# P and Q are the roots of 3x*2-12x+6. Find 1/p*2 - 1/q*2 ?

Mar 24, 2018

$\frac{1}{p} ^ 2 - \frac{1}{q} ^ 2 = 2 \sqrt{2}$.....$\left(p < q\right)$.
Hint: ${\left(x - y\right)}^{2} = {x}^{2} + {y}^{2} - 2 x y = {x}^{2} + 2 x y + {y}^{2} - 4 x y$
$\implies {\left(x - y\right)}^{2} = {\left(x + y\right)}^{2} - 4 x y$
please use '^' instead of ' * ' . $i . e . {x}^{2} \to$x^2 and not x*2

#### Explanation:

$3 {x}^{2} - 12 x + 6 = 0$.

Comparing with $a {x}^{2} + b x + c = 0$,we get

$a = 3 , b = - 12 \mathmr{and} c = 6$

If the roots of this equn. are $p \mathmr{and} q$, then

$p + q = - \frac{b}{a} \mathmr{and} p q = \frac{c}{a}$

$i . e . p + q = - \frac{- 12}{3} = 4 \mathmr{and} p q = \frac{6}{3} = 2$
Now,
$\frac{1}{p} ^ 2 - \frac{1}{q} ^ 2 = \frac{{q}^{2} - {p}^{2}}{{p}^{2} {q}^{2}} = \frac{\left(q + p\right) \left(q - p\right)}{p q} ^ 2$,....$\left(p < q\right)$

=>1/p^2-1/q^2=((4)sqrt((q-p)^2))/2^2=sqrt((q-p)^2

=>1/p^2-1/q^2=sqrt((q+p)^2-4pq)=sqrt(4^2-4(2)

$\implies \frac{1}{p} ^ 2 - \frac{1}{q} ^ 2 = \sqrt{16 - 8} = \sqrt{8} = 2 \sqrt{2}$....$\left(p < q\right)$

Mar 24, 2018

$3 {x}^{2} - 12 x + 6 = 0$
$\implies {x}^{2} - 4 x + 2 = 0$

Roots, $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot 2}}{2}$

$x = \frac{4 \pm \sqrt{8}}{2} = \frac{4 \pm 2 \sqrt{2}}{2}$

$x = \left(2 \pm 2 \sqrt{2}\right)$

To find, $\frac{1}{p} ^ 2 - \frac{1}{q} ^ 2$

$\implies \left(\frac{1}{p} + \frac{1}{q}\right) \left(\frac{1}{p} - \frac{1}{q}\right)$

$\implies \left(\frac{1}{2 + 2 \sqrt{2}} + \frac{1}{2 - 2 \sqrt{2}}\right) \left(\frac{1}{2 + 2 \sqrt{2}} - \frac{1}{2 - 2 \sqrt{2}}\right)$

$\implies \left(\frac{\left(2 - 2 \sqrt{2}\right) + \left(2 + 2 \sqrt{2}\right)}{\left(2 - 2 \sqrt{2}\right) \left(2 + 2 \sqrt{2}\right)}\right) \left(\frac{\left(2 - 2 \sqrt{2}\right) - \left(2 + 2 \sqrt{2}\right)}{\left(2 - 2 \sqrt{2}\right) \left(2 + 2 \sqrt{2}\right)}\right)$

$\implies \left(\frac{\left(2 + 2\right)}{\left(2 - 2 \sqrt{2}\right) \left(2 + 2 \sqrt{2}\right)}\right) \left(\frac{\left(- 2 \sqrt{2} - 2 \sqrt{2}\right)}{\left(2 - 2 \sqrt{2}\right) \left(2 + 2 \sqrt{2}\right)}\right)$

$\implies \left(\frac{4 \left(- 4 \sqrt{2}\right)}{\left(4 - 8\right)} ^ 2\right)$

$\implies \left(\frac{4 \left(- 4 \sqrt{2}\right)}{- 4} ^ 2\right)$

$\implies \left(- \sqrt{2}\right)$