Math, asked by pushkard941, 30 days ago

Find the values of a and b if: 3+V2 = a + b12 3-12​

Answers

Answered by rahulpachowal
1

Answer:

a = 2 and b = -1

Step-by-step explanation:

Given: \frac{\sqrt{3}-1}{\sqrt{3}+1}=a+b\sqrt{3}

3

+1

3

−1

=a+b

3

To find: vale of a & b

We find value of a & b by rationalizing the denominator of LHS and then equating with RHS

Consider,

LHS

=\frac{\sqrt{3}-1}{\sqrt{3}+1}=

3

+1

3

−1

=\frac{\sqrt{3}-1}{\sqrt{3}+1}\times\frac{\sqrt{3}-1}{\sqrt{3}-1}=

3

+1

3

−1

×

3

−1

3

−1

=\frac{(\sqrt{3}-1)^2}{(\sqrt{3}+1)(\sqrt{3}-1)}=

(

3

+1)(

3

−1)

(

3

−1)

2

=\frac{(\sqrt{3})^2+(1)^2-2\sqrt{3}}{(\sqrt{3})^2-(1)^2}=

(

3

)

2

−(1)

2

(

3

)

2

+(1)

2

−2

3

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

3−1

3+1−2

3

=\frac{4}{2}-\frac{2\sqrt{3}}{2}=

2

4

2

2

3

=2-\sqrt{3}=2−

3

Now equating with RHS = a + b√3

we get a = 2 7 b = -1

Therefore, a = 2 and b = -1

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