Chapter 1.2, Problem 39HP

To determine

To write an solve an equation

Answer to Problem 39HP

  $x=6$ is the solution.

Explanation of Solution

Given information:

  $JK=x^2-4x$

  $KL=3x-2$

  $JL=28$

Calculation:

According to question K is the point between J and L .

So,

  $JK+KL=JL$

Substituting values,

  $x^2-4x+3x-2=28$

Solving,

  $x^2-x-2=28$

Now subtracting 28 from both sides,

  $x^2-x-30=0$

Factorising,

  $x^2-6x+5x-30=0$

Now taking out common,

  $x(x-6)+5(x-6)=0$

Taking common again,

  $(x+5)(x-6)=0$

For equation to be o,

  $(x+5)=0$ OR $(x-6)=0$

Solving,

  $x=-5$ and $x=6$

  $x=-5$ is not used as lines cannot be in negative.

Hence, $x=6$ is the solution.