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What are the zeros of the function defined by y = x2 + 6x + 5?

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To find the zeros of the function y = x^2 + 6x + 5, we need to solve the equation for x when y is equal to zero.

Setting y = 0, the equation becomes:
0 = x^2 + 6x + 5

To solve this quadratic equation, we can factor it or use the quadratic formula. In this case, let's use the quadratic formula, which states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by:

x = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = 1, b = 6, and c = 5. Substituting these values into the quadratic formula, we have:

x = (-(6) ± √((6)^2 - 4(1)(5))) / (2(1))

Simplifying further:

x = (-6 ± √(36 - 20)) / 2
x = (-6 ± √16) / 2
x = (-6 ± 4) / 2

This gives us two possible solutions:

x = (-6 + 4) / 2 = -1
x = (-6 - 4) / 2 = -5

Therefore, the zeros of the function y = x^2 + 6x + 5 are x = -1 and x = -5.
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User Dhawalk
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