a An absolute value measures an expression's distance from a midpoint on a number line. Since distance cannot be negative, the absolute value of a number is always non-negative.
|5x+8| ≥ - 4
Regardless of the number we substitute for x, the absolute value of x and the absolute value of 5x+8 will always be non-negative. Therefore, it will be greater than - 4. Since any value of x satisfies the inequality, the solution is all real numbers.
Test a value from each interval to see if it satisfies the original inequality.
Step 1
We will start by solving the related equation.
x^2+x-20=0
⇕
1x^2+ 1x+( - 20)=0
We see above that a= 1, b= 1, and c= - 20. Let's substitute these values into the Quadratic Formula to solve the equation.
Now we can calculate the first root using the positive sign and the second root using the negative sign.
x=- 1 ± 9/2
x_1=- 1 + 9/2
x_2=- 1 - 9/2
x_1=8/2
x_2=- 10/2
x_1= 4
x_2= - 5
Step 2
The solutions of the related equation are 4 and - 5. Let's plot them on a number line. Since the original is a strict inequality, the points will be open.
Step 3
Finally, we must test a value from each interval to see if it satisfies the original inequality. Let's choose a value from the first interval, x < - 5. For simplicity, we will choose x=- 6.
Since x=- 6 did not produce a true statement, the interval x < - 5 is not part of the solution. Similarly, we can test the other two intervals.
Interval
Test Value
Statement
Is It Part of the Solution?
- 5 < x < 4
0
- 20 < 0 âś“
Yes
x > 4
5
10 ≮ 0 *
No
We can now write the solution and show it on a number line.
- 5 < x < 4
c We will use the Quadratic Formula to solve the given quadratic equation.
ax^2+ bx+ c=0 ⇕ x=- b± sqrt(b^2-4 a c)/2 aLet's start by rewriting the equation so all of the terms are on the left-hand side.
2x^2-6x=- 5 ⇔ 2x^2-6x+ 5=0
Now we can identify the values of a, b, and c.
2x^2-6x+5=0 ⇕ 2x^2+( - 6)x+ 5=0
We see that a= 2, b= - 6, and c= 5. Let's substitute these values into the Quadratic Formula.
Oops! The square root of a negative number is undefined for real numbers! As we can see, using the Quadratic Formula for the given equation resulted in a contradiction. Thus, there are no real solutions.
For further explanation, notice that the left-hand side of the equation is a quadratic function. Let's graph it.
The solutions of the given equation are x-intercepts of this function. Notice that this parabola lies above the x-axis, thus it does not have x-intercepts. Therefore there are no real solutions of the given equation.
d This equation would be much easier to solve if it had no fractions. We can start solving by changing this equation to a simpler equivalent equation by eliminating fractions. To do this, we will multiply both sides of the equation by 9, which is the lowest common denominator.
To solve an equation, we should first gather all of the variable terms on one side of the equation and all of the constant terms on the other side using the Properties of Equality.
