Maths Doubt As!!

[ZaqZainab]

I have got trouble with completing the square i have just started my As and i don't understand when my teacher explains Here let me do a question i want some one to tell me where i am wrong and how should i continue
For example- p^2+14p-38=0
= p^2+14p+(7)^2-(7)^2-38=0
= (p+7)^2-87=0
what next? is it =(p+7)^2=87
=p+7=root of 87
=p=root of 87-7
is it all is this the answer?
Another example- 8x^2+16x=42
8x^2+16x-42=0
8x^2+16x+(8)^2-(8)^2-42=0
(x+8)^2-22=0
x+8=root of 22
x=root of 22-8
Is that all please help me out its just the starting of AS and it is already so confusing

 

[oohunteroo]

the second question:
8x^2?
are you sure the coefficent of x^2 is 8?

 

[Saad Mughal]

I have got trouble with completing the square i have just started my As and i don't understand when my teacher explains Here let me do a question i want some one to tell me where i am wrong and how should i continue
For example- p^2+14p-38=0
= p^2+14p+(7)^2-(7)^2-38=0
= (p+7)^2-87=0
what next? is it =(p+7)^2=87
=p+7=root of 87
=p=root of 87-7
is it all is this the answer?
Another example- 8x^2+16x=42
8x^2+16x-42=0
8x^2+16x+(8)^2-(8)^2-42=0
(x+8)^2-22=0
x+8=root of 22
x=root of 22-8
Is that all please help me out its just the starting of AS and it is already so confusing

There is another, more simpler method, I'll guide you through the examples step by step:-

Example 1:
(p^2 + 14p) - 38 = 0 (Take out the common value in the first two terms, in this case 1 is the common value).
(p^2 + 14p + 49 - 49) - 38 = 0 (Last term = (Middle term/2)^2)
(p^2 + 14p + 49) - 87 = 0 (Take the negative squared value outside the bracket)
(p+7)^2 - 87 = 0
(p+7)^2 = 87
(p+7) = ±√87
p = ±√87 - 7
p = √87 - 7 or p= -√87 - 7

Example 2:
8x^2+16x = 42
8x^2+16x-42 = 0
8(x^2+2x) - 42 = 0
8(x^2+2x+1-1) -42 = 0
8(x+1)^2 - 8 - 42 = 0
8(x+1)^2 - 50 = 0
8(x+1)^2 = 50
(x+1)^2 = 50/8
(x+1)^2 = 25/4
(x+1) = ±√(25/4)
(x+1) = ±(5/2)
x = ±(5/2) - 1
x = 5/2 - 1 or x = -5/2 - 1
x = 3/2 , x = -7/2

This is much like your method but the brackets and taking out the common value help in making the method less prone to errors. Also, always remember to make sure that whenever you take the square root of a number, you add the ± sign before it, this ensures that you get two answers rather than one (all quadratic equations have two roots).

 

[ZaqZainab]

the second question:
8x^2?
are you sure the coefficent of x^2 is 8?

yeah

 

[ZaqZainab]

There is another, more simpler method, I'll guide you through the examples step by step:-

Example 1:
(p^2 + 14p) - 38 = 0 (Take out the common value in the first two terms, in this case 1 is the common value).
(p^2 + 14p + 49 - 49) - 38 = 0 (Last term = (Middle term/2)^2)
(p^2 + 14p + 49) - 87 = 0 (Take the negative squared value outside the bracket)
(p+7)^2 - 87 = 0
(p+7)^2 = 87
(p+7) = ±√87
p = ±√87 - 7
p = √87 - 7 or p= -√87 - 7

Example 2:
8x^2+16x = 42
8x^2+16x-42 = 0
8(x^2+2x) - 42 = 0
8(x^2+2x+1-1) -42 = 0
8(x+1)^2 - 8 - 42 = 0
8(x+1)^2 - 50 = 0
8(x+1)^2 = 50
(x+1)^2 = 50/8
(x+1)^2 = 25/4
(x+1) = ±√(25/4)
(x+1) = ±(5/2)
x = ±(5/2) - 1
x = 5/2 - 1 or x = -5/2 - 1
x = 3/2 , x = -7/2

This is much like your method but the brackets and taking out the common value help in making the method less prone to errors. Also, always remember to make sure that whenever you take the square root of a number, you add the ± sign before it, this ensures that you get two answers rather than one (all quadratic equations have two roots).

thank you so much