$ \Rightarrow a^2 x^2-2 a b x+b^2+b^2 x^2-2 b c x+c^2=0$
$ \Rightarrow(a x-b)^2+(b x-c)^2=0 $
$ \Rightarrow a x-b=0, \quad b x-c=0 $
$ \Rightarrow a+b>c \quad b+c>a \quad \quad c+a>b$
$a+a x>b x $
$a+a x > a x^2 $
$ x^2-x-1 < 0$
$a x+b x > a $
$ a x+a x^2 > a $
$ x^2+x-1 > 0$
$a x^2+a>a x$
$x^2-x+1>0$
always true
$\frac{1-\sqrt{5}}{2}<\mathrm{x}<\frac{1+\sqrt{5}}{2}$
$\mathrm{x}<\frac{-1-\sqrt{5}}{2}, \text { or } \mathrm{x}>\frac{-1+\sqrt{5}}{2} $
$\Rightarrow \frac{\sqrt{5}-1}{2} < x < \frac{\sqrt{5}+1}{2}$
$\Rightarrow \alpha=\frac{\sqrt{5}-1}{2}, \beta=\frac{\sqrt{5}+1}{2}$
$12\left(\alpha^2+\beta^2\right)=12\left(\frac{(\sqrt{5}-1)^2+(\sqrt{5}+1)^2}{4}\right)=36$
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| વર્ગ: | $0-6$ | $6-12$ | $12-18$ | $18-24$ | $24-30$ |
| આવૃતિ: | $a$ | $b$ | $12$ | $9$ | $5$ |
જો મધ્યક $=\frac{309}{22}$ અને મધ્યસ્થ $=14$, હોય તો $(a-b)^{2}$ ની કિમંત મેળવો.
$(i)R$ નો પ્રદેશ ................ અને $(ii)R$ નો વિસ્તાર ..........