Solve each of the quadratic equations.
PT. 1
3x = 0.5x2
A. x = -6 or x = 0
B. x = -4 or x = 3
C. x = -2
D. x = 1.5 x = 0 or x = 6

PT.2
0 = 5x2 - 2x + 6
A. x= 1±3i/2
B. x=1±√11/5
C. x=1±I√29/5

Answers

Answer 1

Answer:

the answer to pt 1 is D. x=1.5 x=0 or x=6

and the answer to part 2 is C. x=1±I√29/5

Step-by-step explanation:

Answer 2

1. The solution of this equation is x = 0 or x = 6 which is option D.

2. The solution of this equation is [tex]x = \dfrac{1 \pm i\sqrt{29}}{5}[/tex] which is option C.

Use the concept of quadratic equation defined as:

The definition of a quadratic as a second-degree polynomial equation demands that at least one squared term must be included. It also goes by the name quadratic equations.

The quadratic equation has the following generic form:

ax² + bx + c = 0

1. The given quadratic equation is,

3x = 0.5x²

Rearranging it,

0.5x² - 3x = 0

Common out x,

x(0.5x - 3) = o

Then we have,

x = 0 or,  

0.5x - 3 = 0

Add 3 on both sides,

0.5x  = 3

Divide both sides by 0.5,

x = 3/0.5

x = 6

Hence the solution of this quadratic equation is,

x = 0 and x = 6 which is option D.

2. The given quadratic equation is:

0 = 5x² - 2x + 6

It can be written as,

5x² - 2x + 6 = 0

Since we know the quadrature formula for ax² + bx + c = 0 is,

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

Here we have,

a = 5

b = -2

c = 6

Substitute values into the formula,

[tex]x = \dfrac{2 \pm \sqrt{4 - 120}}{10}\\\\x = \dfrac{1 \pm i\sqrt{29}}{5} \text{ }\text{ \text{ }} \text{\text{ }\ since } \sqrt{-1} = i[/tex]

Hence, option C is correct.

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The complete question is:

Solve each of the quadratic equations.

PT. 1

3x = 0.5x²

A. x = -6 or x = 0

B. x = -4 or x = 3

C. x = -2 or x = 1.5

D. x = 0 or x = 6

PT.2

0 = 5x² - 2x + 6

A.  [tex]x = \dfrac{1 \pm 3i}{5}[/tex]

B. [tex]x = \dfrac{1 \pm i\sqrt{11}}{5}[/tex]  

C. [tex]x = \dfrac{1 \pm i\sqrt{29}}{5}[/tex]


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