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1 year ago

Find the value of a' and 'b' if 2/ 7+3√5 =a-b√5

Mathematics

2 answers:

Whitepunk [10]1 year ago

4 0

Answer:

$a=\frac{7}{2} \ \ ,\ \ \ b=\frac{3}{2}$

Step-by-step explanation:

$\frac{2}{7+3\sqrt{5} } =\frac{2\times \left( 7-3\sqrt{5} \right) }{\left( 7+3\sqrt{5} \right) \left( 7-3\sqrt{5} \right) }$

$=\frac{2\times \left( 7-3\sqrt{5} \right) }{7^{2}-\left( 3\sqrt{5} \right)^2 }$

$=\frac{2\times \left( 7-3\sqrt{5} \right) }{49-45 }$

$=\frac{2\times \left( 7-3\sqrt{5} \right) }{4 }$

$=\frac{7-3\sqrt{5} }{2 }$

$=\frac{7}{2} -\frac{3}{2} \sqrt{5}$

Then

a = 7/2

b = 3/2

ycow [4]1 year ago

3 0

Step-by-step explanation:

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3 years ago

Solve for x. x3=−1000 Enter your answer in the box. x =

levacccp [35]

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3 years ago

Read 2 more answers

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Sergio [31]

Ali's solution is incorrect
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3 years ago

Let f(x)=x+a/x+b such that f(f(1)=0 & f(2)=-3 then a&b resctivily are.

Yuliya22 [10]

Answer:

The value of a = - 1 , And b = $\frac{ - 7 }{3}$

Step-by-step explanation:

Given as :

Function f(x) = $\frac{(x + a)}{(x + b)}$

And f(f(1)) = 0 And f(2) = - 3

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or, - 3 = $\frac{(2 + a)}{(2 + b)}$

I.e 2 + a = - 6 - 3b

Or, a + 3b = - 8 ....... 1

Again f(f(1)) = 0

So, $\frac{(1 + a)}{(1 + b)}$ = 0

Or, 1 + a = 0

∴ a = - 1

So , put htis value of a i n eq 1 , we get value of b

So , - 1 + 3b = - 8

Or, 3b = - 7

∴ b = $\frac{ - 7 }{3}$

Hence The value of a = - 1 , And b = $\frac{ - 7 }{3}$ Answer

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3 years ago

Find the value of a' and 'b' if 2/ 7+3√5 =a-b√5​

Find the value of a' and 'b' if 2/ 7+3√5 =a-b√5