Answer
405k+ views
Hint: The given expression is ${x^2} - 9x + 18$. We can see that the coefficient of $x$ is $ - 9$ and constant term is $18$. We can express these numbers as sum and product respectively using the numbers six and three. Then we can rearrange the terms to get the factors. Since the given equation is second degree, we will get two linear (first degree) factors.
Complete step-by-step solution:
We are given the equation, ${x^2} - 9x + 18$.
We can split the term $9x$.
Since $9 = 6 + 3$, we have,
${x^2} - 9x + 18 = {x^2} - (6 + 3)x + 18$
Opening brackets we get,
${x^2} - 9x + 18 = {x^2} - 6x - 3x + 18$
Now we can see the constant term is $18$.
Also $18 = 6 \times 3$
So we can rewrite the equation as,
${x^2} - 9x + 18 = {x^2} - 6x - 3x + 6 \times 3$
Now we can take $x$ common from the first two terms and $ - 3$ common from the last two terms of the right side.
Thus we have,
${x^2} - 9x + 18 = x(x - 6) - 3(x - 6)$
So we get a common factor $x - 6$.
This gives,
${x^2} - 9x + 18 = (x - 6)(x - 3)$
Thus we had factorised the given expression.
For the equation ${x^2} - 9x + 18$, $x - 6$ and $x - 3$ are the factors.
Note: There is another method to solve the equation. It is called completing the square method.
For, the given equation can be rewritten as ${x^2} - 9x = - 18$
Now add the square of half of the coefficient of $x$ on both sides.
Here the coefficient of $x$ is $ - 9$. Half of $ - 9$ is $ - \dfrac{9}{2}$.
Squaring we get $\dfrac{{81}}{4}$.
Adding this on both sides we get,
${x^2} - 9x + \dfrac{{81}}{4} = \dfrac{{81}}{4} - 18$
$ \Rightarrow {x^2} - 9x + \dfrac{{81}}{4} = \dfrac{{81 - 72}}{4} = \dfrac{9}{4}$
We can observe that the left side is in the form ${a^2} - 2ab + {b^2}$ form which is equal to ${(a - b)^2}$.
So we get,
${x^2} - 9x + \dfrac{{81}}{4} = {(x - \dfrac{9}{2})^2}$
And also $\dfrac{9}{4} = {( \pm \dfrac{3}{2})^2}$.
So taking roots on both sides we can write,
$x - \dfrac{9}{2} = \pm \dfrac{3}{2}$
This gives $x = \dfrac{9}{2} + \dfrac{3}{2} = \dfrac{{12}}{2} = 6$ or $x = \dfrac{9}{2} - \dfrac{3}{2} = \dfrac{6}{2} = 3$
So the factors are $x - 6$ and $x - 3$.
Complete step-by-step solution:
We are given the equation, ${x^2} - 9x + 18$.
We can split the term $9x$.
Since $9 = 6 + 3$, we have,
${x^2} - 9x + 18 = {x^2} - (6 + 3)x + 18$
Opening brackets we get,
${x^2} - 9x + 18 = {x^2} - 6x - 3x + 18$
Now we can see the constant term is $18$.
Also $18 = 6 \times 3$
So we can rewrite the equation as,
${x^2} - 9x + 18 = {x^2} - 6x - 3x + 6 \times 3$
Now we can take $x$ common from the first two terms and $ - 3$ common from the last two terms of the right side.
Thus we have,
${x^2} - 9x + 18 = x(x - 6) - 3(x - 6)$
So we get a common factor $x - 6$.
This gives,
${x^2} - 9x + 18 = (x - 6)(x - 3)$
Thus we had factorised the given expression.
For the equation ${x^2} - 9x + 18$, $x - 6$ and $x - 3$ are the factors.
Note: There is another method to solve the equation. It is called completing the square method.
For, the given equation can be rewritten as ${x^2} - 9x = - 18$
Now add the square of half of the coefficient of $x$ on both sides.
Here the coefficient of $x$ is $ - 9$. Half of $ - 9$ is $ - \dfrac{9}{2}$.
Squaring we get $\dfrac{{81}}{4}$.
Adding this on both sides we get,
${x^2} - 9x + \dfrac{{81}}{4} = \dfrac{{81}}{4} - 18$
$ \Rightarrow {x^2} - 9x + \dfrac{{81}}{4} = \dfrac{{81 - 72}}{4} = \dfrac{9}{4}$
We can observe that the left side is in the form ${a^2} - 2ab + {b^2}$ form which is equal to ${(a - b)^2}$.
So we get,
${x^2} - 9x + \dfrac{{81}}{4} = {(x - \dfrac{9}{2})^2}$
And also $\dfrac{9}{4} = {( \pm \dfrac{3}{2})^2}$.
So taking roots on both sides we can write,
$x - \dfrac{9}{2} = \pm \dfrac{3}{2}$
This gives $x = \dfrac{9}{2} + \dfrac{3}{2} = \dfrac{{12}}{2} = 6$ or $x = \dfrac{9}{2} - \dfrac{3}{2} = \dfrac{6}{2} = 3$
So the factors are $x - 6$ and $x - 3$.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)