How would you find the zeroes for this?
x^2+3x-9
The zeroes of the function or "roots" are not integers or fractions in this case. Use the quadratic equation.
The roots are
[-3 +/- sqrt (9+36)]/2 = -4.854 and 1.854
can't you partial factor
x^2+3x+9
x^2+3x=0
x(x+3)=0
x= 0 or x=-3
what am I doing wrong?
The equation you wrote was x^2+3x-9, not x^2+3x+9. Neither factors in the manner you have shown. You can't just throw out the 9, as you have done.
Then when is partial factoring used?
To find the zeroes of a quadratic equation, we need to solve for the values of x where the equation equals zero. In other words, we are looking for the x-values when the equation crosses the x-axis.
For the quadratic equation x^2 + 3x - 9, we can find the zeroes by setting the equation equal to zero:
x^2 + 3x - 9 = 0
There are a few methods to solve this equation and find the zeroes. One common method is factoring, but this equation may not be easily factorable. In such cases, we can use the quadratic formula.
The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the zeroes (or roots) can be found using the formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, a = 1, b = 3, and c = -9. Plugging these values into the quadratic formula, we get:
x = (-3 ± √(3^2 - 4(1)(-9))) / (2(1))
Simplifying further, we have:
x = (-3 ± √(9 + 36)) / 2
x = (-3 ± √45) / 2
Since the square root of 45 is not a perfect square, we cannot simplify it any further. Therefore, the zeroes of the quadratic equation x^2 + 3x - 9 are:
x = (-3 + √45) / 2 and x = (-3 - √45) / 2