Solve the polynomial inequality. Write the answer in...

Key Concepts

Quadratic Equations

Quadratic equations are second-degree polynomial equations in the form ax² + bx + c = 0. They form the basis of many algebraic problems and can be solved using methods such as factoring, completing the square, or applying the quadratic formula to find their roots.

Polynomial Inequalities

Polynomial inequalities involve determining the set of variable values for which a polynomial expression is either greater than, less than, greater than or equal to, or less than or equal to zero. Solving these inequalities typically includes finding the roots of the corresponding equality to divide the number line into intervals that can then be tested for the inequality's conditions.

Quadratic Formula

The quadratic formula, given by x = (-b ± ?(b² - 4ac)) / (2a), is a general method for finding the roots of any quadratic equation. It is particularly useful when a quadratic does not factor easily, and it highlights the role of the discriminant in determining the nature of the solutions.

Sign Analysis

Sign analysis involves testing the intervals defined by the roots of a polynomial to determine the sign (positive or negative) of the expression within each interval. This step is crucial in solving polynomial inequalities, as it helps identify the continuous ranges where the inequality holds.

Interval Notation

Interval notation is a concise way to express subsets of the real numbers, particularly when describing the solution sets of inequalities. Parentheses indicate that endpoints are not included, while brackets indicate that endpoints are included, allowing for clear representation of continuous ranges that satisfy an inequality.

Transcript

00:01 Here we are given with the inequality as x square plus 3x less than equal to 6.

00:08 We can rewrite this equation as x square plus 3x minus 6 less than equal to 0.

00:16 Now to find out the solution of this equation, we are using quadratic formula, which is x is equal to minus b plus x is equal to minus b plus under root b squared.

00:30 Minus 4 ac upon 2a.

00:34 So we can find out the values of a, b, c by comparing the given equation by general equation which is ax squared plus bx plus c is equal to 0.

00:44 Now i am just substituting the values and finding out the solution for x.

00:48 X will be equal to minus 3 plus minus b square minus 4 ac will be equal to 33.

00:56 I am just writing out 33 over here upon 2.

00:59 So there will be 2 values for x which are x x is equal to minus 3 minus root 33 over 2 and minus 3 plus root 33 over 2.

01:14 These 2's value are estimated as minus 4 .37 and 1 .37.

01:23 So we are plotting these numbers on number line but these boundary points.

01:28 Here we have minus 4 .37 or we can say minus 3 minus root 3.

01:33 33 by 2 and here we have 1 .37 which can be said as minus 3 plus root 33 by 2.

01:41 Now we have 3 intervals we are taking 1 point from each interval, we are taking minus 6 from first interval, we are taking minus 4 from second interval and we are taking 2 from the third interval...

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