Question

if a=√2+1/√2-1 and b=√2-1/√2+1,then find the value of a²+b²-4ab

Answer 1

Answer:

$Value \: of \:a^{2}+b^{2}-4ab=30$

Step-by-step explanation:

$i)a = \frac{\sqrt{2}+1}{\sqrt{2}-1}$

/*Rationalising the denominator ,we get

$=\frac{(\sqrt{2}+1)(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)}$

$=\frac{(\sqrt{2}+1)^{2}}{(\sqrt{2})^{2}-1^{2}}$

/*By algebraic identities:

i) (x+y)²=x²+y²+2xy

ii) (x-y)(x+y)=x²-y² */

$=\frac{2+1+2\sqrt{2}}{2-1}$

$=\frac{3+2\sqrt{2}}{1}$

$=3+2\sqrt{2}\:---(1)$

$ii)b = \frac{\sqrt{2}-1}{\sqrt{2}+1}$

/*Rationalising the denominator ,we get

$=\frac{(\sqrt{2}-1)(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)}$

$=\frac{(\sqrt{2}-1)^{2}}{(\sqrt{2})^{2}-1^{2}}$

/*By algebraic identities:

i) (x-y)²=x²+y²-2xy

2xyii) (x-y)(x+y)=x²-y² */

$=\frac{2+1-2\sqrt{2}}{2-1}$

$=\frac{3-2\sqrt{2}}{1}$

$=3-2\sqrt{2}\:---(2)$

$iii) a-b\\=3+2\sqrt{2}-(3-2\sqrt{2})\\=3+2\sqrt{2}-3+2\sqrt{2})\\=4\sqrt{2}\:--(3)$

$iv)ab \\= (3+2\sqrt{2})(3-2\sqrt{2})\\=3^{2}-(2\sqrt{2})^{2}\\=9-8\\=1\:---(4)$

$Now,\\a^{2}+b^{2}-4ab\\=(a^{2}+b^{2}-2ab)-2ab\\=(a-b)^{2}-2ab\\=(4\sqrt{2})^{2}-2\times 1$

/* From (3) and (4)*/

$=32-2$

$=30$

Therefore,

$Value \: of \:a^{2}+b^{2}-4ab=30$

•••♪

Answer 2

Answer:

U can see above its good explanation