Question

Coordinates of a point P are $(a,b)$ where $a$ is a root of the equation

$x^2+x-42=0$

and $b$ is an integral root of the equation
$x^2+ax+a^2-37=0$.

The coordinates of P can be

• A. $(6,4)$
• B. $(-7,4)$
• C. $(-7,3)$
• D. $(6,-3)$

Answer 1

Answer: (B), (C)

$(-7,3)$
$(-7,4)$

$a^2+a-42=0\Rightarrow a=-7$ or $a=6.$
Since b is a root of $x^2+ax+a^2-37=0$
For $a=-7,$ we have $x^2-7x+49-37=0$
$\Rightarrow x^2-7x+12=0\Rightarrow x=4,3,so,b=4$ or $3.$
So the coordinates of P can be $(-7,4)$ or $(-7,3)$,

For $a=6,$ we have $x^2+6x-1=0$ which does not give an integral value, so $a\ne 6.$