CMAT 2015 Solved Question Paper

(b) We have, y = 5${\mathit{x}}^2$ + 24x - 28 ............(i)
and y = 3${\mathit{x}}^2$ - 3x - 118 ....................(ii)
From Eq. (i). y = 5${\mathit{x}}^2$ + 24x - 28 put in Eq. (ii), we get
5${\mathit{x}}^2$ + 24x - 28 = 3${\mathit{x}}^2$ - 3x - 118
⇒ 2${\mathit{x}}^2$ + 27x = - 90
⇒ 2${\mathit{x}}^2$ + 27x + 90 = 0
⇒ 2${\mathit{x}}^2$ + 15x +12x + 90 = 0
⇒ x(2x + 15)+6(2x + 15) = 0
⇒ (2x + 15) (x + 6 ) = 0
∴ x = $-\frac{15}{2}$ , - 6
Put x = $-\frac{15}{2}$ in Eq. (ii), we get
y = 3 × $\frac{225}{4}-3\times -\left(\frac{15}{2}\right)-118$
$=\frac{675}{4}+\frac{45}{2}-118$
$=\frac{293}{4}$
Now, put x = - 6 in Eq. (ii), we get
y = 36 x 3 - 3 x (-6 ) - 118
= 108 + 18 -118
= 8
So, the coordinate pairs are (-6 , 8 ) and $\left(-\frac{15}{2},\frac{293}{4}\right)$