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2 years ago

Solve x squared - 12x + 36 = 0 using the Quadratic formula.

Mathematics

2 answers:

dezoksy [38]2 years ago

5 0

Answer:

The value of x = 6

Step-by-step explanation:

Points to remember

Solution of a quadratic equation ax² + bx + c = 0

x = [-b ± √(b² - 4ac)]/2a

It is given that,

x² -12x + 36 = 0

To find the solution of given equation

Here a = 1, b = -12 and c = 36

x = [-b ± √(b² - 4ac)]/2a

= [--12 ± √((-12)² - 4*1*36)]/2*1

= [12 ± √(144 - 144)]/2

= 12/2 = 6

The value of x = 6

KatRina [158]2 years ago

3 0

Answer:

x = 6

Step-by-step explanation:

Equation: x² - 12x + 36 = 0

Quadratic formula: ax² + bx + c = 0 --> x = (-b ± √(b² - 4ac))/2a

Substitute: x = (12 ± √(12² - 4 * 36))/2

Multiply: x = (12 ± √(144 - 144))/2

Subtract: x = (12 ± 0)/2

Solve: x = 6

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Don works in home construction. He determined that the function t(A)=144A/9.252 gives the total number of 9 in. tiles needed to

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Answer:

6,226 tiles of 9 in are needed to cover an area of 400 square feet

Step-by-step explanation:

Let

A -----> an area in square feet

t(A) ---> the total number of 9 in. tiles needed

we have

$t(A)=\frac{144A}{9.252}$

For A=400 ft^2

substitute in the equation and solve for t

$t(400)=\frac{144(400)}{9.252}$

$t(400)=6,226\ tiles$

That means -----> 6,226 tiles of 9 in are needed to cover an area of 400 square feet

6 0

3 years ago

Patrick has just finished building a pen for his new dog. The pen is 3 feet wider than it is long. He also built a doghouse to p

zalisa [80]

General Idea:

(i) Assign variable for the unknown that we need to find

(ii) Sketch a diagram to help us visualize the problem

(iii) Write the mathematical equation representing the description given.

(iv) Solve the equation by substitution method. Solving means finding the values of the variables which will make both the equation TRUE

Applying the concept:

Given: x represents the length of the pen and y represents the area of the doghouse

Statement 1: "The pen is 3 feet wider than it is long"

$Length \; of\; the \; pen = x\\ Width \; of\; the\; pen=x+3$

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Statement 2: "He also built a doghouse to put in the pen which has a perimeter that is equal to the area of its base"

$Area \; of\; the\; Dog \; house=y\\ Perimeter \; of\; Dog\; house=y$

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Statement 3: "After putting the doghouse in the pen, he calculates that the dog will have 178 square feet of space to run around inside the pen."

$Area \; of \; the\; Pen - Area \;of \;the\;Dog \;House=\;Space\;inside\;Pen\\ \\ x \cdot (x+3)-y=178\\ Distributing \;x\;in\;the\;left\;side\;of\;the\;equation\\ \\ x^2+3x-y=178\Rightarrow\; 1^{st}\; Equation\\$

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Statement 4: "The perimeter of the pen is 3 times greater than the perimeter of the doghouse."

$Perimeter\; of\; the\; Pen=3\; \cdot \; Perimeter\; of\; the\; Dog\; House\\ \\ 2(x \; + \; x+3)=3 \cdot y\\ Combine\; like\; terms\; inside\; the\; parenthesis\\ \\ 2(2x+3)=3y\\ Distribute\; 2\; in\; the\; left\; side\; of\; the\; equation\\ \\ 4x+6=3y\\ Subtract \; 6\; and \; 3y\; on\; both\; sides\; of\; the\; equation\\ \\ 4x+6-3y-6=3y-3y-6\\ Combine\; like\; terms\\ \\ 4x-3y=-6 \Rightarrow \; \; 2^{nd}\; Equation\\$