Completing the square with rational numbers?

Question

Please could somebody walk me through this example as I'm totally confused and talking to the book just doesn't help!?

Question

Use the method of CTS to write 2x^2 - 5x + 11 in the form a(x+b)^2 +c, where a, b and c are rational numbers.

Solution

First take out 2 as a factor.

2x^2 - 5x + 11 = 2[x^2 - 5/2x + 11/2]

Now complete the square on the contents of the square bracket,

= 2[(x - 5/4)^2 - 25/16 + 11/2]

= 2[(x - 5/4)^2 + 63/16]

= 2(x - 5/4)^2 + 63/8.

Any help greatly appreciated! :o)

Answer 1

2x^2-5x+11
2(x^2-5/2x+11/2)
2(x^2-2*x*5/4+25/16-25/16+11)
2(x-5/4)^2+2(88-25)/16
2[(x-5/4)^2-{3i(7)^1/2}^2
2[x-5/4+3i(7)^1/2][x-5/4-3i(7)^1/2]
this involves imaginary nos
i don't know if you are familaiar with those

Answer 2

2x^2 - 5x + 11
(2x^2 - 5x) + 11
2(x^2 - (5/2)x) + 11

find half of (5/2), which is (5/4), then square it, which would be (25/16).

2(x^2 - (5/2)x + (25/16) - (25/16)) + 11
2((x - (5/4))^2 - (25/16)) + 11

now multiply everything inside the () and you get

2(x - (5/4))^2 - (50/16) + 11

reduce the (50/16)

2(x - (5/4))^2 - (25/8) + 11
2(x - (5/4))^2 - (25/8) + (88/8)
2(x - (5/4))^2 + ((-25 + 88)/8)
2(x - (5/4))^2 + (63/8)

a = 2
b = (-5/4)
c = (63/8)

Answer 3

to solve the equation of the second degree in an unknown x
first be sure that it contains no frations, no brackets and it is equal to zero
for example:
if the given equation is: (1/2) x^2 - (1/3)x = 16
then it is better to multiply both sides of the equation by 6, the equation will be:
3x^2 - 2x = 96
then equate the equation to zero, that is it becomes:
3x^2 - 2x - 96 = 0

also if it is given that:
x(x - 3) =1
then expand the brackets and equate to zero, the equation will be in the form:
x^2 - 3x - 1= 0

Solving the quadratic equations can be made in many ways, the following one of these ways:

given that the equation is:
ax^2 + bx + c= 0, then ...
First: transform the constant term c to the right hand side
that is the equation becomes:
ax^2 + bx = -c
second: multiply both sides of the equation by 4*a
that is the equation becomes:
4a^2*x^2 + 4abx = -4ac
Third: add b^2 to the both sides of the equation
that the equation becomes:
4a^2*x^2 + 4abx + b^2= b^2 - 4ac
Fourth: Now the left hand side is a perfest square, so you can write the
equation as follows:
(2ax + b)^2 = b^2 - 4ac
Fifth: Extract the square root of both sides, the equation will be:
(2ax + b) = + or - sqrt(b^2 - 4ac)
sixth: add (-b) to both side then divide by (2a) you get the solution

For example consider the equation:
2x^2 - 5x - 3 = 0
First: transform the costant term 3) to the right hand side, by adding 8 to
both sides: you will get:
2x^2 - 5x = 3

Second: Multiply both sides by (4a = 4*2 = 8) you will get:
16x^2 - 40x = 24

Third: Add (b^2 =(-5)^2 = 25 ) to both sides
the equation will be:
16x^2 - 40x + 25 = 49

Forth: Factorize the left hand side as a perect square:
(4x - 5)^2 = 49

Fifth: Extract the square root of both sides, you will get either
4x - 5 = 7 or 4x - 5 = -7

sixth: Add 5 to both sides then divide by 4, you will get:
x = 3 or x = -2/3
Best wishes

Answer 4

to end the sq. you're taking 0.5 the x co-effectual and the sq. it. subsequent you upload this and to compensate you need to additionally subtract it. right here the co-effectual of x = -13/3 Halve it = -13/6 sq. it (-13/6)^2 =169/36 upload this and subtract it The addition would be secure in the sq.. this is amazingly complicated with fractions - right that's an occasion devoid of fractions x^2 + 6x + a million = 0 x^2+6x + 3^2 -3^2 +a million =0 (x+3)^2 - 9+a million =0 (x+3)^2 - 8 = 0

Answer 5

Ask your teacher to walk you through it