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What are the zeros of the quadratic equation #x^2+5x=-6#?

Answer 1

Zeros at #x=-2# and #x=-3#

#x^2+5x=-6# #hArrcolor(white)("XXX")x^x+5x+6=0#

#hArrcolor(white)("XXX")(x+2)(x+3)=0#

either #color(white)("XXX")(x+2)=0color(white)("XX")rarrcolor(white)("XX")x=-2# or #color(white)("XXX")(x+3)=0color(white)("XX")rarrcolor(white)("XX")x=-3#

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Answer 2

Savannah Anderson

To find the zeros of the quadratic equation (x^2 + 5x = -6), we first rewrite it in the standard form (ax^2 + bx + c = 0). Then, we can use the quadratic formula:

[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}]

Comparing (x^2 + 5x = -6) with (ax^2 + bx + c = 0), we have (a = 1), (b = 5), and (c = -6).

Substituting these values into the quadratic formula, we get:

[x = \frac{-5 \pm \sqrt{(5)^2 - 4(1)(-6)}}{2(1)}]

[x = \frac{-5 \pm \sqrt{25 + 24}}{2}]

[x = \frac{-5 \pm \sqrt{49}}{2}]

[x = \frac{-5 \pm 7}{2}]

Now, we have two solutions:

[x_1 = \frac{-5 + 7}{2} = 1]

[x_2 = \frac{-5 - 7}{2} = -6]

Therefore, the zeros of the quadratic equation (x^2 + 5x = -6) are (x = 1) and (x = -6).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.