Question 126

Two quadratic equations are given below. Solve these equations and select the appropriate option

$$x^{2} - 15x +26 = 0$$
$$y^{2} + 9y - 22= 0$$

Solution

$$x^{2} - 15x +26 = 0$$
$$(x-2)(x-13)=0$$
X = 2 or 13

$$y^{2} + 9y - 22= 0$$
$$(y+11)(y-2) = 0$$
Y= 2 or -11

Therefore, x is greater than or equal to y. option A.

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