Please teach me how to complete the square.?

Question

Im a C1 A level student, but I just cant get my hang around it!!!

Please do both of these qus: (just copy and paste the ² symbol from here for your answer).

1) 2x² + 3x -5

2) -x² +6x +9

Especially for the first one, please show EVERY single step.

Thank you vvvvv much for your help =D
Early Happy New Year!!!

Answer 1

The general method for 'completing the square' when given a quadratic such as ax^2 + bx + c is:
ax^2 + bx + c = a(x + b/2a)^2 + (c - b^2/4a)

Just memorise the damn thing (or scratch it on your pencil case or whatever). You plug the numbers in and the roots pop out.

Answer 2

1) 2x² + 3x - 5

You need to have x^2 on its own, so divide out by the 2 :
= 2[x^2 + (3/2)x - 5/2]

Now you need to add a constant to x^2 + (3/2)x to make
it a square. This constant is formed by first taking half of
the value of the constant that x is multiplied by.
This constant is 3/2, so half of that is 3/4. The second
step is to then square it, which gives (3/4)^2 (or 9/16).
So now we add (3/4)^2 in the brackets, but note that it
is also necessary to subtract it as well, otherwise the
original expression will be changed from its true value.
So adding and subtracting :

= 2[x^2 + (3/2)x + (3/4)^2 - 5/2 - 9/16]

Note how I left it as (3/4)^2, so we can more easily see
what the eventual square will be, but I subtracted the
same number as 9/16, so that that can more easily be
summed with the -5/2 .
We now have :

= 2[(x + 3/4)^2 - 49/16]

Finally, multiply back by the 2 :

= 2(x + 3/4)^2 - 49/8

That has now completed the square, but it's a good idea
to put it in the standard vertex form, which is : f(x) =
a(x - k)^2 + h. Two things for which this is noteworthy, are :

(i) if 'a' is positive, the graph's concavity faces upwards
while if 'a' is negative, the concavity faces downwards.

(ii) The vertex of the graph is the point (k, h).

To change your expression to this form, just needs a
little sign change to : f(x) = 2[x - (-3/4)]^2 - 49/8.
Now the vertex is seen to be the point (-3/4, -49/8)
and the graph's concavity faces upwards because
'a' (= 2) is positive.
--------------------
2) -x^2 + 6x + 9

Here, we take out the minus sign (which is really -1) :
= -1(x^2 - 6x - 9)

Next, half of 6 is 3, and 3^2 = 9,
so we add and subtract 9 inside the brackets :
= -1(x^2 - 6x + 9 - 9 - 9)

Form the square and sum the constants :
= -1[(x - 3)^2 - 18]

Multiply back by the -1 :
= -1(x - 3)^2 + 18

Now let f(x) = -1(x - 3)^2 + 18.
This is already in standard form, so vertex is (3, 18)
and the graph's concavity faces downwards because
'a' (= -1) is negative.

Answer 3

Find the x-intercepts of y = 4x2 – 2x – 5.
First off, remember that finding the x-intercepts means setting y equal to zero and solving for the x-values, so this question is really asking you to "Solve 4x2 – 2x – 5 = 0".

This is the original problem. 4x2 – 2x – 5 = 0
Move the loose number over to the other side. 4x2 – 2x = 5
Divide through by whatever is multiplied on the squared term.
Take half of the coefficient (don't forget the sign!) of the x-term, and square it. Add this square to both sides of the equation.

Convert the left-hand side to squared form, and simplify the right-hand side. (This is where you use that sign that you kept track of earlier.)

Square-root both sides, remembering the "±" on the right-hand side. Simplify as necessary.
Solve for "x =".
Remember that the "±" means that you have two values for x.

1) 2x^2 + 3x - 5
2x^2 + 3x = 5
Divide all terms by 2
x^2 + 3/2x = 5/2
Get 1/2 of the second term or divide it by two
3/2 divided by 2 = 3/4
Then square it ==> 3/4 ^2 or 3/4 x 3/4 = 9/16
Add this number to the left and right sides
x^2 + 3/2x + 9/16 = 5/2 + 9/16
(x + 3/4)^2 = 49/16
Square both sides
x + 3/4 = +/- 7/4
x = + 7/4 - 3/4 ===> 4/4 or 1
x = - 7/4- 3/4 = -10/4 or -5/2 or - 2 1/2 or - 2.5
=====================
2) - x^2 + 6x + 9
-x^2 + 6x = -9
Multiply both sides by -1
x^2 - 6x = 9
Divide - 6 by 2, square it and add the answer to both sides
-6/2 = - 3 ==== -3^2 = 9
x^2 - 6x + 9 = 9 + 9
(x - 3)^2 = 18
Square both sides
x - 3 = +/- 3 sqrt 2
x = 3 sqrt2 + 3 and
x = - 3 sqrt2 + 3

Answer 4

Hi,


1) 2x² + 3x -5 = 0
In this expression, A = 2, B = 3, and C = -5, To solve by completing the square:

a Move the constant to the other side of the equation.

2x² + 3x = 5

b Divide all terms by the coefficient of A. Fractions are OK.

2x²/2 + 3x/2 = 5/2
x² + (3/2)x = 5/2

c Multiply ½ times the coefficient of x. Square the result and add this number to both sides of the equation.

x² + (3/2)x = 5/2

½ * 3/2 = 3/4 <= now square this
(3/4)² = 9/16 <= add this to both sides of the equation

x² + (3/2)x + 9/16 = 5/2 + 9/16

d Factor the left side into a binomial squared. It will always start with the variable, the sign in front of the x term and the number that was squared up in part c. Combine terms on the right side by getting their common denominator.

x² + (3/2)x + 9/16 = 5/2 + 9/16
(x + 3/4)² = 49/16

e Take the square root of each side of the equation. On the left side, the square on the binomial will disappear. On the right side, put a "±" in front of the radical. Simplify the radical if possible.

(x + 3/4)² = 49/16
.._______......._____
√(x + 3/4)² = √49/16

x + 3/4 = ±7/4

f Now solve for x. When possible combine like terms on the right side.

x + 3/4 = ±7/4
x = - 3/4 ± 7/4
x = - 3/4 + 7/4 or x = - 3/4 - 7/4
x = 1 or -5/2 <=== ANSWERS


2) -x² +6x +9 = 0
In this expression, A = -1, B = 6, and C = 9, To solve by completing the square:

a Move the constant to the other side of the equation.

-x² + 6x = -9

b Divide all terms by the coefficient of A. Fractions are OK.

-x²/-1 + 6x/-1 = -9/-1
x² - 6x = 9

c Multiply ½ times the coefficient of x. Square the result and add this number to both sides of the equation.

x² - 6x = 9
½ * -6 = -3 <= now square this
(-3)² = 9 <= add this to both sides of the equation

x² - 6x + 9= 9 + 9

d Factor the left side into a binomial squared. It will always start with the variable, the sign in front of the x term and the number that was squared up in part c. Combine terms on the right side.

x² - 6x + 9 = 18
(x - 3)² = 18

e Take the square root of each side of the equation. On the left side, the square on the binomial will disappear. On the right side, put a "±" in front of the radical. Simplify the radical if possible.

(x - 3)² = 18
..______........__
√(x - 3)² ...= √18
...............__
x - 3 = ±√18

f Now solve for x. Simplify the radical on the right side.

..............__
x - 3 = ±√18

................_
x - 3 = ±3√2

..............._
x = 3 ± 3√2 <== ANSWER

I hope those help!! :-)

Answer 5

sometimes it is better to look for interger solutions the following way first:

(Mx + n)(Px + q) = MPx^2 + (nP + qM) + nq

you want to solve for the values of M,P, n and q

so A = MP, B = (nP + qM), and C = nq for the equation
A^2 + Bx + C

AC = MPnq = 2*(-5) = -10
B = nP + qM = 6
what two numbers have a product -10 but sum to 6,

there are no intergers that do this, so I will complete the square

x^2 + (3/2)x - 5/2

x^2 + (3/2)x + 9/16 - 9/16 -5/2

(x + 3/2)^2 - 49/16

please check for errors

Answer 6

2x² + 3x -5
= 2(x² + 3x/2 - 5/2)
= 2(x² + 2x(3/4) + (3/4)² - 49/16)
= 2((x+(3/4))² - (7/4)²)
= 2(x+ 3/4 + 7/4)(x + 3/4 - 7/4)
= 2(x + 5/2)(x - 1)
= (2x+5)(x-1)

-x² +6x +9
= -(x² - 6x - 9)
= -(x² - 6x + 9 - 18)
= -((x-3)² - (2√3)²)
= -(x-3+2√3)(x-3-2√3)

Answer 7

Question 1
2 [ x ² + (3/2) x - 5/2 ]
2 [ (x ² + (3/2) x + 9/16) - 9/16 - 5/2 ]
2 [ (x + (3/2) x + 9/16) - 49/16 ]
2 [ (x + (3/4) )² - 49 /16 ]
2 [ (x + (3/4) )² - (7/4)² ]

Question 2
- [ x ² - 6 x - 9 ]
- [ (x ² - 6 x + 9) - 18 ]
- [ (x - 3) ² - (3√2) ² ]

Answer 8

1)2x²+3x -5

2(x²+3/2x + (3/4)² ) -5 -9/8
2(x+3/4)²-49/8


2) -x²+8x+9
16+9 - (x²-8x+4²)
25 - (x-4)²

Easy way :-

You Know that your answer will be in Form of (x+A)²+B
i will do the first one .

(x+A)²+B ≡ 2x²+3x-5
(x²+2Ax+A²)+B ≡ 2x² + 3x - 5
x²+2Ax+A²+B ≡ 2(x²+3/2x - 5/2)
Forget about the two .

2Ax ≡ 3/2x
A=3/4


A²+B ≡ -5/2
2(3/4)²+B=-5/2
B=-49/16

Write the equation :-
2(x+3/4)²-49/16)
2(x+3/4)²-49/8

Answer 9

this is by practice, cant teach you on internet.
2x^2 + 3x -- 5 = 2x^2 + 5x -- 2x -- 5 = x(2x--5) --1(2x--5) = (2x--5)(x--1)
--x^2 + 6x + 9 = (9 + 6x + x^2) -- (x√2)^2 = (3+x)^2--(x√2)^2
= (3 + x + x√2)(3 + x -- x√2)