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Q: How do you Evaluate x-2 for x equals -3?

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x2 + 6x + 12 = 0 x2 + 6x + 9 = -3 (x + 3)2 = -3 x + 3 = ± √-3 x = -3 ± i√3

x2 + 7x + 9 = 3 ∴ x2 + 7x + 6 = 0 ∴ (x + 6)(x + 1) = 0 ∴ x ∈ {-6, -1}

3 6

The answer is x = 5. There is nothing to evaluate there!

x2 - 8x + 15 = 0 can be factored as (x - 3)(x - 5) = 0 : This holds true when one or both of the terms in brackets equals zero. When x - 3 = 0 then x = 3 When x - 5 = 0 then x = 5.

Related questions

-4

-x2 = x - 6 x2 + x - 6 = 0 (x - 2)(x + 3) = 0 x ∈ {-3, 2}

3

x2 + 2x = 3 x2 + 2x - 3 = 0 (x + 3)(x - 1) = 0 x ∈ {-3, 1}

x2 - x - 6 = (x - 3)(x + 2) = 0; whence, x = 3 or -2.

Interpreting the question as "evaluate f(x) = -x2 + 5 for x = 4". This is the same as asking "what is f(4)". To evaluate f(4) just replace each instance of the x in the definition of f(x) by the number 4: f(4) = -42 + 5. The result is -11.

If: x2 = 3 Then: x = square root of 3

x2-2x-3 = 0 (x+1)(x-3) = 0 x = -1 or x = 3

x2 + 2x - 15 = (x - 3)(x + 5)

2x = x2 + 4x - 3 x2 + 2x - 3 = 0 (x - 1)(x - 2) = 0

-7

x2+3=124 x2=121 x= square root(121) x=11

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