Question

plzz solve above 2 question
plzz do it fast plzz
and don't spam
i will report u​

Answer 1

GIVEN :-

• (3 - √5)/(3 + 2√5) = a√5 - b

TO FIND :-

• The value of a and b.

SOLUTION :-

→ (3 - √5)/(3 + 2√5) = a√5 - b

→[( 3 - √5)(3 - 2√5)]\[(3 + 2√5)((3 - 2√5)] = a√5- b

★ Using identity :- (a + b) (a - b) = a² - b².

→ [(9 - 6√5 - 3√5 + 10)/(3)² - (2√5)²] = a√5 - b

→ [( 9 + 10 - 9√5)\(9 - √25)] = a√5 - b

→ [(19 - 9√5)\(9 - 20)] = a√5 - b

→ [(19 - 9√5)/(-11)] = a√5 - b

★ Transfer (-) to numerator.

→ [(-19 - 9√5)/11] = a√5 - b

★ Arrange The L.H.S according to R.H.S.

→ (-9√5 - 19)/11 = a√5 - b

→ (-9/11)√5 - (19/11) = a√5 - b

★ comparing both side,

→ a = (-9/11) , b = (19/11)

❏ Hence value of a is (-9/11) and b is (19/11)

Answer 2

Given :-

$\rm\to\frac{3 -\sqrt{5}}{3 + 2 \sqrt{5}} = a \sqrt{5} - b$

To Find :-

• Value of a & b

Solution :-

$\rm\frac{3 -\sqrt{5}}{3 + 2 \sqrt{5}} = a \sqrt{5} - b \\ \\ \implies\rm \frac{(3x - \sqrt{5}) \: (3 - 2 \sqrt{5})}{(3 + 2 \sqrt{5}) \: (3 - 2 \sqrt{5}) } = a \sqrt{5} - b$

By using identities,

• a² - b² = (a+ b) (a - b)

$\implies\rm \frac{9 - 6 \sqrt{5} - 3 \sqrt{5} + 10 }{ {3}^{2} - { (2\sqrt{5}) }^{2} } = a \sqrt{5} - b \\ \\ \implies\rm \frac{9 + 10 - 9 \sqrt{5} }{9 - \sqrt{25} } = a \sqrt{5} - b \\ \\ \implies\rm \frac{19 - 9 \sqrt{5} }{9 - 20} = a \sqrt{5} - b \\ \\ \implies\rm \frac{ - 19 - 9 \sqrt{5} }{11} = a \sqrt{5} - b$

Arrange the L.H.S according to R.H.S,

$\implies\rm \frac{ - 9 \sqrt{5} - 19 }{11} = a \sqrt{5} - b \\ \\ \implies\rm (\frac{ - 9}{11}) \: \sqrt{5} - \frac{19}{11} = a \sqrt{5} - b$

Comparing both sides,

• $\rm a = \frac{-9}{11}, \: b = \frac{19}{11}$

$\green{\bigstar \: \sf Hence, the \: value \: of \: a = \frac{-9}{11} \: and \: b = \frac{19}{11}}$