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4 votes
What are the real zeros for the function
f(x)=x^2-3x-28

2 Answers

4 votes

x^2-3x-28=0\\ x^2-7x+4x-28=0\\ x(x-7)+4(x-7)=0\\ (x+4)(x-7)=0\\ x=-4 \vee x=7
answered
User Menno Jongerius
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Factor the equation: (x-7)(x+4) Solve for X: X-7=0 x+4=0 X=7 x=-4 A zero is when the function crosses the x-axis which is solved by setting the function equal to zero.
answered
User Qwattash
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8.2k points

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