Practicing Success
The number of positive integral solutions of $x^2+9<(x+3)^2 < 8x + 25$, is |
2 3 4 5 |
5 |
We have, $x^2+9<(x+3)^2 < 8x + 25$ $⇒x^2+9<x^2+6x+9 <8x + 25$ $⇒x^2+9<x^2+6x+9$ and $x^2+6x+9 <8x+25$ $⇒6x>0$ and $x^2 -2x-16 <0$ $x^2-2x-16=0$ so $D=\sqrt{2^2+4×16}=2\sqrt{17}$ $x = \frac{2±2\sqrt{17}}{2}=1+\sqrt{17}$ for $(x^2-2x-16)<0$ $⇒ 1-\sqrt{17}<x<1+\sqrt{17}$ $⇒0<x<1+ \sqrt{17}⇒x=1,2,3,4,5$ |