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serg [7]
3 years ago
14

What are the roots of the quadratic function f(b)=b^2-75

Mathematics
1 answer:
Ivahew [28]3 years ago
5 0
Set f(b) = b^2 - 75 = to 0 and solve for b:

b^2 = 75
b^2 = 25(3)
b = plus or minus sqrt(25[3]) = plus or minus sqrt(25)*sqrt(3) = plus or minus 5 sqrt(3).
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8 0
3 years ago
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Paul [167]

Step-by-step explanation:

\large\underline{\sf{Solution-}}

Given expression is

\rm :\longmapsto\:\dfrac{1}{ \sqrt{3}  +  \sqrt{5}  -  \sqrt{2} }

can be re-arranged as

\rm :\longmapsto\:\dfrac{1}{ \sqrt{3}   -   \sqrt{2}   +  \sqrt{5} }

\rm \:  =  \: \dfrac{1}{( \sqrt{3}  -  \sqrt{2} ) +  \sqrt{5} }

On rationalizing the denominator, we get

\rm \:  =  \: \dfrac{1}{( \sqrt{3}  -  \sqrt{2} ) +  \sqrt{5} }  \times \dfrac{( \sqrt{3}  -  \sqrt{2} ) -  \sqrt{5} }{( \sqrt{3}  -  \sqrt{2} ) -  \sqrt{5} }

We know,

\rm :\longmapsto\:\boxed{\tt{ (x + y)(x - y) =  {x}^{2} -  {y}^{2} \: }}

So, using this, we get

\rm \:  =  \: \dfrac{ \sqrt{3} -  \sqrt{2}   -  \sqrt{5} }{ {( \sqrt{3}  -  \sqrt{2} )}^{2}  -  {( \sqrt{5}) }^{2} }

\rm \:  =  \: \dfrac{ \sqrt{3} -  \sqrt{2}   -  \sqrt{5} }{3 + 2 - 2 \sqrt{6}   - 5}

\rm \:  =  \: \dfrac{ \sqrt{3} -  \sqrt{2}   -  \sqrt{5} }{5 - 2 \sqrt{6}   - 5}

\rm \:  =  \: \dfrac{ \sqrt{3} -  \sqrt{2}   -  \sqrt{5} }{ - 2 \sqrt{6}}

\rm \:  =  \: \dfrac{ - ( -  \sqrt{3} +  \sqrt{2}  + \sqrt{5}) }{ - 2 \sqrt{6}}

\rm \:  =  \: \dfrac{-  \sqrt{3} +  \sqrt{2}  + \sqrt{5}}{2 \sqrt{6}}

On rationalizing the denominator, we get

\rm \:  =  \: \dfrac{-  \sqrt{3} +  \sqrt{2}  + \sqrt{5}}{2 \sqrt{6}}  \times \dfrac{ \sqrt{6} }{ \sqrt{6} }

\rm \:  =  \: \dfrac{-  \sqrt{18} +  \sqrt{12}  + \sqrt{30}}{2  \times 6}

\rm \:  =  \: \dfrac{-  \sqrt{3 \times 3 \times 2} +  \sqrt{2 \times 2 \times 3}  + \sqrt{30}}{12}

\rm \:  =  \: \dfrac{-  3\sqrt{2} + 2 \sqrt{3}   + \sqrt{30}}{12}

Hence,

\boxed{\tt{ \rm \dfrac{1}{ \sqrt{3}  +  \sqrt{5}  -  \sqrt{2} } =\dfrac{-  \sqrt{3 \times 3 \times 2} +  \sqrt{2 \times 2 \times 3}  + \sqrt{30}}{12}}}

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<h3><u>More Identities to </u><u>know:</u></h3>

\purple{\boxed{\tt{  {(x  -  y)}^{2} =  {x}^{2} - 2xy +  {y}^{2}}}}

\purple{\boxed{\tt{  {(x   +   y)}^{2} =  {x}^{2} + 2xy +  {y}^{2}}}}

\purple{\boxed{\tt{  {(x   +   y)}^{3} =  {x}^{3} + 3xy(x + y) +  {y}^{3}}}}

\purple{\boxed{\tt{  {(x - y)}^{3} =  {x}^{3} - 3xy(x  -  y) -  {y}^{3}}}}

\pink{\boxed{\tt{  {(x + y)}^{2} +  {(x - y)}^{2} = 2( {x}^{2} +  {y}^{2})}}}

\pink{\boxed{\tt{  {(x + y)}^{2}  -  {(x - y)}^{2} = 4xy}}}

6 0
3 years ago
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