Solution - Solving quadratic inequalities using the quadratic formula

The coefficients of our inequality, $$10s^2+1s-49>0$$, are:

$$a$$ = 10

$$b$$ = 1

$$c$$ = -49

To find the roots of a quadratic equation, plug its coefficients ($$a$$, $$b$$ and $$c$$ ) into the quadratic formula:

$$s=\left(-b\pm sqrt\left(b^2-4ac\right)\right)/\left(2a\right)$$

$$s=\left(-1\pm sqrt\left(1^2-4*10*-49\right)\right)/\left(2*10\right)$$

$$s=\left(-1\pm sqrt\left(1-4*10*-49\right)\right)/\left(2*10\right)$$

$$s=\left(-1\pm sqrt\left(1-40*-49\right)\right)/\left(2*10\right)$$

$$s=\left(-1\pm sqrt\left(1--1960\right)\right)/\left(2*10\right)$$

$$s=\left(-1\pm sqrt\left(1+1960\right)\right)/\left(2*10\right)$$

$$s=\left(-1\pm sqrt\left(1961\right)\right)/\left(2*10\right)$$

$$s=\left(-1\pm sqrt\left(1961\right)\right)/\left(20\right)$$

to get the result:

$$s=\left(-1\pm sqrt\left(1961\right)\right)/20$$

Simplify $$1961$$ by finding its prime factors:

The prime factorization of $$1961$$ is $$37\cdot 53$$

$$s=\left(-1\pm sqrt\left(1961\right)\right)/20$$

The ± means two roots are possible.

Separate the equations:
$$s_1=\left(-1+sqrt\left(1961\right)\right)/20$$ and $$s_2=\left(-1-sqrt\left(1961\right)\right)/20$$

$$s_1=\left(-1+sqrt\left(1961\right)\right)/20$$

$$s_1=\left(-1+sqrt\left(1961\right)\right)/20$$

$$s_1=\left(-1+44.283\right)/20$$

$$s_1=\left(-1+44.283\right)/20$$

$$s_1=\left(43.283\right)/20$$

$$s_1=\frac{43.283}{20}$$

$$s_1=2.164$$

$$s_2=\left(-1-sqrt\left(1961\right)\right)/20$$

$$s_2=\left(-1-44.283\right)/20$$

$$s_2=\left(-1-44.283\right)/20$$

$$s_2=\left(-45.283\right)/20$$

$$s_2=-\frac{45.283}{20}$$

$$s_2=-2.264$$

To find the intervals of a quadratic inequality, we start by finding its parabola.

The roots of the parabola (where it meets the x-axis) are: -2.264, 2.164.

Since the $$a$$ coefficient is positive ($$a$$=10), this is a "positive" quadratic inequality and the parabola points upward, like a smile!

If the inequality sign is ≤ or ≥ , then the intervals include the roots and we use a solid line. If the inequality sign is < or > the intervals do not include the roots and we use a dotted line.

Since $$10s^2+1s-49>0$$ has a $$>$$ inequality sign, we look for the parabola intervals that are above the x-axis.

Solution: $$$$

Interval notation: $$$$