A kosarad üres
Solving equations with factorisation
In our algebraic studies we often experience that when ordering different expressions it can be useful to factorise them. This method can also be used efficiently when solving equations.
Example 1
Let us solve the following equation on the set of rational numbers:
\latex{(x+2)(2x+2)(3x-6)(4x-3)=0}.
Solution
There is a product on the left side of the equation. Its value is 0 if and only if one of the factors is 0. It means the need for examining four cases:
if \latex{x+2=0}, then \latex{x=-2};
if \latex{x+2=0}, then \latex{x=-2};
if \latex{2x+2=0}, then \latex{x=-1};
if \latex{3x-6=0}, then \latex{x=2};
if \latex{4x-3=0}, then \latex{x=\frac{3}{4}}.
So four distinct rational numbers will be the solutions of the equation. We can distinguish them as follows:
\latex{x_1=-2, \; x_2=-1, \; x_3=2, \;x_4=\frac{3}{4}}.
\latex{x_1=-2, \; x_2=-1, \; x_3=2, \;x_4=\frac{3}{4}}.
By checking we can easily make sure of the correctness of the roots.
Example 2
Let us solve the below equation on the set of real numbers:
\latex{(2x+7)(x+2)-(x-5)(x+2)=(2x-3)(x+2)}.
Solution
We can see that two-term products are on both sides of the equation. If we expand the products, then we would get an equation which contains quadratic terms too. But we might find it too difficult to solve at this stage.
Instead, let us realise that the factor \latex{(x + 2)} appears in each term. In such a case we often hear the tip to divide through by this factor. It can be done but only after careful examination. We have to be careful not to do an operation which does not have a meaning. Namely, it is not possible to divide by an expression the value of which can be equal to 0. When carrying out such a division we can often lose a root, which means that we can give the solutions of the equation only partially.
We can eliminate this error if we solve the exercise according to the following method.
Solution I
Let us factor out the common factor from the two terms on the left side:
\latex{(x+2)\left[(2x+7)-(x-5)\right]=(2x-3)(x+2)},
\latex{(x+2)\left[(2x+7)-(x-5)\right]=(2x-3)(x+2)},
\latex{(x+2)(x+12)=(2x-3)(x+2)}.
Then we rearrange the equation so that one side is equal to 0. Then we factor out factor \latex{(x+2)}:\latex{(x+2)\left[(x+12)-(2x-3)\right]=0},
\latex{(x+2)(-x+15)=0}.
This equation will be true if \latex{(x+2)=0}, i.e. \latex{x=-2}, of if \latex{-x+15=0}, i.e. \latex{x=15}.
Thus the roots of the equation: \latex{x_1 = –2} and \latex{x_2 = 15}. We can make sure of the correctness of these when checking.
Solution II
If we still choose the operation of dividing out by the unknown factor, then we can do it as follows.
After bringing the equation to the
After bringing the equation to the
\latex{(x+2)(x+12)=(2x-3)(x+2)}
form, let us first consider the case when \latex{x+2=0}. In this case both sides will be equal to 0, i.e. \latex{x_1 = –2} is a solution of the equation.
In the case when \latex{x+2\neq 0}, we can divide both sides by it, and thus we get the following equation:
In the case when \latex{x+2\neq 0}, we can divide both sides by it, and thus we get the following equation:
\latex{x+12=2x-3}.
After rearranging and solving it \latex{x_2 = 15} is resulting, which is also a solution.
Example 3
Let us solve the below equation if the fundamental set is the set of real numbers:
\latex{(2x+6)(x+2)+(x+3)(2x-1)=(3x+9)(x-2)}.
Solution
We can realise that \latex{2x+6=2(x+3)} and \latex{3x+9=3(x+3)}. So we can bring the equation to the following form:
\latex{2(x+3)(x+2)+(x+3)(2x-1)=3(x+3)(x-2)},
\latex{(x+3)(x+9)=0}.
\latex{2(x+3)(x+2)+(x+3)(2x-1)=3(x+3)(x-2)},
\latex{(x+3)\left[2(x+2)+(2x-1)\right]=3(x+3)(x-2)},
\latex{(x+3)(4x+3)=3(x+3)(x-2)}.
When rearranging the right side so that it equals 0 and again factoring out the identical factors the following equation results:\latex{(x+3)(x+9)=0}.
The value of the product on the left side will be 0 if and only if the value of one of the factors is 0. I.e. \latex{x + 3 = 0} or \latex{x + 9 = 0}. From this the solutions of the equation will be as follows: \latex{x_1 = –3} and \latex{x_2 = –9}, we can easily make sure of the correctness of these when checking.
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It happens very often that the appointed exercises are not in this form. While we are solving the exercise we have to find those rearrangements which help to get the roots too.
Example 4
Let us solve the following equation on the set of real numbers:
\latex{x^3-4x^2+3x=0}.
Solution
While solving the equation we try to factorise the expression on the left side of the equation.
We can factor out variable \latex{ x } from each term, and then we get the following equation:
\latex{x(x^2-4x+3)=0}.
Then we split the second factor further. Let us group the appearing terms in the below way:
We can factor out variable \latex{ x } from each term, and then we get the following equation:
\latex{x(x^2-4x+3)=0}.
\latex{x(x^2-x-3x+3)=0},
\latex{x\left[x(x-1)-3(x-1)\right]=0}.
By factoring out factor \latex{(x-1)} from the square brackets we factorise the left side of the equation:\latex{x(x-1)(x-3)=0}.
The roots of this equation based on the previous notes: \latex{x_1 = 0, x_2 = 1, x_3 = 3}. We can make sure of the correctness of these when checking.factorisation
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Another way of factorising the quadratic expression \latex{x^2 - 4x + 3} also offers. And this is the method of completing the square which consists of the following steps.
\latex{x^2-4x+3=(x-2)^2-4+3=(x-2)^2-1}.
\latex{a^2-b^2=(a-b)(a+b)},
- Based on the first two terms we create the square of a binomial:
\latex{x^2-4x+3=(x-2)^2-4+3=(x-2)^2-1}.
- The resulting expression can be thought of as the difference of two square numbers, and as such it can be factorised according to identity:
\latex{a^2-b^2=(a-b)(a+b)},
\latex{(x-2)^2-1=(x-2-1)(x-2+1)=(x-3)(x-1)}
From here we can finish the solution similarly as above.
Exercises
{{exercise_number}}. The product of four consecutive whole numbers is 0. Give these numbers.
{{exercise_number}}. Solve the following equations.
- \latex{(x-4)(3x+6)(5-10x)(12x+48)=0}
- \latex{2x(6-2x)\frac{(5-4x)}{4}=0}
- \latex{x\frac{(12-x)(2x+3)}{x-12}(3x-8)=0}
- \latex{(5x-4)\frac{12x+6}{x+\dfrac{1}{2}}=0}
- \latex{(x-4)(3x+6)+(x-4)(12x+48)=0}
- \latex{2x(6-2x)+\frac{(5-4x)x}{4}=0}
- \latex{x\frac{(12-x)(2x+3)}{2}-x(x-12)(3x-8)=0}
- \latex{(5x-4)-\frac{(12x+6)(5x-4)}{\dfrac{1}{2}}=0}
{{exercise_number}}. Solve the following equations.
- \latex{(x+4)(3x+6)+(5-10x)(3x+6)=6x+12}
- \latex{2x(6-2x)-\frac{x(5+4x)}{4}=x(2x-2)}
- \latex{x\frac{(12-x)(2x+3)}{(x-12)}-(2x+3)(3x-8)=(2x-3)(x+2)}
- \latex{\left(\frac{5}{12}x+\frac{1}{3}\right)\frac{12x+6}{x+\dfrac{1}{2}}+x^2=0}
Quiz
Where is the error in the following train of thought?
\latex{a^2-a^2=a^2-a^2}
\latex{a(a-a)=(a+a)(a-a)} /simpifying by \latex{(a-a)}:
\latex{a=2a}, therefore
\latex{1=2}.
