Compute the value of the discriminant and give the number of real solutions to the quadratic equation?

Question

9x^2+6+1=0

Discriminant:
Number of real solutions:

3x=-1
3x/3=(-1)/3
Am I headed the right direction?

Answer 1

You will recall the quadratic formula:

x=(-b±√(b²-4ac))/2a

This gives you all the solutions for the quadratic equation. The quantity inside the square root:

b²-4ac

is called the discriminant. It is easily seen that if this quantity is positive, the quadratic has two distinct real solutions. Conversely, if it is negative, the solutions of the quadratic are both non-real complex numbers. Finally, if it is zero, both roots of the quadratic equation are the same and thus there is only one real solution.

The discriminant in your case is:

6²-4(9)1 = 36-36 = 0

So your equation has one real solution.

Edit: I'm assuming you meant 9x²+6x+1=0. If you actually meant to put 6+1, then you have no real solutions (the solutions would be ±i√7/3, which are not real)

Answer 2

You may have typed the equation incorrectly.
1) If it is indeed 9x^2+6+1=0, then x^2=-7/9, and the solutions are the two imaginary numbers x=-sqrt(7)/3i and x=+sqrt(7)/3i, where i = sqrt(-1). Note that the i is in the numerator.
2) If you meant to write the equation as 9x^2+6x+1=0, then the discriminate is 36-(4)(9)(1) which =0. Therefore the discriminate is zero, and the double solution is -1/3. This comes from the fact that the quadratic equation Ax^2+Bx+C=0 has a discriminate of B^2-4(A)(C) and a solution x={-B +/- sqrt[B^2-4(A)(C)]}/2A).

Answer 3

if the equation is as you have written it. It would be false no solutions, but if you made a miss print and the equation is

9x^2+6x+1 = 0 then your on the right track

you can factor

(3x+1)(3x+1) = 0 solve for x

x = -1/3

even with the quadratic formula it would be

x = -b (+, -) sqrtb^2-4ac/2a
x = (-6+sqrt6^2-(4)(9)(1))/((2)(9))
x = -1/3

x = (-6-sqrt6^2-(4)(9)(1))/((2)(9))
x = -1/3