Question

If a =1/7-4√3 and b=1/7+4√3, find the value of :-
(a) a(square) +b(square)​

Answer 1

Answer:

ⒶⓃⓈⓌⒺⓇ ༉✪❥ :

$\huge \fcolorbox{navy}{black}{ \bf194}$

Step-by-step explanation:

❖ Given :

$\bf \pink{ a = \frac{1}{7 - 4 \sqrt{3} } \: and \: b = \frac{1}{7 + 4 \sqrt{3} } }$

❖ To Find :

$\bold{ \blue{{a}^{2} + {b}^{2} }}$

❖ Solution:

Let's first rationalise a and b separately,

$\bf \: a = \frac{1}{7 - 4 \sqrt{3} } \times \frac{7 + 4 \sqrt{3} }{7 + 4 \sqrt{3} }$

$\bf = \frac{7 + 4 \sqrt{3} }{(7 - 4 \sqrt{3})(7 + 4 \sqrt{3} )}$

For denominator, using the identity:

$\boxed{ \bold \red{ (x + y)(x - y) = {x}^{2} - {y}^{2} }}$

$= \frac{7 + 4 \sqrt{3} }{ {7}^{2} - {(4 \sqrt{3} )}^{2} }$

$= \frac{7 + 4 \sqrt{3} }{49 - 48}$

$\bf \purple{a = } \bold \purple{7 + 4 \sqrt{3} }$

Similarly,

$\bf \purple{b = 7 - 4 \sqrt{3} }$

Now, we know an identity,

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

Therefore,

$\boxed{ \bf \orange{ {x}^{2} + {y}^{2} = (x + y) ^{2} - 2xy }}$

Here,

☞ x = a = 7+4√3 and y = b = 7-4√3

Substituting values in the above equation,, we get,

$\bf {a}^{2} + {b}^{2} = (7 + 4 \sqrt{3} + 7 - 4 \sqrt{3} )^{2} - 2(7 + 4 \sqrt{3} )(7 - 4 \sqrt{3} )$

$= ({14})^{2} - 2(1) \\ = 196 - 2 \\ = 194$

Here, we got our answer!

Thanks!

Answer 2

Answer:

Derive the equation of motion v = u +at, graphically.

answer is in picture