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Question Number 187980 by Mingma last updated on 24/Feb/23

Answered by Rasheed.Sindhi last updated on 24/Feb/23

$$\bullet x^2+bx+c=0$$$$\alpha +\beta =-b,\alpha \beta =c$$$$\bullet x^2+bx-c=0$$$$\gamma +\delta =-b,\gamma \delta =-c$$$$\alpha +\beta =\gamma +\delta ,\alpha \beta =-\gamma \delta$$$${(\alpha +\beta )}^2={\alpha }^2+{\beta }^2+2\alpha \beta$$$${\alpha }^2+{\beta }^2=b^2-2c$$$${(\gamma +\delta )}^2={\gamma }^2+{\delta }^2+2\gamma \delta$$$${\gamma }^2+{\delta }^2=b^2+2c$$$${\alpha }^2+{\beta }^2+{\gamma }^2+{\delta }^2=2b^2$$$${(\alpha +\beta )}^2-2\alpha \beta +{(\gamma +\delta )}^2-2\gamma \delta =2b^2$$$${(\alpha +\beta )}^2-\cancel{2\left(c\right)}+{(\gamma +\delta )}^2-\cancel{2(-c)}=2b^2$$$${(\alpha +\beta )}^2+{(\gamma +\delta )}^2=2b^2$$$$2b^2=p^2+q^2$$

Commented by Mingma last updated on 24/Feb/23

Well explained, sir!

Commented by Mingma last updated on 24/Feb/23

Can you show the last part?

Commented by Rasheed.Sindhi last updated on 24/Feb/23

$$\mathcal{T}hisisjust2b^2=b^2+b^2:)$$

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