thanks for the help!

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

Answer 1

ok
the general ec. is

x=[ -b±√(b²-4ac)]/2a
for #1, a=1, coef. of x²
b= -4, coef. of x
c= -5, indep. term.

then apply the gral. formula

x= [-(-4)±√[(-4)²-4(1)(-5)]]/2(1)

x=[4±√(16+20)]/2

x=[4±√36]/2=[4±6]/2

x1= [4+6]/2=10/2=5
x2= [4-6]/2= -2/2=-1

sol. x1=5; x2= -1

Applying the same method for the others you'll get:

2. x1= 5; x2= -3
3. x1= 3; x2= -1
4. x1= -5; x2=1

Answer 2

1.x^2-4x-5=0
=>x^2+x-5x-5=0
=>x(x+1)-5(x+1)=0
=>(x+1)(x-5)=0
Therefore either x+1=0 or x-5=0
If x+1=0,then x=-1
If x-5=0,then x=5
x= -1 or 5 ans
2) z^2-2z-15=0
=>z^2+3z-5z-15=0
=>z(z+3)-5(z+3)=0
=>(z+3)(z-5)=0
Therefore z= -3 or 5
3)y^2=2y+3
=>y^2-2y-3=0
=>y^2+y-3y-3=0
=>y(y+1)-3(y+1)=0
=>(y+1)(y-3)=0
Therefore y= -1 or 3
4.r^2=5-4r
=r^2+4r-5=0
=>r^2+5r-r-5=0
=>r(r+5)-1(r+5)=0
=>(r+5)(r-1)=0
Therefore r=-5 or 1
When a quadratic equation cannot be factored,there is another way to find the value of the variable.
The quadratic equation is essentially in the form of ax^2+bx+c=0,where "a" is the coeffeicient of x^2,"b" is the coefficient of x and c is the constant or numeral.
Let us take the first problem
it isx^2-4x-5=0
Here a=1,b= -4 and c=-5
The values of x is obtained ny the formula { -b+/-sqrt(b^2-4ac)}/2a
Hence x= {4+/-sqrt(16+20)}/2
=(4+/-6)/2
=2+/-3
=2+3 or 2-3
=5 or -1
Incidentally,who told you that quadratic equations cannot be solved by factoring.only when we fail to find proper factors ,then only we adopt the later process.

Answer 3

The simplest way to solve a quadratic equation is to factor it and then solve for the variable. The other way is to use the formula for the roots of a quadratic equation of the form
ax^2 + bx +c = 0 are:

{-b + sqrt(b^2 - 4ac)}/2a and {-b - sqrt(b^2 - 4ac)}/2a

For your problems the a, b and c are:

1. a = 1, b = -4, c = -5
2. a = 1, b = -2, c = -15
3. a = 1, b = -2, c = -3
4. a = 1, b = 4, c = -5

To learn more about quadratic equations refer to this link:

http://en.wikipedia.org/wiki/Quadratic_equation

Answer 4

You can use the quadratic equation to solve these, which is
(-b +/- sqrt b^2-4ac)/2a.
In any quadratic, ax^2+bx+c.
So, for #1...
a=1, b=-4, c=-5 so to solve...
(-(-4)+/- sqrt (-4)^2 - 4(1)(-5))/2(1)
4 +/- sqrt 16+20 / 2
4 +/- sqrt 36 / 2
(4 +/- 6)/2
2 +/- 3
x=5 or x=-1

Use the same formula to solve for all the other ones.

Answer 5

solution of an equation of the form
ax^2+bx+c=0 is

x= (1/2a)*{ -b+/-(b^2-4ac)^(1/2)}

Answer 6

1. http://www.webmath.com/cgi-bin/quadform.cgi?a=1&b=-4&c=-5&back=quadform.html

2. http://www.webmath.com/cgi-bin/quadform.cgi?a=1&b=-2&c=-15&back=quadform.html

3. http://www.webmath.com/cgi-bin/quadform.cgi?a=1&b=-2&c=-3&back=quadform.html

4. http://www.webmath.com/cgi-bin/quadform.cgi?a=1&b=-5&c=4&back=quadform.html

Answer 7

[-b +- sqrt(b^2-4ac)]/2a
1. [4+-sqrt(16+20)]/2
2. [-2+-sqrt(4+60)]/2
3. [-2+-sqrt(4+12)]/2
4. [-4+-sqrt(16+20)]/2