Question
Find two consecutive odd positive integers, sum of whose squares is $970.$
✓
Answer
Let two consecutive positive integers be $x$ and $x+2$
$\text { A.T.Q. }(x)^2+(x+2)^2=970$
$\Rightarrow x^2+x^2+4 x+4-970=0$
$\Rightarrow 2 x^2+4 x-966=0$
$\Rightarrow x^2+2 x-483=0$
$\Rightarrow x^2+23 x-21 x-483=0$
$\Rightarrow x(x+23)-21(x+23)=0$
$\Rightarrow(x-21)(x+23)=0$
Either $x-21=0$ or $x+23=0$
$x=21 \text { or } x=-23 \text { (rejected being }-ve)$
As integers should be + ve
$x=21 \text { and } x+2=21+2=23$
Hence integers are $21, 23$
$\text { A.T.Q. }(x)^2+(x+2)^2=970$
$\Rightarrow x^2+x^2+4 x+4-970=0$
$\Rightarrow 2 x^2+4 x-966=0$
$\Rightarrow x^2+2 x-483=0$
$\Rightarrow x^2+23 x-21 x-483=0$
$\Rightarrow x(x+23)-21(x+23)=0$
$\Rightarrow(x-21)(x+23)=0$
Either $x-21=0$ or $x+23=0$
$x=21 \text { or } x=-23 \text { (rejected being }-ve)$
As integers should be + ve
$x=21 \text { and } x+2=21+2=23$
Hence integers are $21, 23$
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