03Gsolvingcubiceqns

= Solving Cubic Equations =


 * Cubic equations can be solved in a similar way to quadratic equations
 * by factorising and then using the Null Factor Law
 * Or If they form a perfect cube,
 * they can be rearranged to make x the subject.


 * Using the Null Factor Law **

... ... Recall that if ... (x + 3)(x – 5) = 0 ... ... Then either ... x + 3 = 0 .. or .. x – 5 = 0 ... ... So ... .... ... .... x = –3 .... or ...... x = 5

... ... The same rule applies with three factors


 * Example 1 **

... ... Solve (x – 5)(x + 3)(3x – 2) = 0


 * Solution:**

... ... ** Solve using the Null Factor Law **

math . \qquad \big( x - 5 \big) \big( x + 3 \big) \big( 3x - 2 \big) = 0 \\ .\\ . \qquad x - 5 = 0 \quad or \quad x + 3 = 0 \quad or \quad 3x - 2 = 0 \\ .\\ . \qquad x = 5 \qquad or \qquad x = -3 \qquad or \qquad x = \dfrac{2}{3} math


 * Example 2 **

... ... Solve 2x 3 – x 2 – 15x + 18 = 0


 * Solution:**

... ... ** First use the factor law to find a factor **

math . \qquad P(1) = 2 (1)^3 - (1)^2 - 15(1) + 18 = 4 \\. \\ . \qquad P(-1) = 2 (-1)^3 - (-1)^2 - 15(-1) + 18 = 30 \\. \\ . \qquad P(2) = 2 (2)^3 - (2)^2 - 15(2) + 18 = 0 math

... ... ** P(2) = 0 ** .. ** so ** .. ** (x – 2) is a factor **

... ... ** Use short division to factorise the cubic, ** ... ... ** introducing by introducing an unknown value of a **

math . \qquad 2x^3 - x^2 - 15x + 18 \\. \\ . \qquad = \big( x - 2 \big) \big( 2x^2 + ax - 9 \big) \qquad \{ \textit{expand} \} \\. \\ . \qquad = 2x^3 + ax^2 - 4x^2 ... \\. \\ . \qquad \qquad \{ \textit{now equate coefficients of } x^2 \} \\. \\ . \qquad -x^2 = ax^2 - 4x^2 \\. \\ . \qquad -1 = a - 4 \\. \\ . \qquad \;\; a = 3 math

... ... ** So we can now factorise the cubic **

math . \qquad 2x^3 - x^2 - 15x + 18 \\. \\ . \qquad = \big( x - 2 \big) \big( 2x^2 + 3x - 9 \big) \\. \\ . \qquad = \big( x - 2 \big) \big( 2x^2 - 3x + 6x - 9 \big) \\. \\ . \qquad = \big( x - 2 \big) \big( x (2x - 3) + 3(2x-3) \big) \\. \\ . \qquad = \big( x - 2 \big) \big( 2x - 3 \big) \big( x + 3 \big) math

... ... ** Now we can return to the equation and solve it using the Null Factor Law **

math . \qquad 2x^3 - x^2 - 15x + 18 = 0 \\. \\ . \qquad \big( x - 2 \big) \big( 2x - 3 \big) \big( x + 3 \big) = 0 \\. \\ . \qquad x - 2 = 0 \quad or \quad 2x - 3 = 0 \quad or \quad x + 3 = 0 \\. \\ . \qquad x = 2 \qquad or \qquad x = \dfrac{3}{2} \qquad or \qquad x = -3 math


 * Solving Equations involving a Perfect Cube **


 * If the equation is in the form a(x – b) 3 + c = 0
 * it can be solved by rearranging to make x the subject.


 * Example 3 **

... ... Solve 2(x – 3) 3 – 16 = 0


 * Solution:**

math . \qquad 2 \big( x - 3 \big)^3 - 16 = 0 \qquad \{ + 16 \} \\.\\ . \qquad 2 \big( x - 3 \big)^3 = 16 \qquad \;\; \{ \div 2 \} \\.\\ . \qquad \; \big( x - 3 \big)^3 =8 \qquad \;\; \{ \textit{cube root} \} \\.\\ . \qquad \;\; x - 3 = \sqrt[3]{8} \\.\\ . \qquad \;\; x - 3 = 2 \\. \\ . \qquad \quad x = 5 math


 * Example 4 **

... ... Solve (x + 5) 3 + 7 = 0


 * Solution:**

math . \qquad \big( x + 5 \big)^3 + 7 = 0 \\. \\ . \qquad \big( x + 5 \big)^3 = -7 \\. \\ . \qquad \; x + 5 = \sqrt[3]{-7} \\.\\ . \qquad \;\; x = -5 - \sqrt[3]{7} math


 * Solving Equations With the Classpad Calculator **


 * Your calculator can be used to solve cubic equations
 * The solve command is in the ACTION menu, ADVANCED submenu
 * Enter the equation, followed by a comma and then the variable to be solved for.


 * In these examples, where //**x **// is the only pronumeral,
 * the Classpad doesn't need the comma and variable to be included.


 * Notice that the cube root of 7 in the last example is shown as 7 to the power of one-third.

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