Let A and B be any number. Let's say you have (5-A)(10-B) = 0.
For this to be true, either (5-A) or (10-B) must be zero for the the answer to equal zero. Essentially what you are saying is 0(10-B)=0 or (10-A)0=0. This let's you solve for A and B.
You can just set each "factor" to zero and solve for the variable.
5-A=0
A=5
10-B=0
B=10
Another "un-factored" example:
x2 - 10x = -25
x2 - 10x + 25 = 0 <--- the factors have to be equal to zero for this to work
You need two numbers that add to negative 10 and multiply to 25.
(-5) + (-5) = -10
(-5) x (-5) = 25
Therefore:
(x-5)(x-5)=0
And x can only be equal to 5, x=5
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Solve the equation - by whatever means available to you: factorising, graphical, numerical approximations, etc.
This is a quadratic equation in the form of: 52 - 25x + 30 = 0 Solve by using the quadratic equation formula: x = 2 or x = 3 Or by factorising: (5x-10)(x-3) = 0, which means that x = 10/5 or x = 3
solve it
If you solve such an equation for "y", you get an equation in the slope-intercept form.
you can only solve for one in an equation so it can equal something