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Question Number 211732 by BaliramKumar last updated on 19/Sep/24

Answered by TonyCWX08 last updated on 19/Sep/24

$$Usingquadraticformula$$$$x=\frac{{sin}^2\theta \pm \sqrt{{sin}^4\theta +4{cos}^2\theta }}{2}$$$$$$$$p=\frac{{sin}^2\theta +\sqrt{{sin}^4\theta +4{cos}^2\theta }}{2}$$$$p^2={\left(\frac{{sin}^2\theta +\sqrt{{sin}^4\theta +4{cos}^2\theta }}{2}\right)}^2$$$$p^2=\frac{{({sin}^2\theta +\sqrt{{sin}^4\theta +4{cos}^2\theta })}^2}{4}$$$$$$$$q=\frac{{sin}^2\theta -\sqrt{{sin}^4\theta +4{cos}^2\theta }}{2}$$$$q^2=\frac{{({sin}^2\theta -\sqrt{{sin}^4\theta +4{cos}^2\theta })}^2}{4}$$$$$$$$p^2+q^2=\frac{{sin}^4\theta +2{sin}^2\theta \sqrt{{sin}^4\theta +4{cos}^2\theta }+{sin}^4\theta +4{cos}^2\theta +{sin}^4\theta -2{sin}^2\theta \sqrt{{sin}^4\theta +4{cos}^2\theta }+{sin}^4\theta +4{cos}^2\theta }{4}$$$$p^2+q^2=\frac{4{sin}^4\theta +8{cos}^2\theta }{4}$$$$p^2+q^2={sin}^4\theta +2{cos}^2\theta$$$$Minimumvalue=1$$

Commented by BaliramKumar last updated on 19/Sep/24

$$firststep4{cos}^2\theta$$

Commented by TonyCWX08 last updated on 19/Sep/24

$$Edited$$

Answered by BHOOPENDRA last updated on 19/Sep/24

$$p+q=\frac{-b}{a}$$$$p.q=c/a$$$${(p+q)}^2=p^2+q^2+2pq$$$$p^2+q^2={(p+q)}^2-2pq$$$$p^2+q^2={sin}^4\theta +2{cos}^2\theta$$$$={(1-{cos}^2\theta )}^2+2{cos}^2\theta$$$$p^2+q^2=(1+{cos}^4\theta )$$$$=0\le {cos}^4\theta \le 1$$$$=0+1\le {cos}^4\theta +1\le 1+1$$$$=1\le {cos}^4\theta +1\le 2$$$$p^2+q^2=1\left(minimumvalue\right)$$$$$$

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