Q: If 5x plus x2 100 then x is not?

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x2 + 5x + 6 = (x + 2)(x + 3)

x2 + 5x + 3 = 9 x2 + 5x - 6 = 0 (x + 6)(x - 1) = 0 x = -6 or x = 1

x2 + 5x - 6x = x2 - x = x(x - 1) which is zero when x = 0 or x = 1

x2+4x-9 = 5x+3 x2+4x-5x-9-3 = 0 x2-x-12 = 0 (x+3)(x-4) = 0 x = -3 or x = 4

x(x+5)

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x2 + 5x + 6 = (x + 2)(x + 3)

x2 + 5x + 3 = 9 x2 + 5x - 6 = 0 (x + 6)(x - 1) = 0 x = -6 or x = 1

x2 - 5x + 6 = (x - 2)(x - 3)

x2+5x+6 = (x+2)(x+3)

x(x2+5x+6)

x2 + 5x - 6x = x2 - x = x(x - 1) which is zero when x = 0 or x = 1

x2+5x-14 = (x+7)(x-2) when factored

x(x+5)

x2+4x-9 = 5x+3 x2+4x-5x-9-3 = 0 x2-x-12 = 0 (x+3)(x-4) = 0 x = -3 or x = 4

-(x - 7)(x + 2)

(x+1)(x+4)

=(x+3)(x+2)