JK is equal to (x squared - 4x). KL is equal to (3x-2). JL=28. How do you find out what x equals? 

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caledon | High School Teacher | (Level 3) Senior Educator

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JK = `x^(2) - 4x` ` `

KL = `3x -2`

JK and KL are colinear segments of JL, which equals 28

JL = `x^2 -4x + 3x - 2`

which simplifies to

`x^2 -x -2`  = 28

which simplifies to

`x^2 -x -30 `  = 0

Then use the quadratic equation to solve;

-b `+-` `sqrt(B^2 -4ac)` / 2a

1`+-` 11 / 2

12/2 and -10/2

X = -5 and 6

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sid-sarfraz | Student, Graduate | (Level 2) Salutatorian

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JK is equal to (x squared - 4x). KL is equal to (3x-2). JL=28. How do you find out what x equals? 

QUESTION:-

JK = x^2 -4x

KL = 3x - 2

JL = 28

x = ?

SOLUTION:-

Now suppose J, K and L are three points forming a straight line. Where K is the mid point. Now according to this we can say that JL is equal to JK plus KL. 

JL = JK + KL

Insert values;

28 = (x^2 -4x) + (3x-2)

28 = x^2 - 4x + 3x - 2

x^2 - 1x - 2 - 28 = 0

x^2 - x - 30 = 0

Now solve through using quadratic equation;

where,

a = 1

b = -1

c = -30

Formula is;

x = {-b +-sqrt (b^2 - 4ac)} / 2a

x = {-(-1) +- sqrt (-1)^2 - 4 (1)(-30)}/2(1)

x = {1+-sqrt (1+120)}/2

x = (1 +- sqrt 121) /2

x = (1 +-11)/ 2

x = (1 + 11)/2       ,          x = (1-11)/2

x = 12/2               ,           x = -10/2

x = 6                    ,           x = -5

Hence Solved!

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glendamaem | Student | (Level 1) Honors

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JK= x^2 - 4x

KL= 3x - 2

JL = 28

28= x^2 -4x + 3x - 2

28= x^2 -x -2

0= x^2 -x -2 -28

0= x^2 -x -30

Substitute values in quadratic equation. (a=1, b=-1, c=-30)

After doing so, the values of x are -5 and 6.

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tonys538 | Student, Undergraduate | (Level 1) Valedictorian

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J, K and L are points on a straight line. The length of JK is equal to x^2 - 4x, KL is equal to (3x-2) and JL=28. Now, as K lies between J and L, JK + KL = JL This gives: x^2 - 4x + 3x - 2 = 28 x^2 - x - 2 = 28 x^2 - x - 30 = 0 x^2 - 6x + 5x - 30 = 0 x(x - 6) + 5(x - 6) = 0 (x - 6)(x + 5) = 0 This gives x = 6 or x = -5 The values of x are -5 and 6
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rachellopez | Student, Grade 12 | (Level 1) Valedictorian

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JK = x^2 - 4x

KL = 3x - 2

JL = 28

With the information given, you can conclude that JK and KL are both collinear segments, meaning they pass through the same straight line (JL). Therefore, you can conclude that JL is equal to the combined equations of JK and KL.

JL = x^2 - 4x + 3x - 2

JL = x^2 - x - 2 = 28

You can make this equation be a quadratic by subtracting 28 from both sides.

x^2 - x - 30 = 0

Now you are able to solve for x using the quadratic equation `(-b+-sqrt(b^(2)(4ac)))/(2a) `.

x = -5 and 6

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