Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

4*(x-5)^2-(36)=0

Step by step solution :

Step 1 :

Equation at the end of step 1 :

4 • (x – 5)2 – 36 = 0

Step 2 :

2.1 Evaluate : (x-5)2 = x2-10x+25

Step 3 :

Pulling out like terms :

3.1 Pull out like factors :

4×2 – 40x + 64 = 4 • (x2 – 10x + 16)

Trying to factor by splitting the middle term

3.2 Factoring x2 – 10x + 16

The first term is, x2 its coefficient is 1 .

The middle term is, -10x its coefficient is -10 .

The last term, “the constant”, is +16

Step-1 : Multiply the coefficient of the first term by the constant 1 • 16 = 16

Step-2 : Find two factors of 16 whose sum equals the coefficient of the middle term, which is -10 .

-16 + -1 = -17

-8 + -2 = -10 That’s it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -8 and -2

x2 – 8x – 2x – 16

Step-4 : Add up the first 2 terms, pulling out like factors :

x • (x-8)

Add up the last 2 terms, pulling out common factors :

2 • (x-8)

Step-5 : Add up the four terms of step 4 :

(x-2) • (x-8)

Which is the desired factorization

Equation at the end of step 3 :

4 • (x – 2) • (x – 8) = 0

Step 4 :

Theory – Roots of a product :

4.1 A product of several terms equals zero.

When a product of two or more terms equals zero, then at least one of the terms must be zero.

We shall now solve each term = 0 separately

In other words, we are going to solve as many equations as there are terms in the product

Any solution of term = 0 solves product = 0 as well.

Equations which are never true :

4.2 Solve : 4 = 0

This equation has no solution.

A a non-zero constant never equals zero.

Solving a Single Variable Equation :

4.3 Solve : x-2 = 0

Add 2 to both sides of the equation :

x = 2

Solving a Single Variable Equation :

4.4 Solve : x-8 = 0

Add 8 to both sides of the equation :

x = 8

Supplement : Solving Quadratic Equation Directly

Solving x2-10x+16 = 0 directly

Earlier we factored this polynomial by splitting the middle term. let us now solve the equation by Completing The Square and by using the Quadratic Formula

Parabola, Finding the Vertex :

5.1 Find the Vertex of y = x2-10x+16

For any parabola,Ax2+Bx+C,the x -coordinate of the vertex is given by -B/(2A) . In our case the x coordinate is 5.0000

Plugging into the parabola formula 5.0000 for x we can calculate the y -coordinate :

y = 1.0 * 5.00 * 5.00 – 10.0 * 5.00 + 16.0

or y = -9.000

Parabola, Graphing Vertex and X-Intercepts :

Root plot for : y = x2-10x+16

Axis of Symmetry (dashed) {x}={ 5.00}

Vertex at {x,y} = { 5.00,-9.00}

x -Intercepts (Roots) :

Root 1 at {x,y} = { 2.00, 0.00}

Root 2 at {x,y} = { 8.00, 0.00}

Solve Quadratic Equation by Completing The Square

5.2 Solving x2-10x+16 = 0 by Completing The Square .

Subtract 16 from both side of the equation :

x2-10x = -16

Now the clever bit: Take the coefficient of x , which is 10 , divide by two, giving 5 , and finally square it giving 25

Add 25 to both sides of the equation :

On the right hand side we have :

-16 + 25 or, (-16/1)+(25/1)

The common denominator of the two fractions is 1 Adding (-16/1)+(25/1) gives 9/1

So adding to both sides we finally get :

x2-10x+25 = 9

Adding 25 has completed the left hand side into a perfect square :

x2-10x+25 =

(x-5) • (x-5) =

(x-5)2

Things which are equal to the same thing are also equal to one another. Since

x2-10x+25 = 9 and

x2-10x+25 = (x-5)2

then, according to the law of transitivity,

(x-5)2 = 9

We’ll refer to this Equation as Eq. #5.2.1

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of

(x-5)2 is

(x-5)2/2 =

(x-5)1 =

x-5

Now, applying the Square Root Principle to Eq. #5.2.1 we get:

x-5 = √ 9

Add 5 to both sides to obtain:

x = 5 + √ 9

Since a square root has two values, one positive and the other negative

x2 – 10x + 16 = 0

has two solutions:

x = 5 + √ 9

or

x = 5 – √ 9

Solve Quadratic Equation using the Quadratic Formula

5.3 Solving x2-10x+16 = 0 by the Quadratic Formula .

According to the Quadratic Formula, x , the solution for Ax2+Bx+C = 0 , where A, B and C are numbers, often called coefficients, is given by :

– B ± √ B2-4AC

x = ————————

2A

In our case, A = 1

B = -10

C = 16

Accordingly, B2 – 4AC =

100 – 64 =

36

Applying the quadratic formula :

10 ± √ 36

x = —————

2

Can √ 36 be simplified ?

Yes! The prime factorization of 36 is

2•2•3•3

To be able to remove something from under the radical, there have to be 2 instances of it (because we are taking a square i.e. second root).

## Answers ( )

Answer:19/2 or 9.5

Step-by-step explanation:Step 1: Simplify both sides of the equation.

4(x−5)(2)=36

8x+−40=36(Distribute)

8x−40=36

Step 2: Add 40 to both sides.

8x−40+40=36+40

8x=76

Step 3: Divide both sides by 8.

8x/8=76/8

x=19/2

Two solutions were found :

x = 8

x = 2

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

4*(x-5)^2-(36)=0

Step by step solution :

Step 1 :

Equation at the end of step 1 :

4 • (x – 5)2 – 36 = 0

Step 2 :

2.1 Evaluate : (x-5)2 = x2-10x+25

Step 3 :

Pulling out like terms :

3.1 Pull out like factors :

4×2 – 40x + 64 = 4 • (x2 – 10x + 16)

Trying to factor by splitting the middle term

3.2 Factoring x2 – 10x + 16

The first term is, x2 its coefficient is 1 .

The middle term is, -10x its coefficient is -10 .

The last term, “the constant”, is +16

Step-1 : Multiply the coefficient of the first term by the constant 1 • 16 = 16

Step-2 : Find two factors of 16 whose sum equals the coefficient of the middle term, which is -10 .

-16 + -1 = -17

-8 + -2 = -10 That’s it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -8 and -2

x2 – 8x – 2x – 16

Step-4 : Add up the first 2 terms, pulling out like factors :

x • (x-8)

Add up the last 2 terms, pulling out common factors :

2 • (x-8)

Step-5 : Add up the four terms of step 4 :

(x-2) • (x-8)

Which is the desired factorization

Equation at the end of step 3 :

4 • (x – 2) • (x – 8) = 0

Step 4 :

Theory – Roots of a product :

4.1 A product of several terms equals zero.

When a product of two or more terms equals zero, then at least one of the terms must be zero.

We shall now solve each term = 0 separately

In other words, we are going to solve as many equations as there are terms in the product

Any solution of term = 0 solves product = 0 as well.

Equations which are never true :

4.2 Solve : 4 = 0

This equation has no solution.

A a non-zero constant never equals zero.

Solving a Single Variable Equation :

4.3 Solve : x-2 = 0

Add 2 to both sides of the equation :

x = 2

Solving a Single Variable Equation :

4.4 Solve : x-8 = 0

Add 8 to both sides of the equation :

x = 8

Supplement : Solving Quadratic Equation Directly

Solving x2-10x+16 = 0 directly

Earlier we factored this polynomial by splitting the middle term. let us now solve the equation by Completing The Square and by using the Quadratic Formula

Parabola, Finding the Vertex :

5.1 Find the Vertex of y = x2-10x+16

For any parabola,Ax2+Bx+C,the x -coordinate of the vertex is given by -B/(2A) . In our case the x coordinate is 5.0000

Plugging into the parabola formula 5.0000 for x we can calculate the y -coordinate :

y = 1.0 * 5.00 * 5.00 – 10.0 * 5.00 + 16.0

or y = -9.000

Parabola, Graphing Vertex and X-Intercepts :

Root plot for : y = x2-10x+16

Axis of Symmetry (dashed) {x}={ 5.00}

Vertex at {x,y} = { 5.00,-9.00}

x -Intercepts (Roots) :

Root 1 at {x,y} = { 2.00, 0.00}

Root 2 at {x,y} = { 8.00, 0.00}

Solve Quadratic Equation by Completing The Square

5.2 Solving x2-10x+16 = 0 by Completing The Square .

Subtract 16 from both side of the equation :

x2-10x = -16

Now the clever bit: Take the coefficient of x , which is 10 , divide by two, giving 5 , and finally square it giving 25

Add 25 to both sides of the equation :

On the right hand side we have :

-16 + 25 or, (-16/1)+(25/1)

The common denominator of the two fractions is 1 Adding (-16/1)+(25/1) gives 9/1

So adding to both sides we finally get :

x2-10x+25 = 9

Adding 25 has completed the left hand side into a perfect square :

x2-10x+25 =

(x-5) • (x-5) =

(x-5)2

Things which are equal to the same thing are also equal to one another. Since

x2-10x+25 = 9 and

x2-10x+25 = (x-5)2

then, according to the law of transitivity,

(x-5)2 = 9

We’ll refer to this Equation as Eq. #5.2.1

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of

(x-5)2 is

(x-5)2/2 =

(x-5)1 =

x-5

Now, applying the Square Root Principle to Eq. #5.2.1 we get:

x-5 = √ 9

Add 5 to both sides to obtain:

x = 5 + √ 9

Since a square root has two values, one positive and the other negative

x2 – 10x + 16 = 0

has two solutions:

x = 5 + √ 9

or

x = 5 – √ 9

Solve Quadratic Equation using the Quadratic Formula

5.3 Solving x2-10x+16 = 0 by the Quadratic Formula .

According to the Quadratic Formula, x , the solution for Ax2+Bx+C = 0 , where A, B and C are numbers, often called coefficients, is given by :

– B ± √ B2-4AC

x = ————————

2A

In our case, A = 1

B = -10

C = 16

Accordingly, B2 – 4AC =

100 – 64 =

36

Applying the quadratic formula :

10 ± √ 36

x = —————

2

Can √ 36 be simplified ?

Yes! The prime factorization of 36 is

2•2•3•3

To be able to remove something from under the radical, there have to be 2 instances of it (because we are taking a square i.e. second root).

√ 36 = √ 2•2•3•3 =2•3•√ 1 =

± 6 • √ 1 =

± 6

So now we are looking at:

x = ( 10 ± 6) / 2

Two real solutions:

x =(10+√36)/2=5+3= 8.000

or:

x =(10-√36)/2=5-3= 2.000

Two solutions were found :

x = 8

x = 2