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kari74 [83]
3 years ago
8

Which system of equations is represented by the matrix below?

Mathematics
1 answer:
Igoryamba3 years ago
7 0

Answer:

C

Step-Cby-step explanation:

The answer is C

it's the only one that makes sense

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Answer:

The two roots of the quadratic equation are

x_1= - \frac{1}{3} \text{ and } x_2= \frac{2}{3}

Step-by-step explanation:

Original quadratic equation is 9x^{2}-3x-2=0

Divide both sides by 9:

x^{2} - \frac{x}{3} - \frac{2}{9}=0

Add \frac{2}{9} to both sides to get rid of the constant on the LHS

x^{2} - \frac{x}{3} - \frac{2}{9}+\frac{2}{9}=\frac{2}{9}  ==> x^{2} - \frac{x}{3}=\frac{2}{9}

Add \frac{1}{36}  to both sides

x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{2}{9} +\frac{1}{36}

This simplifies to

x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{1}{4}

Noting that (a + b)² = a² + 2ab + b²

If we set a = x and b = \frac{1}{6}\right) we can see that

\left(x - \frac{1}{6}\right)^2 = x^2 - 2.x. (-\frac{1}{6}) + \frac{1}{36} = x^{2} - \frac{x}{3}+\frac{1}{36}

So

\left(x - \frac{1}{6}\right)^2=\frac{1}{4}

Taking square roots on both sides

\left(x - \frac{1}{6}\right)^2= \pm\frac{1}{4}

So the two roots or solutions of the equation are

x - \frac{1}{6}=-\sqrt{\frac{1}{4}}  and x - \frac{1}{6}=\sqrt{\frac{1}{4}}

\sqrt{\frac{1}{4}} = \frac{1}{2}

So the two roots are

x_1=\frac{1}{6} - \frac{1}{2} = -\frac{1}{3}

and

x_2=\frac{1}{6} + \frac{1}{2} = \frac{2}{3}

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Step-by-step explanation:

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