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Mashcka [7]
3 years ago
11

Using the Quadratic Formula, which of the following is the solution to the quadratic equation x2 + 2x - 3 = 0?

Mathematics
1 answer:
saw5 [17]3 years ago
6 0

Answer: x=1 or x=-3

Step-by-step explanation:

Quadratic formula is -b+-\sqrt{b^{2}-4ac } /2

a=1 b=2 c=-3

Substitute

(-2+-\sqrt{2^{2}-4(1)(-3) }) /2\\(-2+-\sqrt{4+12}) /12\\(-2+-\sqrt{16}) /2\\(-2+-4) /2\\x=\frac{-2+4}{2} \\x=\frac{-2-4}{2} \\\\\\Simplify\\x=\frac{2}{2} \\\\x=\frac{-6}{2} \\\\Simplify\\x=1\\x=-3

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Three terms of an arithmetic sequence are shown below. Which recursive formula defines the sequence?: f(1) = 6, f(4) = 12, f(7)
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Rom4ik [11]

Answer:

a) \sqrt{61 - 24 \sqrt{5} }  =  - 4  + 3 \sqrt{5}

b)( \sqrt{ ( {c}^{2}   -  1) ({b}^{2}    -  1) } - {2 \sqrt{bc} }) (\sqrt{ ( {c}^{2}   -  1) ({b}^{2}    -  1) }  + {2 \sqrt{bc}  } )

c) \frac{ \sqrt{9 - 4 \sqrt{5} } }{2 -  \sqrt{5} }  =   - 1

Step-by-step explanation:

We want to simplify

\sqrt{61 - 24 \sqrt{5} }

Let :

\sqrt{61 - 24 \sqrt{5} }  = a - b \sqrt{5}

Square both sides of the equation:

(\sqrt{61 - 24 \sqrt{5} } )^{2}  =  ({a - b \sqrt{5} })^{2}

Expand the RHS;

61 - 24 \sqrt{5} =  {a}^{2}  - 2ab \sqrt{5}  + 5 {b}^{2}

Compare coefficients on both sides:

{a}^{2}  + 5 {b}^{2}  = 61 -  -  - (1)

- 24 =  - 2ab \\ ab = 12 \\ b =  \frac{12}{b}  -  -  -( 2)

Solve the equations simultaneously,

\frac{144}{ {b}^{2} }  + 5 {b}^{2}  = 61

5 {b}^{4}  - 61 {b}^{2}  + 144 = 0

Solve the quadratic equation in b²

{b}^{2}  = 9 \: or \:  {b}^{2}  =  \frac{16}{5}

This implies that:

b =  \pm3 \: or \: b =  \pm  \frac{4 \sqrt{5} }{5}

When b=-3,

a =  - 4

Therefore

\sqrt{61 - 24 \sqrt{5} }  =  - 4  + 3 \sqrt{5}

We want to rewrite as a product:

{b}^{2}  {c}^{2}  - 4bc -  {b}^{2}  -  {c}^{2}  + 1

as a product:

We rearrange to get:

{b}^{2}  {c}^{2}   -  {b}^{2}  -  {c}^{2}  + 1- 4bc

We factor to get:

{b}^{2} ( {c}^{2}   -  1)  -  ({c}^{2}   -  1)- 4bc

Factor again to get;

( {c}^{2}   -  1) ({b}^{2}   -  1)- 4bc

We rewrite as difference of two squares:

(\sqrt{( {c}^{2}   -  1) ({b}^{2}   -  1) })^{2} - ( {2 \sqrt{bc} })^{2}

We factor the difference of square further to get;

( \sqrt{ ( {c}^{2}   -  1) ({b}^{2}    -  1) } - {2 \sqrt{bc} }) (\sqrt{ ( {c}^{2}   -  1) ({b}^{2}    -  1) }  + {2 \sqrt{bc}  } )

c) We want to compute:

\frac{ \sqrt{9 - 4 \sqrt{5} } }{2 -  \sqrt{5} }

Let the numerator,

\sqrt{9 - 4 \sqrt{5} }  = a - b \sqrt{5}

Square both sides of the equation;

9 - 4 \sqrt{5}  =  {a}^{2}  - 2ab \sqrt{5}  + 5 {b}^{2}

Compare coefficients in both equations;

{a}^{2}  + 5 {b}^{2}  = 9 -  -  - (1)

and

- 2ab =  - 4 \\ ab = 2 \\ a =  \frac{2}{b}  -  -  -  - (2)

Put equation (2) in (1) and solve;

\frac{4}{ {b}^{2} }  + 5 {b}^{2}  = 9

5 {b}^{4}   - 9 {b}^{2}  + 4 = 0

b =  \pm1

When b=-1, a=-2

This means that:

\sqrt{9 - 4 \sqrt{5} }  =  - 2 +  \sqrt{5}

This implies that:

\frac{ \sqrt{9 - 4 \sqrt{5} } }{2 -  \sqrt{5} }  =  \frac{ - 2 +  \sqrt{5} }{2 -  \sqrt{5} }  =  \frac{ - (2 -  \sqrt{5)} }{2 -  \sqrt{5} }  =  - 1

3 0
3 years ago
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