Question

solve the following

1. solve the following inequality for X
2x^2 +5x <3

2. without solving the equation 2x^2 +5x =3, how would I know that it has two rational solutions.

3. 3^2(x+1) - 8.3^x+1 = 1

4. log (x+1) + log x =1

4.1. show that the above equation will have a irrational root if the base is changed to 2, and a rational root if the base is changed to 4

5. the sum of squares of two consecutive even intigers is 1 252. find the intigers

6. suppose x, 2x+1 and 3x+2 are consecutive terms in a geometric sequence. calculate x and hence find the other two terms. how can you be sure the answer is correct?

7. The sizes of angles of a triangle is in proportion 1:2:3. find the size of each angle.

Answer 1

1) 2x²+5x < 3

i.e., 2x²+5x-3 < 3-3

i.e., 2x²+6x-x-3 < 0

i.e., (2x-1)(x+3) < 0

i.e., (x+3)(x-1/2) < 0

Computation of signs :

	x<-3	x=-3	-3<x<1/2	x=1/2	x>1/2
(x-1/2)	-	-	-	0	+
(x+3)	-	0	+	+	+
(x+3)(x-1/2)	+	0	-	0	+

From the table we get that the range that satisfy the required condition is : -3<x<1/2.

2) From the above table we get, the equation 2x²+5x = 3 has solutions at x = -3 and x = 1/2.

Therefore, the equation has two rational solutions x = -3 and x = 1/2.

3) Given, 3^2(x+1)-8*3^x+1 = 1

Let y = 3^x+1. Then the equation becomes, y²-8y = 1

i.e., y²-8y-1 = 0

i.e., y =

i.e., y =

i.e., 3^x+1 =

i.e., x =

i.e., x =    [Since is undefined]

i.e., x =

4) Case I : log₂(x+1)+log₂(x) = 1

i.e., log₂[x(x+1)] = 1

i.e., x(x+1) = 2¹

i.e., x²+x = 2

i.e., x²+x-2 = 0

i.e., x²+2x-x-2 = 0

i.e., (x+2)(x-1) = 0

i.e., x = -2, 1

Therefore, the equation has rational roots when the base is 2.

Case II : log₄(x+1)+log₄(x) = 1

i.e., log₄[x(x+1)] = 1

i.e., x(x+1) = 4¹

i.e., x²+x = 4

i.e., x²+x-4 = 0

i.e., x =

Therefore, the equation has irrational roots when the base is 4.

[I think in the question, there is some mistake. It should be "show that the above equation will have a irrational root if the base is changed to 4, and a rational root if the base is changed to 2"]

5) Let two consecutive integers be 2x and 2x+2.

Then by given conditions, we have, (2x)²+(2x+2)² = 1252

i.e., 4x²+4(x+1)² = 1252

i.e., x²+(x+1)² = 313

i.e., 2x²+2x+1 = 313

i.e., 2x²+2x-312 = 0

i.e., x²+x-156 = 0

i.e., x²+13x-12x-156 = 0

i.e., (x+13)(x-12) = 0

i.e., x = -13, 12

Therefore, the positive integers are 24 and 26 and negative integers are -26 and -24.

6) By the given conditions, x/(2x+1) = (2x+1)/(3x+2)

i.e., x(3x+2) = (2x+1)²

i.e., 3x²+2x = 4x²+4x+1

i.e., x²+2x+1 = 0

i.e., (x+1)² = 0

i.e., x+1 = 0

i.e., x = -1

Therefore, first term is -1.

And, other terms are : 2x+1 = 2(-1)+1 = -1 and 3x+2 = 3(-1)+2 = -1

Hence, the three consecutive terms are -1, -1 and -1, where the common ratio is 1.

7) Let x, 2x, 3x be the sizes of the angles of the given triangle.

Then, by the given conditions, we have, x+2x+3x = 180

i.e., 6x = 180

i.e., x = 180/6

i.e., x = 30

Therefore, the sizes of the angles are ,i.e., .