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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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<h2> StartFraction 7 over 10 EndFraction x + 2 and one-half y + 6</h2>

Step-by-step explanation:

Given the expression 2(\frac{3x}{5}+2\frac{3y}{4}-\frac{x}{4}-1 \frac{1}{2}y+3)

To simplify the expression, we need to first collect the like terms of the functions in parentheses as shown;

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