Question

${p}^{2} = 4 \sqrt{5} + 9$
what is the value of P? ​

Answer 1

Answer:

Verification of answer :-

To verify your answer put p = 12 in p/2+2=8. If you get RHS = LHS then your answer is correct.

Step-by-step explanation:

Thank you

Hope it helps to you

Answer 2

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

As it is given that,

$\rm \: {p}^{2} = 4 \sqrt{5} + 9$

can be re-arranged as

$\rm \: {p}^{2} = 9 + 4 \sqrt{5}$

can be rewritten as

$\rm \: {p}^{2} = 5 + 4 + 4 \sqrt{5}$

$\rm \: {p}^{2} = {( \sqrt{5} )}^{2} + {2}^{2} + 4 \sqrt{5}$

$\rm \: {p}^{2} = {( \sqrt{5} )}^{2} + {2}^{2} + 2 \times \sqrt{5} \times 2$

We know,

$\boxed{ \rm{ \: {x}^{2} + {y}^{2} + 2xy = {(x + y)}^{2} \: }}$

So, using this identity, we get

$\rm \: {p}^{2} = {( \sqrt{5} + 2)}^{2}$

$\rm\implies \:p \: = \: \pm \: ( \sqrt{5} + 2)$

$\rm\implies \:p \: = \: \sqrt{5} + 2 \: \: \: or \: \: \: - \sqrt{5} - 2$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$