1. Write the equation in a way that the constants are in the right side while the terms with x are on the left.
x^2 – 6x = 5
2. Make sure that the coefficient of the x^2 term is 1.
x^2 – 6x = 5
3. Adding a term to both sides that will complete the square in the left side. This is done by dividing the coefficient of the x term by 2 and squaring it. Note: The same amount should be added to the right side to balance the equation.
x^2 – 6x + 9 = 5 + 9
(x-3)^2 = 14
Therefore, 9 should be added to both sides to complete the square.
First of all, it is better to note that x2 is not equal to x² .In mathematics x2 is equal to 2nd x or 2x i.e. 2 times x=2*x=x*2.
I don't know what this question means ! But if it means to just confine x2 and -6x in a square or to make them part of a square then it could be done as follows.
Assume x2 as x² as per your needs as case may be.
Now use the identity
(a² - 2a×b + b²) =(a - b )²
Here , in LHS
a appears twice as in a² and in 2a×b.
So for given expression (x2 – 6x = 5) or (x² – 6x = 5) or (x² - 6x -5 =0)
Assume x=a
Then,
6x =2*x*3 =2*a*b
=> 3 = b
Thus
(x² - 6x -5 =0)
=> x² - 2*x*3 + 3² -3² -5 =0
=> ( x² - 2*x*3 + 3² ) - 3² -5 =0
=> (x - 3)² - 3² = 5
=> (x - 3)² = 5 + 3²
=> (x - 3)² = 5 + 9 = 14 = ( ±√(14) )²
Now comparing
(x² – 6x = 5) and (x - 3)² = 5 + 9,
it can be said that number 9 should be added to both sides of (x² – 6x = 5) or (x² – 6x) = ( 5 ) to make ( x² ) and (-6x ) as the constituting parts of the square (x-3)² .
9
Step-by-step explanation:
Given
x² - 6x = 5
To complete the square
add ( half the coefficient of the x- term )² to both sides
x² + 2(- 3)x + 9 = 5 + 9, that is
(x - 3)² = 14
9 should be added to complete the square
Step-by-step explanation:
1. Write the equation in a way that the constants are in the right side while the terms with x are on the left.
x^2 – 6x = 5
2. Make sure that the coefficient of the x^2 term is 1.
x^2 – 6x = 5
3. Adding a term to both sides that will complete the square in the left side. This is done by dividing the coefficient of the x term by 2 and squaring it. Note: The same amount should be added to the right side to balance the equation.
x^2 – 6x + 9 = 5 + 9
(x-3)^2 = 14
Therefore, 9 should be added to both sides to complete the square.
9
Step-by-step explanation:
I think 9 should be added to both sides.
cofficient of x = 6
half of it = 6/2 = 3
square the 3
to give 3 squared = 3*3 9
9
Step-by-step explanation:
To make it a Perfect Square Trinomial, you can square root 9 and multiply it to get 6, therefore 9 is correct
Your answer is 0.625hope this : b
First of all, it is better to note that x2 is not equal to x² .In mathematics x2 is equal to 2nd x or 2x i.e. 2 times x=2*x=x*2.
I don't know what this question means ! But if it means to just confine x2 and -6x in a square or to make them part of a square then it could be done as follows.
Assume x2 as x² as per your needs as case may be.
Now use the identity
(a² - 2a×b + b²) =(a - b )²
Here , in LHS
a appears twice as in a² and in 2a×b.
So for given expression (x2 – 6x = 5) or (x² – 6x = 5) or (x² - 6x -5 =0)
Assume x=a
Then,
6x =2*x*3 =2*a*b
=> 3 = b
Thus
(x² - 6x -5 =0)
=> x² - 2*x*3 + 3² -3² -5 =0
=> ( x² - 2*x*3 + 3² ) - 3² -5 =0
=> (x - 3)² - 3² = 5
=> (x - 3)² = 5 + 3²
=> (x - 3)² = 5 + 9 = 14 = ( ±√(14) )²
Now comparing
(x² – 6x = 5) and (x - 3)² = 5 + 9,
it can be said that number 9 should be added to both sides of (x² – 6x = 5) or (x² – 6x) = ( 5 ) to make ( x² ) and (-6x ) as the constituting parts of the square (x-3)² .
I hope it helps.
Thank you for patiently reading it.
9
Step-by-step explanation:
x^2 - 6x = 5
Take the coefficient of x
-6
Divide it by 2
-6/2 = -3
Square it
(-3)^2 =9
Add 9 to each side