Solve quadratic inequalities with calculations Algebra I

Recall that:

• If f\left(x\right)=ax^{2}+bx+c has x_{1} and x_{2} as roots, then f has the sign of a outside x_{1} and x_{2}, and the opposite sign of a inside x_{1} and x_{2}.
• If f\left(x\right)=ax^{2}+bx+c has one double root, then f has the sign of a on the entire domain of real numbers.
• If f\left(x\right)=ax^{2}+bx+c has no root, then f has the sign of a on the entire domain of real numbers.

We want to solve:

2x^2-3x-1 \gt 4

First write the inequality in standard form:

2x^{2}-3x-5 \gt 0

Calculate the roots:

x_{1}=\dfrac{-\left(-3\right)-\sqrt{\left(-3\right)^{2}-4\left(2\right)\left(-5\right)}}{2\left(2\right)}=\dfrac{3-\sqrt{49}}{4}=-1

x_{2}=\dfrac{-\left(-3\right)+\sqrt{\left(-3\right)^{2}-4\left(2\right)\left(-5\right)}}{2\left(2\right)}=\dfrac{3+\sqrt{49}}{4}=\dfrac{5}{2}

Since a=2 is positive, we can conclude that 2x^{2}-3x-5 \gt 0 outside x_{1} and x_{2}.

In other words, x is a solution of the inequality if and only if:

x \lt -1 or x \gt \dfrac{5}{2}

Recall that:

• If f\left(x\right)=ax^{2}+bx+c has x_{1} and x_{2} as roots, then f has the sign of a outside x_{1} and x_{2}, and the opposite sign of a inside x_{1} and x_{2}.
• If f\left(x\right)=ax^{2}+bx+c has one double root, then f has the sign of a on the entire domain of real numbers.
• If f\left(x\right)=ax^{2}+bx+c has no root, then f has the sign of a on the entire domain of real numbers.

We want to solve:

3x^2+6x-7 \leqslant x+1

First, write the inequality in standard form:

3x^{2}+5x-8 \leqslant 0

Calculate the roots:

x_{1}=\dfrac{-5-\sqrt{\left(5\right)^{2}-4\left(3\right)\left(-8\right)}}{2\left(3\right)}=\dfrac{-5-\sqrt{121}}{6}=-\dfrac{8}{3}

x_{2}=\dfrac{-5+\sqrt{\left(5\right)^{2}-4\left(3\right)\left(-8\right)}}{2\left(3\right)}=\dfrac{-5+\sqrt{121}}{6}=1

Since a=3 is positive, we can conclude that 3x^{2}+5x-8 \leqslant 0 inside x_{1} and x_{2}.

In other words, x is a solution of the inequality if and only if:

-\dfrac{8}{3}\leqslant x\leqslant1

Recall that:

• If f\left(x\right)=ax^{2}+bx+c has x_{1} and x_{2} as roots, then f has the sign of a outside x_{1} and x_{2}, and the opposite sign of a inside x_{1} and x_{2}.
• If f\left(x\right)=ax^{2}+bx+c has one double root, then f has the sign of a on the entire domain of real numbers.
• If f\left(x\right)=ax^{2}+bx+c has no root, then f has the sign of a on the entire domain of real numbers.

We want to solve:

\left(4x-3\right)\left(x+5\right)\geqslant 3x^{2}+10x-7

First, write the inequality in standard form:

4x^{2}+17x-15\geqslant3x^{2}+10x-7

x^{2}+7x-8 \geqslant 0

Calculate the roots:

x_{1}=\dfrac{-7-\sqrt{\left(7\right)^{2}-4\left(1\right)\left(-8\right)}}{2\left(1\right)}=\dfrac{-7-\sqrt{81}}{2}=-8

x_{2}=\dfrac{-7+\sqrt{\left(7\right)^{2}-4\left(1\right)\left(-8\right)}}{2\left(1\right)}=\dfrac{-7+\sqrt{81}}{2}=1

Since a=1 is positive, we can conclude that x^{2}+7x-8 \geqslant 0 outside x_{1} and x_{2}.

In other words, x is a solution of the inequality if and only if:

x\leqslant-8 or x\geqslant1

Recall that:

• If f\left(x\right)=ax^{2}+bx+c has x_{1} and x_{2} as roots, then f has the sign of a outside x_{1} and x_{2}, and the opposite sign of a inside x_{1} and x_{2}.
• If f\left(x\right)=ax^{2}+bx+c has one double root, then f has the sign of a on the entire domain of real numbers.
• If f\left(x\right)=ax^{2}+bx+c has no root, then f has the sign of a on the entire domain of real numbers.

We want to solve:

4 \gt 3x^{2}-2x-4

First, write the inequality in standard form:

3x^{2}-2x-8 \lt 0

Calculate the roots:

x_{1}=\dfrac{-\left(-2\right)-\sqrt{\left(-2\right)^{2}-4\left(3\right)\left(-8\right)}}{2\left(3\right)}=\dfrac{2-\sqrt{100}}{6}=-\dfrac{4}{3}

x_{2}=\dfrac{-\left(-2\right)+\sqrt{\left(-2\right)^{2}-4\left(3\right)\left(-8\right)}}{2\left(3\right)}=\dfrac{2+\sqrt{100}}{6}=2

Since a=3 is positive, we can conclude that 3x^{2}-2x-8 \lt 0 inside x_{1} and x_{2}.

In other words, x is a solution of the inequality if and only if:

-\dfrac{4}{3} \lt x \lt 2

Recall that:

• If f\left(x\right)=ax^{2}+bx+c has x_{1} and x_{2} as roots, then f has the sign of a outside x_{1} and x_{2}, and the opposite sign of a inside x_{1} and x_{2}.
• If f\left(x\right)=ax^{2}+bx+c has one double root, then f has the sign of a on the entire domain of real numbers.
• If f\left(x\right)=ax^{2}+bx+c has no root, then f has the sign of a on the entire domain of real numbers.

We want to solve:

x^{2}-9 \lt 0

Calculate the roots:

x_{1}=\dfrac{-0-\sqrt{\left(0\right)^{2}-4\left(1\right)\left(-9\right)}}{2\left(1\right)}=\dfrac{0-\sqrt{36}}{2}=-3

x_{2}=\dfrac{-0+\sqrt{\left(0\right)^{2}-4\left(1\right)\left(-9\right)}}{2\left(1\right)}=\dfrac{0+\sqrt{36}}{2}=3

Since a=1 is positive, we can conclude that x^{2}-9 \lt 0 inside x_{1} and x_{2}.

In other words, x is a solution of the inequality if and only if:

-3 \lt x \lt 3

Recall that:

• If f\left(x\right)=ax^{2}+bx+c has x_{1} and x_{2} as roots, then f has the sign of a outside x_{1} and x_{2}, and the opposite sign of a inside x_{1} and x_{2}.
• If f\left(x\right)=ax^{2}+bx+c has one double root, then f has the sign of a on the entire domain of real numbers.
• If f\left(x\right)=ax^{2}+bx+c has no root, then f has the sign of a on the entire domain of real numbers.

We want to solve:

7x^{2}+x+20 \geqslant 9

First, write the inequality in standard form:

7x^{2}+x+11 \geqslant 0

Calculate the roots:

x_{1}=\dfrac{-1-\sqrt{\left(1\right)^{2}-4\left(7\right)\left(11\right)}}{2\left(7\right)}=\dfrac{-1-\sqrt{-307}}{14} which is not a real number.

x_{2}=\dfrac{-1+\sqrt{\left(1\right)^{2}-4\left(7\right)\left(11\right)}}{2\left(7\right)}=\dfrac{-1+\sqrt{-307}}{14} which is not a real number.

Therefore, 7x^{2}+x+11 has no real roots.

Since a=7 is positive, we can conclude that 7x^{2}+x+11 \geqslant 0 on the entire domain of real numbers.

In other words, x is a solution of the inequality if and only if:

x\in\mathbb{R}

Recall that:

• If f\left(x\right)=ax^{2}+bx+c has x_{1} and x_{2} as roots, then f has the sign of a outside x_{1} and x_{2}, and the opposite sign of a inside x_{1} and x_{2}.
• If f\left(x\right)=ax^{2}+bx+c has one double root, then f has the sign of a on the entire domain of real numbers.
• If f\left(x\right)=ax^{2}+bx+c has no root, then f has the sign of a on the entire domain of real numbers.

We want to solve:

-5x^{2}+2x-10 \gt 6

First we have to write the inequality in standard form:

-5x^{2}+2x-16 \gt 0

Calculate the roots:

x_{1}=\dfrac{-2-\sqrt{\left(2\right)^{2}-4\left(-5\right)\left(-16\right)}}{2\left(-5\right)}=\dfrac{-2-\sqrt{-316}}{-10} which is not a real number.

x_{2}=\dfrac{-2+\sqrt{\left(2\right)^{2}-4\left(-5\right)\left(-16\right)}}{2\left(-5\right)}=\dfrac{-2+\sqrt{-316}}{-10} which is not a real number.

So, -5x^{2}+2x-16 has no real roots.

Since a=-5 is negative, we conclude that there are no real numbers for which -5x^{2}+2x-16 \gt 0

In other words, the solution of the inequality is the empty set.