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hoa [83]
3 years ago
7

Quincy uses the quadratic formula to solve for the values of x in a quadratic equation. He finds the solution, in simplest radic

al form, to be x = \frac{-3± \sqrt{-19}}{2}.
Which best describes how many real number solutions the equation has?

A. Zero, because the discriminant is negative.
B. Zero, because the discriminant is not a perfect square.
C. One, because the negative and the minus cancel each other out.
D. Two, because of the ± symbol.
Mathematics
2 answers:
Alexxx [7]3 years ago
8 0

Answer:

<h2>A. Zero, because the discriminant is negative. </h2>

Step-by-step explanation:

I just did it.

tatiyna3 years ago
8 0

Answer:

A. Zero, because the discriminant is negative.

Step-by-step explanation:

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

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Consider the initial value problem:
Inessa [10]

Answer:

\left \{ {{u'=v} \atop {v'=5v-6u-5\sin(2t)}} \right. \\u(0)=-5;u'(0)=2

Step-by-step explanation:

The given initial value problem is;

y''-5y'+6y=-5\sin(2t)---(1)\\y(0)=-5,y'(0)=2

Let

u(t)=y(t)----(2)\\v(t)=y'(t)----(3)

Differentiating both sides of equation (1) with respect to t, we obtain:

u'(t)=y'(t)---(4)\\ \\\implies u'(t)=v(t)---(5)

Differentiating both sides of equation (2) with respect to t gives:

v'(t)=y''(t)----(6)

From equation (1),

y''=5y'-6y-5\sin(2t)\\y''=5v(t)-6u(t)-5\sin(2t)\\\implies v'(t)=5v(t)-6u(t)-5\sin(2t)----(7)

Putting t=0 into equation (2) yields

u(0)=y(0)\\\implies u(0)=-5

Also putting t=0 into equation (3)

u'(0)=y'(0)\\u'(0)=2

The system of first order equations is:

\left \{ {{u'=v} \atop {v'=5v-6u-5\sin(2t)}} \right. \\u(0)=-5;u'(0)=2

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