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Find the zeros of this function f (x)=x^2-25
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You are just being asked to treat the function as a quadratic equation and solve for \[x\]
that is it
Actually thats just factoring, you have to find the values of \[x\] that 'zero' the function.
True, so, after you factor, get x alone for each x in the "()" i.e. (x-5)(x+5) x-5=0 x=5 I think you can get the other one
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\[ x^2-25\qquad\text{is of the form}\\ a^2-b^2=(a+b)(a-b)\\\text{we can write}\\ x^2-25^2=(x)^2-(5)^2\\=(x+5)(x-5) \\\text{to find roots, set it equal to } 0\\ (x+5)(x-5)=0\\\text{i.e., either}\\ x+5=0\qquad{\rm or}\qquad x-5=0\\ \implies \boxed{x=-5}\qquad{\rm or}\qquad \boxed{x=5} \]
kapeesh?
At least let the man think, @electrokid .
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