Solve each equation by factoring

Question

A college student invested part of a $25,000 inheritance at 7% interest and the rest at 6%.  If his annual interest is $1,670 how much did he invest at 6%?  If I told you the answer is $8,000, in your own words, using complete sentences, explain how you would solve the problem.

Solve each equation by factoring.

8.  x² + 8x + 15 = 0

Complete the square to make each binomial a perfect square triangle.

30. x² - 12x

Use the quadratic formula to solve each equation.

54.  6x² + x - 2= 0

58. 3x² + 18x + 15= 0

Solve each problem.

4. Geometric problem: A rectangle is 5 times as long as it is wide.  If the area is 125 square feet,         find its perimeter.

6.  Geometric problem:  The base of a triangle is one-third as long as its height.  If the area of the triangle is 24 square meters, how long is its base?

Perform all operations.  Give all answers in a + bi form.

14.  (- 7 + 2i) + (2 - 8i)

16.   (11 + 2i) - (13 - 5i )

18.  (5 + ) - (23i - 32)

Solve each inequality, graph the solution set and write each answer in interval notation.

14.  -2X + 4 ? 6

22.  3(x + 2)  ≤ (2(x + 5)

54.  x² - 13x + 12 ≤ 0

1.  A college student invested part of a $25,000 inheritance at 7% interest and the rest at 6%.  If his annual interest is $1,670 how much did he invest at 6%?  If I told you the answer is $8,000, in your own words, using complete sentences, explain how you would solve the problem.

è Let us assume that x be the amount invested at rate of 6%.

Then the amount invested at 7% is 25000-x

So the total annual interest

=7% of (25000-x) + 6% of x

=7/100 of (25000-x) + 6/100 of x

=1750-7x/100+6x/100

=1750-x/100

By the problem

1750-x/100=1670

So, x/100=1750-1670

Or,x/100=80

Or,x=8000

So your answer was right

He invested $8000 at 6%

Solve each equation by factoring.

8.  x² + 8x + 15 = 0

èx^2+3x+5x+15=0

èx(x+3)+5(x+3)=0

è(x+3)(x+5)=0

So either x+3=0 or x+5=0

Therefore x= -3 or x= -5

Complete the square to make each binomial a perfect square triangle.

30. x² - 12x

è(x)^2-2*x*6

So for making a perfect square we need to add (6)^2 with the above expression

So now (x)^2-12x+(6)^2

=(x-6)^2 (a perfect square)

Use the quadratic formula to solve each equation.

54.  6x² + x - 2= 0

èx= {-b ±(b^2-4ac)^1/2}/2a

here a=6 b=1 c= -2

so b^2-4ac=1+48=49

therefore (b^2-4ac)^1/2=7

so, x= (-1±7)/2

from the above expression we get x=3, -4

58. 3x² + 18x + 15= 0

èx= {-b ±(b^2-4ac)^1/2}/2a

here a=3 b=18 c=15

so b^2-4ac=324-180=144

therefore (b^2-4ac)^1/2=12

so, x=(-18±12)/6

Solve each problem.

4. Geometric problem: A rectangle is 5 times as long as it is wide.  If the area is 125 square feet,         find its perimeter.

èLet us assume that the breath of the rectangle is x

So the length is 5x

Therefore perimeter of the rectangle is 2(x+5x)=12x

By the problem x*5x=125

                        èx*x=25

                       èx=5

So perimeter is 12x=12*5=60 feet

6.  Geometric problem:  The base of a triangle is one-third as long as its height.  If the area of the triangle is 24 square meters, how long is its base?

èLet us assume that base of the triangle is x

So the height of the triangle is 3x

We know that the area of a triangle is ½*base*height

Given that area of the triangle is 24 square meter

So ½*x*3x=24

    è3*x*x=48

    èx*x=16

   èx=4

So base of the triangle is 4 meter

Perform all operations.  Give all answers in a + bi form.

14.  (- 7 + 2i) + (2 - 8i)

   è(-7+2)+(2i-8i)

   è -5-6i

16.   (11 + 2i) - (13 - 5i )

     è(11-13)+(2i+5i)

     è-2+7i

18.  (5 + ) - (23i - 32)

      è(5+8i)-(23i-32)

     è(5+32)+(8i-23i)

     è37-15i

Solve each inequality, graph the solution set and write each answer in interval notation.

14.  -2X + 4 < 6

     è-2x+4-6<6-6  subtract 6 from both sides

     è-2x-2<0         simplify

     è-2x-2+2<0+2 add 2 with both sides

     è-2x<2           simplify

     è-2x/-2<2/-2  divide both sides by -2. reverse the direction of the inequality

     èx>-1

This inequality is true for any value of x that are greater than -1. In interval notation the solution is (-1,∞). The graph of the solution is sketched below.

 

22.  3(x + 2)  ≤ 2(x + 5)

     è3x+6 ≤ 2x+10

     è3x+6-2x ≤ 2x+10-2x subtract 2x from both sides

     è x+6 ≤ 10                    simplify

     èx+6-10 ≤ 10-10         subtract 10 from both sides

     èx-4≤0                        simplify

     èx-4+4≤0+4               add 4 with both sides

     èx≤4

This inequality is true for any value of x that are less than or equal to 4. In interval notation the solution is (-∞,4]. The bracket at 4 indicates that 4 is in the set. The graph of the solution is sketched below.

 

 

54.  x² - 13x + 12 ≤ 0

     èx^2-x-12x+12≤0     factorization

     èx(x-1)-12(x-1)≤0

     è(x-1)(x-12)≤0

     For equal to 0 the solution will be

x-1=0 or x-12=0

x-1=0

or,x-1+1=0+1 add 1 with both sides

or, x=1 simplify

similarly

x=12

and for the negative value

either x-1<0 and x-12>0

or x-1>0 and x-12<0

 

but we can see that if x-1<0 then x-12 can not be greater than 0

so we can not take this inequality

so our inequalities will be x-1>0 and x-12<0

so for x-1>0

 èx-1+1>0+1 add 1 with both sides

 èx>1             simplify

And for

x-12<0

èx-12+12<0+12 add 12 with both sides

èx<12                simplify

Now merging x=1 and x>1 we get x≥1

And merging x=12 and x<12 we get x≤12

So the solution for the x is

1≤x≤12

This inequality is true for any value of x that are less than or equal to 12 and greater than or equal to 1. In interval notation the solution is [1,12]. The bracket at 1 and 12 indicates that 1 and 12 are in the set. The graph of the solution is sketched below