M7MBA wrote:Is x^2 - 10x > - 20?

(1) 3 < x < 7

(2) 2 < x < 9

The OA is the option A.

How can I show that the correct answer here is the option A? I ask for some help. Please.

Hello M7MBA.

I would solve it as follows: $$x^2-10x>-20$$ $$\Leftrightarrow\ \ \ x^2-10x+20>0$$ Now, we solve the equation to find the roots as follows: $$x=\frac{-\left(-10\right)\pm\sqrt{\left(-10\right)^2-4\left(1\right)\left(20\right)}}{2}=\frac{10\pm\sqrt{100-80}}{2}=\frac{10\pm\sqrt{20}}{2}$$ $$\Rightarrow\ x=\frac{10\pm2\sqrt{5}}{2}=5\pm\sqrt{5}$$ $$\Rightarrow\ x_1=5+\sqrt{5}\approx7.23.$$ $$\Rightarrow\ x_2=5-\sqrt{5}\approx2.76.$$ Now, the quadratic equation will have three different intervals, where it will have different signs.

The intervals are: ( -infinity, 2.76 ), (2.76, 7.23) and (7.23, infinity).

We are interested in knowing where the quadratic equation is positive. To find it, let's evaluate x=4 in the quadratic equation, then we will get $$4^2-10\left(4\right)+20=16-40+20=-4<0.$$ Hence, the given equation is negative between 2.76 and 7.23.

(1) 3 < x < 7

Since this interval is contained in (2.76, 7.23), then we can say that x^2 - 10x > - 20 is false always for 3 < x < 7. Hence, this statement is SUFFICIENT.

(2) 2 < x < 9

In this interval, there are values that are contained in the interval (2.76, 7.23), where the equation is positive, but also in the given interval, there are values that lie in the intervals ( -infinity, 2.76 ) or (7.23, infinity), where the equation is positive.

Since, we cannot get a unique answer here, then this statement is SUFFICIENT.

Therefore, the correct answer is the DS question is the option A.

I really hope this helps you.