Question

Which point satisfies the linear quadratic system y=x+3 and y=5-x${}^2$?

• A. (-2,1)
• B. (2,1)
• C. (-1,2)
• D. (4,-1)

Answer 1

The correct option is A (-2,1)

On solving both,
$x+3=5-x^2$
$\Rightarrow x^2+x-2=0$
$\Rightarrow x^2+2x-x-2=0$
$\Rightarrow x(x+2)x-1(x+2)=0$
$\Rightarrow (x+2)(x-1)=0$
$\Rightarrow x=-2,1$
When, $x=-2,y=-2+3=1$, point is (-2,1)
When, $x=1,y=1+3=4$, point is (1,4)
Option A is correct.