Solving an Equation Involving an Absolute Value Find all solutions of the equation algebraically

Solving an Equation Involving an Absolute Value Find all...

Key Concepts

Absolute Value Function

An absolute value function returns the non-negative magnitude of a number regardless of its sign. It defines the expression |a| such that it is a if a is non-negative and -a if a is negative. This property is essential for understanding how equations involving absolute values split into distinct cases based on the sign of the input.

Case Analysis

Case analysis is a method where the absolute value expression is broken down into two cases: one where the expression inside is non-negative and another where it is negative. By handling each scenario separately, one can convert the absolute value equation into simpler equations that can be solved individually.

Quadratic Equations

Quadratic equations, which take the standard form ax² + bx + c = 0, are solved using methods such as factoring, completing the square, or the quadratic formula. These equations often arise when squaring or setting up equations from the case analysis of absolute value expressions.

Checking for Extraneous Solutions

When solving equations that involve functions with restrictions, such as absolute values, it is important to verify potential solutions in the original equation. This ensures that any solutions that may have been introduced during the algebraic manipulation, and that do not satisfy the original domain restrictions, are excluded.

Transcript

00:01 It is given that the equation modulus of x plus 1 is equal to be x squared minus 5.

00:08 Now the absolute value of x plus 1 is equal to plus minus x square minus 5.

00:15 Now either x plus 1 is equal to x square minus 5 or x plus 1 is equal to minus of x square minus 5.

00:25 Now from here we have x square minus x minus five minus one is equal to 0.

00:32 So we have the quadratic equation xkere minus x minus 6 is equal to 0.

00:38 Now from here we have x plus 1 is equal to be minus x square plus 5.

00:43 So we have x care plus x plus 1 minus 5 is equal to 0.

00:50 So we have the quadratic equation.

00:52 We have x square plus x minus 4 is equal to 0 now by simplifying and factorizing this quadratic equation we have the factors x minus 3 multiplied with x plus 2 is equal to 0 now by factorizing this equation now the value of x from this quadratic equation by using the quadratic formula we have minus b plus minus square root of b square minus 4 a c upon 2a.

01:23 Now we're putting the value of a, b, and c by comparing it with the standard quality equation, we have the value of x is equal to minus 1 plus minus scale root of 17 upon 2.

01:34 Now from here we have two values of x, that is 3 and minus 2.

01:39 Now we get four values of x.

01:41 Now we need to check which of the following are the solutions of our equation.

01:47 Now, firstly, put x is equal to 3.

01:51 Now put x.

01:52 Is equal to 3 in the equation we have left -hand side of the equation will equal to modulus of 3 plus 1 that is equal to modulus of 4 models of 4 is equal to 4 now the right -and -side of the equation will reduce this to 3 square minus 5 which is equal to 9 minus 5 which is equal to 4 so now we can see that left -and -side is equal to right -and -side so the value of x is equal to 3 is the solution of the given equation now put x is equal to minus 2.

02:24 In the equation, we have left -hand side is equal to be modulus of minus 2 plus 1 which is equal to be the modulus of minus 1...

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