( 2 + (√2) )^8 = (√A) + (√B) Find: A−B = ?


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Question Number 215439 by hardmath last updated on 06/Jan/25

${(2 + \sqrt{2})}^{8} = \sqrt{A} + \sqrt{B}$ $Find : A - B = ?$
	Answered by Rasheed.Sindhi last updated on 07/Jan/25
	$If z is a binomial surd and n is natural (z^n )^(−) =(z^(−) )^n Is this correct? ( 2 + (√2) )^8 = (√A) + (√B) ...(i) ⇒( 2 − (√2) )^8 = (√A) − (√B) ...(ii) (i)×(ii): ( 2 + (√2) )^8 ( 2 − (√2) )^8 =A−B {(2+(√2) )(2+(√2) )}^8 =A−B (4−2)^8 =A−B A−B=256$
	$I f z i s a b i n o m i a l s u r d a n d n i s$ $n a t u r a l$ $\overset{―}{(z^{n})} = {(\overset{―}{z})}^{n}$ $I s t h i s c o r r e c t ?$ ${(2 + \sqrt{2})}^{8} = \sqrt{A} + \sqrt{B} . . . (i)$ $\Rightarrow {(2 - \sqrt{2})}^{8} = \sqrt{A} - \sqrt{B} . . . (i i)$ $(i) \times (i i) :$ ${(2 + \sqrt{2})}^{8} {(2 - \sqrt{2})}^{8} = A - B$ ${(2 + \sqrt{2}) (2 + \sqrt{2})}^{8} = A - B$ ${(4 - 2)}^{8} = A - B$ $A - B = 256$
	Commented by TonyCWX08 last updated on 07/Jan/25

	$Y e a h .$
	Commented by mr W last updated on 07/Jan/25

	$r i g h t!$ $g e n e r a l l y f o r a, b, c, n \in N :$ ${(a + b \sqrt{c})}^{n}$ $= \underset{k = 0}{\sum^{n}} C_{k}^{n} a^{n - k} {(b \sqrt{c})}^{k}$ $= \underset{r = 0}{\sum^{⌊ \frac{n}{2} ⌋}} C_{2 r}^{n} a^{n - 2 r} {(b \sqrt{c})}^{2 r} + \underset{r = 0}{\sum^{⌊ \frac{n}{2} ⌋ - 1}} C_{2 r + 1}^{n} a^{n - 2 r + 1} {(b \sqrt{c})}^{2 r + 1}$ $= \underset{r = 0}{\sum^{⌊ \frac{n}{2} ⌋}} C_{2 r}^{n} a^{n - 2 r} b^{2 r} c^{r} + \underset{r = 0}{\sum^{⌊ \frac{n}{2} ⌋ - 1}} C_{2 r + 1}^{n} a^{n - 2 r + 1} b^{2 r + 1} c^{r} \sqrt{c}$ $= \underset{r = 0}{\sum^{⌊ \frac{n}{2} ⌋}} C_{2 r}^{n} a^{n - 2 r} b^{2 r} c^{r} + (\underset{r = 0}{\sum^{⌊ \frac{n}{2} ⌋ - 1}} C_{2 r + 1}^{n} a^{n - 2 r + 1} b^{2 r + 1} c^{r}) \sqrt{c}$ $= A + B \sqrt{c}$ $s i m i l a r l y$ ${(a - b \sqrt{c})}^{n}$ $= \underset{r = 0}{\sum^{⌊ \frac{n}{2} ⌋}} C_{2 r}^{n} a^{n - 2 r} b^{2 r} c^{r} - (\underset{r = 0}{\sum^{⌊ \frac{n}{2} ⌋ - 1}} C_{2 r + 1}^{n} a^{n - 2 r + 1} b^{2 r + 1} c^{r}) \sqrt{c}$ $= A - B \sqrt{c}$
	Commented by Rasheed.Sindhi last updated on 07/Jan/25

	$Than k s sirs!$
	Answered by TonyCWX08 last updated on 07/Jan/25

	${(2 + \sqrt{2})}^{2} = 4 + 4 \sqrt{2} + 2 = 6 + 4 \sqrt{2}$ ${(6 + 4 \sqrt{2})}^{2} = 36 + 48 \sqrt{2} + 32 = 68 + 48 \sqrt{2}$ ${(68 + 48 \sqrt{2})}^{2} = 4624 + 6528 \sqrt{2} + 4608 = 9232 + 6528 \sqrt{2}$ $= \sqrt{9232^{2}} + \sqrt{2 {(6528)}^{2}}$ $A - B = 9232^{2} - 2 {(6528)}^{2} = 256$
	Commented by hardmath last updated on 07/Jan/25

	${(2 + \sqrt{2})}^{8} . . .$
	Commented by TonyCWX08 last updated on 07/Jan/25

	${(2 + \sqrt{2})}^{8} = {({({(2 + \sqrt{2})}^{2})}^{2})}^{2}$
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