Category Archives: Algebra

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 zero. This will result to: 2x2 + 7x – 4 = 0

The new equation assumes the quadratic equation similar to ax2 + bx + c = 0 where ac is a product of two values; mn which when added result to coefficient be; c =mn, b = m+n. Thus expanding this equation we find that ac = 2 * -4 = -8, two numbers when multiplied result to -8 and when added result to 7 are 8 and -1. Thus the expanded equation is:

2x2 + 8x – x – 4 = 0

Factoring out the common values the new equation is

(2x2 + 8x) –(x+ 4) = 0

2x(x+4) -1(x+4) = 0

(2x-1)(x+4)= 0

Using zero factor property, the values of x will be

2x – 1 = 0 or x + 4 = 0

For 2x – 1 = 0

2x = 1

x = ½

For

x + 4 = 0

x = -4

Checking

When x = 1/2

2x2 + 7x=4

2 *(1/2)2 + 7 *1/2 = 4

2/4 + 7/2 = 4

½ + 7/2 = 4

(1 +7)/2 = 4

8/2 = 4

4 = 4

When x = -4

2x2 + 7x= 4

2 *-42 + 7 * -4 = 4

2 *16 – 28 =4

32 – 28 = 4

4 = 4

 

See also Intermediate Algebra: Week 4 Discussion

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

10 = 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