Math, asked by satyamkumar433pdn3xf, 6 months ago

t²-32t - 2336=0
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Answers

Answered by unmana53
1

Answer:

The first term is, t2 its coefficient is 1 .

The middle term is, -32t its coefficient is -32 .

The last term, "the constant", is -160

Step-1 : Multiply the coefficient of the first term by the constant 1 • -160 = -160

Step-2 : Find two factors of -160 whose sum equals the coefficient of the middle term, which is -32 .

-160 + 1 = -159

-80 + 2 = -78

-40 + 4 = -36

-32 + 5 = -27

-20 + 8 = -12

-16 + 10 = -6

-10 + 16 = 6

-8 + 20 = 12

-5 + 32 = 27

-4 + 40 = 36

-2 + 80 = 78

-1 + 160 = 159

Observation : No two such factors can be found !!

Conclusion : Trinomial can not be factored

Equation at the end of step

1

:

t2 - 32t - 160 = 0

STEP

2

:

Parabola, Finding the Vertex

2.1 Find the Vertex of y = t2-32t-160

Parabolas have a highest or a lowest point called the Vertex . Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) . We know this even before plotting "y" because the coefficient of the first term, 1 , is positive (greater than zero).

Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.

Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.

For any parabola,At2+Bt+C,the t -coordinate of the vertex is given by -B/(2A) . In our case the t coordinate is 16.0000

Plugging into the parabola formula 16.0000 for t we can calculate the y -coordinate :

y = 1.0 * 16.00 * 16.00 - 32.0 * 16.00 - 160.0

or y = -416.000

Parabola, Graphing Vertex and X-Intercepts :

Root plot for : y = t2-32t-160

Axis of Symmetry (dashed) {t}={16.00}

Vertex at {t,y} = {16.00,-416.00}

t -Intercepts (Roots) :

Root 1 at {t,y} = {-4.40, 0.00}

Root 2 at {t,y} = {36.40, 0.00}

Solve Quadratic Equation by Completing The Square

2.2 Solving t2-32t-160 = 0 by Completing The Square .

Add 160 to both side of the equation :

t2-32t = 160

Now the clever bit: Take the coefficient of t , which is 32 , divide by two, giving 16 , and finally square it giving 256

Add 256 to both sides of the equation :

On the right hand side we have :

160 + 256 or, (160/1)+(256/1)

The common denominator of the two fractions is 1 Adding (160/1)+(256/1) gives 416/1

So adding to both sides we finally get :

t2-32t+256 = 416

Adding 256 has completed the left hand side into a perfect square :

t2-32t+256 =

(t-16) • (t-16) =

(t-16)2

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

t2-32t+256 = 416 and

t2-32t+256 = (t-16)2

then, according to the law of transitivity,

(t-16)2 = 416

We'll refer to this Equation as Eq. #2.2.1

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

Note that the square root of

(t-16)2 is

(t-16)2/2 =

(t-16)1 =

t-16

Now, applying the Square Root Principle to Eq. #2.2.1 we get:

t-16 = √ 416

Add 16 to both sides to obtain:

t = 16 + √ 416

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

t2 - 32t - 160 = 0

has two solutions:

t = 16 + √ 416

or

t = 16 - √ 416

Solve Quadratic Equation using the Quadratic Formula

2.3 Solving t2-32t-160 = 0 by the Quadratic Formula .

According to the Quadratic Formula, t , the solution for At2+Bt+C = 0 , where A, B and C are numbers, often called coefficients, is given by :

- B ± √ B2-4AC

t = ————————

2A

In our case, A = 1

B = -32

C = -160

Accordingly, B2 - 4AC =

1024 - (-640) =

1664

Applying the quadratic formula :

32 ± √ 1664

t = ——————

2

Can √ 1664 be simplified ?

Yes! The prime factorization of 1664 is

2•2•2•2•2•2•2•13

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).

√ 1664 = √ 2•2•2•2•2•2•2•13 =2•2•2•√ 26 =

± 8 • √ 26

√ 26 , rounded to 4 decimal digits, is 5.0990

So now we are looking at:

t = ( 32 ± 8 • 5.099 ) / 2

Two real solutions:

t =(32+√1664)/2=16+4√ 26 = 36.396

or:

t =(32-√1664)/2=16-4√ 26 = -4.396

Two solutions were found :

t =(32-√1664)/2=16-4√ 26 = -4.396

t =(32+√1664)/2=16+4√ 26 = 36.396

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