-9x^2-6x-1= 0?

Question

Looking for the discriminate:
and number of real solutions:
This question has me confused

So far:

-(9x^2+6x+1)=0
x=(-6-(6^2-4*9*1)Ý2)/(2*9)

my answer is -1/3 for the discriminate
Am I doing this corectly?

Solutions:

x=-2*3/(2*3^2),x=-2*3/(2*3^2)
x=-1/3^(2-1),x=-1/3^(2-1)
x=-1/3,x=-1/3

Answer 1

I think I get zero for the discriminant (b² - 4ac) meaning 1 real root...

When you have a quadratic equation in the form:
ax² + bx + c = 0

The quadratic equation says:
x = [ -b +/- sqrt(b² - 4ac) ] / 2a

The portion under the sqrt sign is called the 'discriminant'. The discriminant is b² - 4ac, not the whole quadratic formula.

If the discriminant is zero, then notice how the +/- will give you the same answer. When the discriminant is zero, there is *one* real root.

If the discriminant is positive, then you'll take the sqrt and the +/- will mean you will have *two* real roots.

Finally, if the discriminant is negative, then you'll have two imaginary (complex) roots.

In your example you have:
a = -9
b = -6
c = -1

The discriminant is b² - 4ac:
D = (-6)² - 4(-9)(-1)
D = 36 - 36
D = 0

So you have *one* real root.

Continuing with the quadratic formula you have:
x = ( -b +/- 0 ) / 2a
x = -b / 2a

In your example:
x = -(-6) / 2(-9)
x = 6 / -18
x = -1/3

So you have *one* real root of x = -1/3.

It it helps to visualize the problem, I've attached a graph of the equation below. You can see it is an upside down parabola that crosses the x-axis at only one point (x = -1/3). Thus it has only one real root.

Answer 2

The above equation can also be written as
9x^2 + 6x + 1 =0
Factorizing
(3x + 1)^2 = 0
so x = -1/3, -1/3

Answer 3

9x^2 + 6x + 1 =0
Factorizing
(3x + 1)^2 = 0
so x = -1/3, -1/3

Answer 4

There is no confusion.
It is simple.
-9x^2-6x-1=0
multiply by -1 the
entire equation
9x^2+6x+1=0
9x^2+3x+3x+1=0
3x[3x+1]+1[3x+1]=0
[3x+1][3x+1]=0
[3x+1]^2=0
x=-1/3,-1/3
the roots are real,rational
and equal since the expression
is a perfect square

Answer 5

what i would do is distribute the negative into the equation which will change all the signs. next plug the coefficients(#s in front of X) into the quadratic equation. the discriminate is the stuff underneath the radical.

-9x^2-6x-1
(6+-sq. root(6^2-4(-9)(-1)))/2(-9)
discriminate: 0
(6+- sq. root(0))/-18
there is one real zero, it is -1/3

Answer 6

discriminate=b^2-4ac
change eqn to
9x^2+6x+1
disc=36-4*9*1=36-36=0

factors are (3x+1)^2
x=-1/3

on the right track. -1/3 is the solution. The discriminate=0

Answer 7

discriminant is b^2 -4ac and = 36-4*9*1 =0 roots are equal and = -1/3
do yu need all this in this minor ques.
-9x^2 -3x -3x -1 =0
-3x(3x+1) - 1(3x+1) =0
(-3x-1)(3x+1)=0
x= -1/3

Answer 8

-9x^2-6x-1= 0

When you mention 'discriminant', you undoubtedly refer to the quadratic formula

[-b +- sqrt(b^2 - 4ac)]/2a

Which is the solution to the general quatratic equation

ax^2 + bx + c = 0

b^2 - 4ac is the discriminant

If it is > 0, two unequalreal roots
if It is = 0, two equal real roots
if it is < 0, two unequal imaginary roots

Use the values of a, b, c in your equation to answer your questions.

Answer 9

3x2 – 7x + 3 =0 a million. First multiply the coefficient of x^2 with the consistent(3). 2. Then get 2 factors of the ensuing term so as that their sum is the coefficient of the middle term(-7). 3. Then replace the middle term coefficient of the above equation with those factors and you will get 4 words rather of three. 4. Distribute those into 2 communities. 5. Take easy component from each and each team after which finally from the words which provide you with the factorization of the given espression. 6. placed each and each component =0 and you will get the fee of x.

Answer 10

I think you are looking for discriminant.

discriminant = b^2 - 4ac (as in ax2+bx+c)

so rewriting your equation you get:
9x^2+6x+1=0

b^2 - 4ac = 36 - 36 = 0

Therefore there is only one real solution.