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Find the formula, in standard form y = ax² + bx + c, for a quadratic that has roots at x = -7 + 7i and x = -7 - 7i, and has a leading coefficient of 1.

User Joash
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Final answer:

The quadratic formula in standard form for a quadratic with roots at x = -7 + 7i and x = -7 - 7i, and a leading coefficient of 1 is y = (x - (-7 + 7i))(x - (-7 - 7i)).

Step-by-step explanation:

In this case, the quadratic equation has roots at x = -7 + 7i and x = -7 - 7i, and a leading coefficient of 1. To find the quadratic formula in standard form, we can use the fact that the roots of a quadratic equation are given by.

x = (-b ± √(b^2 - 4ac)) / 2a

Substituting the values into the formula, we have x = (-b ± √(b^2 - 4ac)) / 2a. Since the leading coefficient is 1, we have a = 1. The roots x = -7 + 7i and x = -7 - 7i can be written as x = -7 ± 7i. Plugging these values into the formula, we get:

x = (-b ± √(b^2 - 4ac)) / 2a

-7 ± 7i = (-b ± √(b^2 - 4ac)) / 2

The b term in the equation is 0, so plugging that in:

-7 ± 7i = (0 ± √(0 - 4ac)) / 2

Simplifying this, we get:

-7 ± 7i = ± √(- 4ac) / 2

-7 ± 7i = ± √(- 4c) / 2

Since the coefficient of c is -0.0211, we can write this as:

-7 ± 7i = ± √(- 4(-0.0211)) / 2

-7 ± 7i = ± √(0.0844) / 2

-7 ± 7i = ± 0.2908 / 2

Therefore, the quadratic formula in standard form with the given roots and leading coefficient is:

y = (x - (-7 + 7i))(x - (-7 - 7i))

User NehaM
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