Question

if p= 1/2-root3 and q=1/root 3 +2 then the value of 7p^2+11pq-7q^2 equals? btw i know this is on a exam in india that i cant talk about ;) but i have submitted already and this is my honest doubt after submitting.

Answer 1

GIVEN :

If $p=\frac{1}{2-\sqrt{3}}$ and $q=\frac{1}{\sqrt{3}+2}$ then the value of $7p^2+11pq-7q^2$

TO FIND :

The  value of $7p^2+11pq-7q^2$

SOLUTION :

Given that the values are  $p=\frac{1}{2-\sqrt{3}}$ and $q=\frac{1}{\sqrt{3}+2}$

From the given value $p=\frac{1}{2-\sqrt{3}}$

Now we can rationalise the denominator by multiplying and dividing by denominator's conjugate we get,

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

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

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

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

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

⇒ $p=2+\sqrt{3}$

Now for $q=\frac{1}{\sqrt{3}+2}$

we can rationalise the denominator by multiplying and dividing by denominator's conjugate we get,

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

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

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

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

$=2-\sqrt{3}$

⇒ $q=2-\sqrt{3}$

Substitute the values in $7p^2+11pq-7q^2$ we get,

$7p^2+11pq-7q^2=7(2+\sqrt{3})^2+11(2+\sqrt{3})(2-\sqrt{3})-7(2-\sqrt{3})^2$

By using the Algebraic identities :

i) $(a+b)^2=a^2+2ab+b^2$

ii) $(a-b)^2=a^2-2ab+b^2$

iii) $(a+b)(a-b)=a^2-b^2$

$=7(2^2+(\sqrt{3})^2+2(2)(\sqrt{3}))+11(2^2-(\sqrt{3})^2)-7(2^2+(\sqrt{3})^2-2(2)(\sqrt{3}))$

$=7(4+3+4\sqrt{3})+11(4-3)-7(4+3-4\sqrt{3})$

$=7(7+4\sqrt{3})+11(4-3)-7(7-4\sqrt{3})$

$=49+28\sqrt{3}+11-49+28\sqrt{3}$

$=56\sqrt{3}+11$

$7p^2+11pq-7q^2=56\sqrt{3}+11$

∴ the value of $7p^2+11pq-7q^2=56\sqrt{3}+11$