Verified answer

the discriminant b^2 -4ac=36+36=72

it is not a perfect square.so the roots will be irrational numbers

x1=[-6+rt(72)]/2=[-6+6rt2]/2

=-3+3rt2=3(rt2-1)

x2=[-6-rt(72)]/2=[-6-6rt2]/2

=-3-3rt2=-3(rt2+1)

I'm not going to solve that equation, but I'll solve this one:

y^2+2y-3=0

use the Quadratic Formula which is:

y = [ -b ± sqrt(b^2 - 4ac) ] / 2a

where a = the # in front of the squared term, b = # in front of the single variable and c = the remaining number with no variable.

so for my example you would do this:

y= [-2 ± sqrt(2^2 - 4*1*(-3)) ] / (2*1)

You can do this by hand:

2^2 = 4 and 4*1*(-3)=(-12) so 4 - (-12) = 16 and the sqrt of 16 = 4

so you have (-2 ± 4)/2

your answers would then be (2/2)=1 or (-6/2) = (-3)

solving that in any calculator would give you the same answers

The tricks to remember: if the "b" term is negative (as in y^2-3y-5) then it would be - (-3) and give you a 3 ± sqrt ....

you cannot take the sqrt of a negative number so if you get one then your equation either is imaginary or you put it in wrong.

You cannot divide by zero so if you have no variable^2 term then you cannot use this formula.

Hope that helps.

I have to ask if you've entered the question correctly because you have a +6x and a -9. For example, when you factor this out (x -3)^2. You can't get a -9, you get a +9. When you factor (x+3)^2 you get a +6x and a +9. This doesn't solve x either way.

For example to solve x^2 +6x -9 =0

(x+3) (x-3) = x^2 -9 (the 3's cancel out) no good

(x+3) (x+3) = x^2 +6x +9 no good

(x-3)(x-3) = x^2 -6x + 9 no good

Since, none of these work then your question must have an error (probably in the signing).

But, let's say one of these did work out (They Don't) but just for kicks..... then

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

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

x+3=0 x-3=0

x= -3 x= +3

This is how you would solve for x.

I hope this helps!

It is hard to write the equations properly using the computer text, but here goes:

With an equation of the form ax^2 + bx +c = 0, use the solution formula for quadratic equations:

x=[-b +/- sqr rt{b^2 - 4ac}]/2a

x equals negative b plus or minus(two solutions) the square root of b squared minus 4 times a times c, all of which is divied by 2 times a.

x = -6 +/- sqr(6^2 -4*1*-9)/2(1) = [-6 +/- sqr(72)]/2 = -3 +/- sqr(18) { the sqr(72) divided by 2 equals the sqr(18)}

the solutions are x = -3 + sqr(18) =1.24 and

x = -3 - sqr(18) =-7.24 (both rounded off to the second decimal place.

For all quadratic equations, there are two situations.

Algorithm

Step 1

Can you find two factors of quadratic equation?

- If the answer is yes, so solve it by factorization

- If the answer is no , so go to step 2.

Step 2

You must apply quadratic formula:

x = -b +/- sqrt ( b^2 - 4ac)/(2a)

to solve this equation: ax^2 + bx + c = 0 (*) => x1 = u1 , x2 =u2

Cases: 1) b^2 -4ac = 0 => 1 real solution

2) b^2 -4ac > 0 => 2 reals roots

3) b^2 -4ac < 0 => imaginary roots

N.B.: Verify carefully your solutions ''u1'' and ''u2'' by substitution

each solution in (*).

I will refer you to excellent web site below. All you must know to

solve properly quadratic equations is there:

- Proof of quadratic equation.

- Factorization

- Square completion

- and so on...

Now, I solve your problem

Step 1 => no factorization, so I go to step 2

Step 2 : We have ax^2 + bx + c =0,

with a = 1

b = 6

c = -9

So x = - 6 +/- sqrt (72)/2

We find two reals roots: x1 = 1.2425

x2 = -7.2425

Verification: ok

x^2+6x-9=0

We know that x^2+6x+9 = (x+3)^2

So by adding 9 and subtracting 9 in the equality we get

x^2 + 6x + 9 - 18 = 0

Which gives (x+3)^2 =18

Which gives x+3 = +sqrt(18) or -sqrt(18)

So x = -3 + sqrt(18) or -3 - sqrt(18)

Another way to solve the problem is applying the Sridhara Acharya's formula to solve quadratic equations.

It says that if ax^2 + bx + c = 0 then

x= [-b +- {sqrt(b^2 - 4ac)}]/2a

Using the formula straightaway,

x^2 + 6x - 9 = 0

a=1 b=6 c= -9

x= [-6 +- sqrt(6^2 + 4*1*9)]/2

By canceling two, we get the same answer as above.

x=3

X^2+6X-9 =0

this is of the form

x^2+bx + c =0

add (b/2) ^2 -(b/2)^2 that is zero

this is because x^2+bx+(b/2)^2=(x+b/2)^2

so x^2+bx = (x+b/2)^2-b^2

comparing with your eqution b = 6 or b/3 = 3

x^2+6x-9 = 0

=>(x+3)^2-9-9 =0

=>(x+3)^2-18 = 0

as x^-a^=0 has roots x = a or -a

=>(x+3) = Sqrt(18) or -sqrt(18) = 3sqrt(2) or -3 sqrt(2)

=> x = -3+3sqrt(2) or -3-3sqrt(2)

x=3

Apparently there are _two_ values of x that satisfy that equation. Going to a handy online function grapher, I see that the graph of this parabola crosses the y axis (y=0) at x= -7.23 (approximately) and x=1.25 (again, approximately).

I hope you get good use out of the online function grapher!

To get exact values, complete the square:

x²+6x+9=18 (adding 18 to both sides).

(x+3)² = 18

x+3 = ±√18

x+3 = ±3√2

x = ±3√2 - 3

which, not too surprisingly, comes out to either -7.2 something or 1.2 something.

Hey. Find Delta = b^2 - 4ac. If delta is negative, then you have imaginary roots. If Delta is 0, you have a single root. If delta is positive you have 2 different real roots

In this case: Delta = 6^2 - 4 . 1 .(-9) = 36+ 36 = 72. Delta is positive, then you can find the 2 roots

x1 =[ -b - V(delta)] / 2

And

x2 = x1 =[ -b - V(delta)] / 2a

I am using V for the square root. You can write too sqrt(delta), this is the way its usually written

Ana