Quincy uses the quadratic formula to solve for the values of x in a quadratic equation. he finds the

8 answers
Question:

Quincy uses the quadratic formula to solve for the values of x in a quadratic equation. he finds the solution, in simplest radical form, to be x = startfraction negative 3 plus or minus startroot negative 19 endroot over 2 endfraction.

Answers

zero, because the discriminant is negative

Step-by-step explanation:

The next question is missing : Which best describes how many real number solutions the equation has?

The quadratic equation gives:

[tex]x = \frac{-3 \pm \sqrt{-19}}{2}[/tex]

The discriminant is equal to -19. A negative discriminant means the quadratic formula has zero real solution.

 The answer is zero because the discriminant is negative.

according to spark notes: There can be 0, 1, or 2 solutions to a quadratic equation, depending on whether the expression inside the square root sign, (b2 - 4ac), is positive, negative, or zero. This expression has a special name: the discriminant.

In this case it is 0, not 2, which was my first and wrong answer

2 solutions

Step-by-step explanation:

A quadratic equation has the formula:

[tex]ax^{2} + bx +c = 0[/tex]

The solution of the equation is given as:

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

The expression: [tex]b^{2}-4ac[/tex] is known as the discriminant given by the symbol D.

Thus, the discriminant D, is given as D = b² - 4ac

There are several conditions to the solution.

If D < 0 the roots are imaginary. They are not real.  

If D = 0, the solution has one real root

If D > 0, the solution as two distinct real roots (negative or positive)

A quadratic equation has only two real roots or solutions.

A quadratic equation in the form of

 a x² + b x + c=0, has

Discriminant = D= b²- 4 ac

x = [tex]\frac{-b \pm \sqrt{D}}{2a}[/tex]

Now coming to roots of a quadratic equation

1. D≥0, both the roots are real i.e both of them may be rational or both of them may be rational.

2. D=0, both the roots are real and equal.

3. D< 0, both the roots are imaginary.

So, out of following options given,

Option A, is not true,

zero, because the discriminant is negative.(It is a true statement if you are talking about real roots but if we consider imaginary roots also then this statement becomes false.) .In the question it is given that  that roots are in radical form that's why this option is incorrect.

. Option B is not true because if Discriminant is not a perfect square then also the quadratic function has two real either rational or irrational roots.

Third option is completely false , it is incorrect statement.[one, because the negative and the minus cancel each other]

Fourth option is true because , in the answer it has been written that the roots are in simplest radical form , The value of D should always be greater than zero,then we look at ± symbol .Then there are two possible real roots.

A. Zero, because the discriminant is negative.

Step-by-step explanation:

Square root of a negative number is not real, hence no real roots

A. Zero, because the discriminant is negative.

Step-by-step explanation:

I just did it.

The answer is zero because the the discriminant is negative

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