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2 years ago

If f(x) = x2, then f(4) = 42 = 16. Find: a. f(1) b. f(-3) c. f(t)

Mathematics

1 answer:

icang [17]2 years ago

4 0

Answer:

Step-by-step explanation:

f(x) = x²

f(4) = 4² = 16

f(1) = 1² = 1

f(-3) = -3² = 9

f(t) = t²

Always wise to us the ^ symbol to represent powers to avoid confusion

4^2 = 4² = 16

-3^2 = 9

t^2

if you are on a PC, the square symbol (²) can be typed by holding down the "alt" key while pressing and releasing "2","5","3" on the keypad, then release "alt". Unfortunately this does not work if you only have the row of numbers on the keyBoard to work with.

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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

2 years ago

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