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zloy xaker [14]
3 years ago
12

Explain why a quadratic equation with a positive discriminant has two real solutions, a quadratic equation with a negative discr

iminant has no real solution, and a quadratic equation with a discriminant of zero has one real solution.
Mathematics
1 answer:
Bad White [126]3 years ago
3 0

Answer:

A quadratic equation can be written as:

a*x^2 + b*x + c = 0.

where a, b and c are real numbers.

The solutions of this equation can be found by the equation:

x = \frac{-b +- \sqrt{b^2 - 4*a*c} }{2*a}

Where the determinant is D = b^2 - 4*a*c.

Now, if D>0

we have the square root of a positive number, which will be equal to a real number.

√D = R

then the solutions are:

x = \frac{-b +- R }{2*a}

Where each sign of R is a different solution for the equation.

If D< 0, we have the square root of a negative number, then we have a complex component:

√D = i*R

x = \frac{-b +- C*i }{2*a}

We have two complex solutions.

If D = 0

√0 = 0

then:

x = \frac{-b +- 0}{2*a} = \frac{-b}{2a}

We have only one real solution (or two equal solutions, depending on how you see it)

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Answer:

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Step-by-step explanation:

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Solve –4(6x + 3) = –12(x + 10). A. x = 2 B. x = 9 C. x = 5 D. x = –5
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answer
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</span>
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