## Intermediate Algebra: Week 4 Discussion

Intermediate Algebra: Week 4 Discussion

Question 1

x^{2 }+ 14x + 49 = 0

This is equivalent to ax^{2} + bx + c = 0

ax^{2} + 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

x^{2 }+ 7x + 7x + 49 = 0

(x^{2 }+ 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

x^{2 }+ 14x +49=0

-7 *-7 + 14 *-7 + 49 = 0

49 -98 + 49 =0

49 + 49 -98 =0

98-98=0

0=0

Question 2

2x^{2 }+ 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: 2x^{2 }+ 7x – 4 = 0

The new equation assumes the quadratic equation similar to ax^{2} + 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:

2x^{2 }+ 8x – x – 4 = 0

Factoring out the common values the new equation is

(2x^{2 }+ 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

2x^{2 }+ 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

2x^{2 }+ 7x= 4

2 *-4^{2} + 7 * -4 = 4

2 *16 – 28 =4

32 – 28 = 4

4 = 4