1.
In the following questions two equations numbered I and
II are given. You have to solve both the equations and
a: if x > y
b: if x ≥ y
c: if x < y
d: if x ≤ y
e: if x = y or the relationship cannot be established.I. $$x^{2}+x-12=0$$
II. $$y^{2}+2y-8=0$$
Write Comment
Comments
By: anil on 05 May 2019 01.36 am
$$x^2+x-12 = 0$$
$$(x-3)(x+4) = 0$$
$$x = -4, 3$$ $$y^2+2y-8 = 0$$
$$(y-2)(y+4) = 0$$
$$y = -4, 2$$ Hence, a relationship can not be established between $$x$$ and $$y$$
Terms And Service:We do not guarantee the accuracy of available data ..We Provide Information On Public Data.. Please consult an expert before using this data for commercial or personal use
$$(x-3)(x+4) = 0$$
$$x = -4, 3$$ $$y^2+2y-8 = 0$$
$$(y-2)(y+4) = 0$$
$$y = -4, 2$$ Hence, a relationship can not be established between $$x$$ and $$y$$