Given x² – 70x + λ = 0 with positive roots a and b where one of the root is less than 10 and $\frac{\lambda }{2}$ and $\frac{\lambda }{3}$ are not integers then find value of $\frac{\sqrt[]{\alpha -1}+\sqrt[]{\beta -1}}{\left|\alpha -\beta \right|}$  is equal to

  $\left|\frac{1}{2}-2i+1\right|=\alpha \left(\frac{1}{2}-2i\right)+\beta \left(1+i\right)$  

$\sqrt[]{\frac{9}{4}+4}=\alpha \left(\frac{1}{2}-2i\right)+\beta \left(1+i\right)$

$\frac{5}{2}=\alpha \left(\frac{1}{2}\right)+\beta +i\left(-2\alpha +\beta \right)$             

$\frac{\alpha }{2}+\beta =\frac{5}{2}$      ...(1)

 –2α + β = 0                    …(2)

Solving (1) and (2)

$\frac{\alpha }{2}+2\alpha =\frac{5}{2}$

$\frac{5}{2}\alpha =\frac{5}{2}$            

a =

$z^2=-\overline{iz}$

$\left|z^2\right|=\left|-\overline{iz}\right|$

$\left|z^2\right|=\left|z\right|$

$\left|z^2\right|-\left|z\right|=0$

$\left|z\right|\left(\left|z\right|-1\right)=0$

|z| = 0 (not acceptable)

|z| = 1

|z|² = 1

z₁ + z₂ = 5

$z_1^3+z_2^3=20+15i$            

  $z_1^3+z_2^3={\left(z_1+z_2\right)}^3-3z_1{z}_2\left(z_1+z_2\right)$           

$z_1^3+z_2^3=125-3z_1\cdot z_2\left(5\right)$            

 ⇒ 20 + 15i = 125 – 15z₁z₂

⇒ 3z₁z₂ = 25 – 4 – 3i

3z₁z₂ = 21– 3i

z₁⋅z₂ = 7 – i

(z₁ + z₂)² = 25

$z_1^2+z_2^2=25-27\left(7-i\right)$     

= 11 + 2i

  ${\left(z_{\text{?}1}^2+z_2^2\right)}^2$   &nb

$x^2-2\left(3k-1\right)x+8k^2-7>0\Rightarrow a>0,\&D<0,$

a = 1 > 0 and D < 0

4 (3k – 1)² – 4 (8k² – 7) < 0

  $\left(9k^2+1-6k-8k^2+7<0\right)$

$k(k-4)-2(k-4)<0$

$k\in \left(2,4\right)$

K = 3

$Given\left|z+5\right|\le 4..........\left(i\right)$

->Represent a circle ${\left(x+5\right)}^2+y^2\le 16$  

$=z\left(1+i\right)+\overline{z}\left(1-i\right)\ge -10..........\left(ii\right)$           

->Represent a line X – y $\ge -5$  

So max |z + 1|² = AQ²

$={\left(-4-2\sqrt[]{2}\right)}^2+8$         

$\begin{array}{l}=32+16\sqrt[]{2}\\ =\alpha +\beta \sqrt[]{2}\end{array}$        

Hence a + b = 48