More solving quadratic equations: the quadratic formula again, what we've all been waiting for. the quadratic formula is pretty
easy to remember, and will always work, even if the solutions to the quadratic equation aren't always real numbers (they can be imaginary - the square roots of negative numbers). so here it is: _______ x = -b ± √b2 - 4ac 2a
1 answer:
You don't have a question in there but I can tell you that the quadratic formula that is shown is not quite correct.
<span>x = -b ± √b2 - 4ac 2a
For one thing, the first 3 terms should ALL be divided by 2a
The square root symbol should be extended from b^2 through 4ac
The division symbol is missing.
So the quadratic formula SHOULD read:
x = [-b +- sq root (b^2 - 4ac) ] / 2a
Also I have attached a graphic of the quadratic formula.
</span>
<span>x = -b ± √b2 - 4ac 2a
For one thing, the first 3 terms should ALL be divided by 2a
The square root symbol should be extended from b^2 through 4ac
The division symbol is missing.
So the quadratic formula SHOULD read:
x = [-b +- sq root (b^2 - 4ac) ] / 2a
Also I have attached a graphic of the quadratic formula.
</span>
You might be interested in
9514 1404 393
Answer:
see below
Step-by-step explanation:
It is easiest to compare the equations when they are written in the same form.
The first set can be written in slope-intercept form.
y = 2x +7
y = 2x +7 . . . . add 2x
These equations are <em>identical</em>, so have infinitely many solutions.
__
The second set can be written in standard form.
y +4x = -5
y +4x = -10
These equations <em>differ only in their constant</em>, so have no solutions.
__
The third set is already written in slope-intercept form. The equations have <em>different slopes</em>, so have exactly one solution.
