ધારો કે, વાસ્તવિક બીજ $\alpha,\beta,\gamma,\delta$,  હોય તો, સમીકરણ $(x – \alpha) (x – \beta) (x – \gamma) (x -\delta  ) = 0$

${x^4} – (\alpha  + \beta  + \gamma  + \delta ){x^3} + (\alpha \beta  + \beta \gamma  + \gamma \delta $

$ + \alpha \delta  + \beta \delta  + \alpha \gamma ){x^2} – (\alpha \beta \gamma  + \beta \gamma \delta $

$ + \alpha \beta \delta  + \alpha \gamma \delta )x + \alpha \beta \gamma \delta  = 0$

${x^4} – \sum \alpha .{x^3} + \sum \alpha \beta .{x^2} – \sum \alpha \beta \gamma .x + \alpha \beta \gamma \delta  = 0$

${x^4} – 4{x^3} + a{x^2} + bx + 1 = 0$ સાથે સરખાવતા,

$\sum \alpha  = 4,\,\sum \alpha \beta \, = a,\,\sum \alpha \beta \gamma  =  – b,\,\alpha \beta \gamma \delta  = 1$

વાસ્તવિક બીજ માટે, $A.M. \geq G.M.$

$\frac{1}{4}(\sum \alpha ) \ge {(\alpha \beta \gamma \delta )^{1/4}};\,$ $\sum \alpha  = 4$

$\therefore \,\,\frac{1}{4}\sum \alpha  = \frac{1}{4} \times 4 = 1$

${(\alpha \beta \gamma \delta )^{1/4}} = 1 \Rightarrow \,\,\alpha \beta \gamma \delta  = 1$ 

$\Sigma \alpha  = 4$ અને $\alpha \beta \gamma \delta  = 1$

$\therefore \,\,\alpha  = \beta  = \gamma  = \delta \,\, = \,\,1$

હવે $,\,\sum \alpha \beta  = a$

$\therefore a = \alpha \beta  + \beta \gamma  + \gamma \delta  + \alpha \delta  + \beta \delta  + \alpha \gamma $

$ = 1 \times 1 + 1 \times 1 + 1 \times 1 + 1 \times 1 + 1 \times 1 + 1 \times 1\,\, = \,\,6$

$ – b = \alpha \beta \gamma  + \alpha \beta \delta  + \alpha \gamma \delta  + \beta \gamma \delta $

$ = 1 \times 1 \times 1 + 1 \times 1 \times 1 + 1 \times 1 \times 1 + 1 \times 1 \times 1$

$ = {(1)^3} + {(1)^3} + {(1)^3} + {(1)^3}$

$ = 1 + 1 + 1 + 1 = 4$

$\therefore \,b =  – 4\,;\,$

$\therefore a = 6\,$ અને $b\,\, = \,\, – 4$