Asked by skuni
find the sets of valuse of x for which 4x^2 + 19x - 5 <= 0
Answers
Answered by
Steve
4x^2+19x-5 = (4x-1)(x+5)
The roots (places where the graph crosses the x-axis) are -5 and 1/4
So, where do you think the value is negative?
The roots (places where the graph crosses the x-axis) are -5 and 1/4
So, where do you think the value is negative?
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