# Solving algebraically Quadratic inequalities with modulus abs $(x^2+5x-6) \gt 5$

I tried to solve this inequality algebraically by opening modulus and doing $x^2+5x-6 \gt5$ or $\lt -5$. I got $2$ solution sets with no overlap. When we solve graphically there are $4$ points of intersection. How do we get the answer algebraically ? thanking you in advance Ashwini

## 2 Answers

We need to solve $$(x^2+5x-6)^2-25>0$$ or $$(x^2+5x-11)(x^2+5x-1)>0$$ or $$\left(x-\frac{-5-\sqrt{69}}{2}\right)\left(x-\frac{-5-\sqrt{29}}{2}\right)\left(x-\frac{-5+\sqrt{29}}{2}\right)\left(x-\frac{-5+\sqrt{69}}{2}\right)>0,$$ which gives the answer: $$\left(-\infty,-\frac{5+\sqrt{69}}{2}\right)\cup\left(-\frac{5+\sqrt{29}}{2},\frac{-5+\sqrt{29}}{2}\right)\cup\left(\frac{-5+\sqrt{69}}{2},+\infty\right).$$

If you look at it graphically, i.e. plotting the quadratic, and look at where the equality is satisfied, you will see that you should get two "solution sets". If you wanted to solve for these intervals algebraically, begin by splitting the problem up. First, you want to set the quadratic equal to 5 and solve, then set it equal to -5 and solve. Setting it equal to 5 will yield two solutions, a and b, where a < b, which we can see that any value greater than b and any value less than a will violate the inequality. Doing the same for -5, we get two values, c and d, and we can see that any values between these two values will also violate the inequality. Putting these results into interval notation shall get you your result.