In this problem, you will use the following version of Cardano's formula for solving the cubic e

Step 1: Substitute ( x = y - frac{6}{3} = y - 2 ) in the given cubic equation ( x^3 - 6x^2 + 144

In this problem, you will use the following version of...

In this problem, you will use the following version of Cardano's formula for solving the cubic equation, which we developed in class: x = ?(?((m/3)³ + (n/2)²) + n/2) - ?(?((m/3)³ + (n/2)²) - n/2) It is for a reduced cubic equation of the form, x³ + mx = n, where n is meant to be positive (although the formula works just as well if n is negative). We will use this formula to solve the cubic equation x³ - 6x² + 144x - 1539 = 0 using Cardano's method. a. First, given a general cubic equation ax³ + bx² + cx + d = 0, we can use the substitution x = y - b/3a to reduce the cubic to a depressed cubic with no squared term. Use the substitution in the case of our specific equation x³ - 6x² + 144x - 1539 = 0 to produce a depressed cubic with no y² term. b. Notice that this can be put into the form y³ + my = n for which Cardano's solution given above works. Finish solving for x using Cardano's formula. Don't forget to solve for x after you've solved for y! c. Now that you've found the real root, factor this out to produce a quadratic and solve the quadratic for the two complex roots.

Transcript

00:01 In this question they given here the cubic equation which is implies that so that is here x -cube minus 6x square plus 144 x minus 15139 is equal to 0 here.

00:23 So based on this equation they're given here so which is three questions now we have to find out.

00:30 So answers to the three questions given three questions.

00:43 So now let come to the answers here.

00:45 But the first question, so before that from the question, xq minus 6x square plus 144x minus 1539 is equal to 0.

00:59 So let us assume that this is the equation number one from this.

01:06 So substitute, which is x.

01:12 Is equal to y minus minus six divided by three which is equal to y plus two so that is implies that after substituting in equation one we will get so y plus two whole cube minus six y plus two whole square plus 144 multiply by y plus two which is minus 15139 is equal to zero here implies that y cube plus 8 plus 6 y square plus 12 y minus 6 y square minus 2y minus 24 y plus 144y plus 288 minus 1539 which is equal to 0 so in this step plus 6 y square and minus 6 y square get cancelled so finally we will get the equation that is y cube plus 132y minus 1267 which is equal to 0 so now in this case let us assume that this is the equation number 2 so here this is the final answer for the first question so now let come to the second one that is b question so comparing equation 2 with the comparing equation 2 with yq plus m y is equal to n.

03:28 After that we get the values that is which is m is equal to 132 comma n is equal to 1267.

03:39 So by using the cardinodes by using cardinodes so here cardinose method that is y is equal to cube root of so it is square root of 132 divided by 3 whole cube plus 112 167 divided by 2 whole square plus 12167 divided by 2 which is minus cube root of square root of so 132 divided by 3 whole cube plus 1267 divided by 2 whole square so minus 12167 divided by 2 whole square so minus 12167 divided by 2 so from this step in this case some numbers are getting cancelled that is here three and so here so here is cube root and here three ones 3 40 4s get cancelled here also 3 ones and 340 4s get cancelled so that now we will get from this step that is equal to cube root of which is square root of so 851 84 place after squaring so 16 -0 50289 divided by 2 plus 1267 divided by 2 which is minus from the cube root of square root of 8 51 8 4 plus 16 0 5289 divided by 2 minus 1267 divided by 2 here from this step we will get here cube root half so 13995 divided by 2 plus 1267 divided by 2 minus cube root half so 1395 divided by 2 minus cube root half so 1395 divided by 2 minus 1267 divided by 2 which is equal to cube root 1.

06:53 Of 2 ,662 divided by 2 minus cube root of 128 divided by 2.

07:01 So in this case we have to separate the cube roots for each term...

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