Mathematics - Quadratic Equation Question with Solution | TestHub

$$y{x}^2-y=ax+1$$

$$y{x}^2-ax-y-1=0$$

For real roots, the discriminant must be non-negative:

$$(-a{)}^2-4(y)(-y-1)\ge 0$$

$$a^2+4y(y+1)\ge 0$$

$$a^2+4y^2+4y\ge 0$$

$$4y^2+4y+a^2\ge 0$$

For this quadratic in $y$ to be always non-negative, its discriminant must be less than or equal to zero:

$$(4{)}^2-4(4)(a^2)\le 0$$

$$16-16a^2\le 0$$

$$16(1-a^2)\le 0$$

$$1-a^2\le 0$$

$$a^2-1\ge 0$$

$$(a-1)(a+1)\ge 0$$

This implies $a\in (-\infty ,-1]\cup [1,\infty )$.

 

Given the constraint $-2\le a\le -1$

 

, the possible integer values for $a$ are $-2,-1$

 

as for a=-1 it has commaon roots in Nr and Denominator. so its range is not R . only -2 is correct and .

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