Intermediate Algebra: Week 4 Discussion

Intermediate Algebra: Week 4 Discussion

Intermediate Algebra: Week 4 Discussion   Intermediate Algebra: Week 4 Discussion Question 1 x2 + 14x + 49 = 0 This is equivalent to ax2 + bx + c = 0 ax2 + bx + c = 0 Where c is attained by multiplying two numbers which when added they result to b coefficient.  The two numbers while multiplied result to 49 and when added result to 14 are 7 and 7 Expanding the equation, this results to x2 + 7x  + 7x + 49 = 0 (x2 + 7x)  + (7x + 49) = 0 Factoring out the common values, this is obtained x (x +7) +7(x + 7) = 0 (x+7)(x+7) = 0 Using the zero factor property, the solution to this equation is: x+7 = 0 x = -7 Checking for accuracy x2 + 14x +49=0 -7 *-7 + 14 *-7 + 49 = 0 49 -98 + 49 =0 49 + 49 -98 =0 98-98=0 0=0 Question 2 2x2 + 7x = 4 We need to create a quadratic equation by shifting 4 to the left side and equating the equation to...
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Intermediate Algebra: Week 3 Discussion

Intermediate Algebra: Week 3 Discussion

  Intermediate Algebra: Week 3 Discussion Question 1 10002/3 This problem will be solved using power exponent rule The first step will be rewriting the number 1000 as a prime to a power. 1000 = 103 Thus the new expression will be (103)2/3 Using power rules the outside exponent will be multiplied with the inner exponent c as follows 3 * 2/3 = 2 Thus, the new exponent will be 2 and the new expression will be 102   = 10 *10 =100 Question 2 (54/46) ½ This problem will be solved using power exponent rule. The power rule will be used in handling the outside component. The inner exponents will be multiplied with outside exponents as follows: 4* ½ and 6 * ½ . The new exponents will be 2 and 3 respectively The new expression will be 52/63 The numerator will be squared; 5 x 5 = 25 while the denominator will be cubed to get 6*6*6 =216 The new fraction will be: 52/63 = 25/216  ...
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