
How do you use algebra $\left( x-1 \right)\left( x+2 \right)=18?$
Answer
447.6k+ views
Hint: In this question the equation we have to solve by using algebra. The equatic is a related quadratic equation. So, first we have to solve the brackets by multiplication then simplify the equation we get it in the standard form; $a{{x}^{2}}+bx+c=0$ After that simply solve it to get the factor. You can also use the quadratic formula for solving the above methods for solving the given equations.
Complete step by step solution:
In the question,
Given equation $\left( x-1 \right)\left( x+2 \right)=18$
Multiply the given equation to get rid of the parenthesis.
$\left( x-1 \right)\left( x+2 \right)=18$
$\therefore {{x}^{2}}+2x-x-2=18$
$\therefore {{x}^{2}}+x-2=18$
The equation we have not suited in the standard form.
Now, subtract it with $18$ form both sides of the equation we get,
${{x}^{2}}+x-2=18$
${{x}^{2}}+x-2-18=18-18$
$\therefore {{x}^{2}}+x-20=0$
Compare the equation with standard form.
$a{{x}^{2}}+bx+c=0$
${{x}^{2}}+x-20=0$
$a=1,b=1,c=-20$
By using the value of $a,b,c$ in above formula
\[x=\dfrac{\left( -1 \right)\pm \sqrt{{{\left( 1 \right)}^{2}}-4\times 1\times \left( -20 \right)}}{2\times 1}\]
$\therefore x=\dfrac{\left( -1 \right)\pm \sqrt{9}}{2}$
$\therefore x=\dfrac{\left( -1 \right)+\sqrt{9}}{2}$ and $x=\dfrac{\left( -1 \right)-\sqrt{9}}{2}$
$x=\dfrac{\left( -1 \right)+\sqrt{1-4\times \left( -20 \right)}}{2}$ or $x=\dfrac{\left( -1 \right)\sqrt{1-4\times \left( -20 \right)}}{2}$
$\therefore x=4$ or $x=-5$
We can also solve it in other ways.
${{x}^{2}}+x-18=18-18$
$\therefore {{x}^{2}}+x-20=0$
${{x}^{2}}+x-20=\left( x+5 \right).\left( x-4 \right)=0$
$\therefore x+5=0$ or $x-4=0$
$\therefore x=-5$ or $x=4$
Hence,After solving the equation $\left( x-1 \right)\left( x+2 \right)=18$ we get the factor $-5$ and $4.$
Additional Information:
Factoring in a polynomial has writing it as a product of two or more polynomials. In the method of factorisation we reduce in the simplest form. The factor can be an integer. A variable or algebraic itself in any type of equation. There are various methods of factorisation; they are factoring out GCF, the grouping method. The difference of square pattern etc. We have several examples of factoring. However for this you should take common factors using distributive property.
As an example: $6{{x}^{2}}+4x=2x\left( 3x+2 \right)$
Note: In the factorisation problem a common mistake is done when we have to expand a binomial. Which is raised to power but students will distribute the exponents. As an example we take ${{\left( x+3 \right)}^{2}}={{x}^{2}}+{{3}^{2}}={{x}^{2}}+9.$ Which is incorrect. Hence you will understand it by the given example. Another root the students will distribute the root but the question is asked to simplify the expression within radical also a square root.
Complete step by step solution:
In the question,
Given equation $\left( x-1 \right)\left( x+2 \right)=18$
Multiply the given equation to get rid of the parenthesis.
$\left( x-1 \right)\left( x+2 \right)=18$
$\therefore {{x}^{2}}+2x-x-2=18$
$\therefore {{x}^{2}}+x-2=18$
The equation we have not suited in the standard form.
Now, subtract it with $18$ form both sides of the equation we get,
${{x}^{2}}+x-2=18$
${{x}^{2}}+x-2-18=18-18$
$\therefore {{x}^{2}}+x-20=0$
Compare the equation with standard form.
$a{{x}^{2}}+bx+c=0$
${{x}^{2}}+x-20=0$
$a=1,b=1,c=-20$
By using the value of $a,b,c$ in above formula
\[x=\dfrac{\left( -1 \right)\pm \sqrt{{{\left( 1 \right)}^{2}}-4\times 1\times \left( -20 \right)}}{2\times 1}\]
$\therefore x=\dfrac{\left( -1 \right)\pm \sqrt{9}}{2}$
$\therefore x=\dfrac{\left( -1 \right)+\sqrt{9}}{2}$ and $x=\dfrac{\left( -1 \right)-\sqrt{9}}{2}$
$x=\dfrac{\left( -1 \right)+\sqrt{1-4\times \left( -20 \right)}}{2}$ or $x=\dfrac{\left( -1 \right)\sqrt{1-4\times \left( -20 \right)}}{2}$
$\therefore x=4$ or $x=-5$
We can also solve it in other ways.
${{x}^{2}}+x-18=18-18$
$\therefore {{x}^{2}}+x-20=0$
${{x}^{2}}+x-20=\left( x+5 \right).\left( x-4 \right)=0$
$\therefore x+5=0$ or $x-4=0$
$\therefore x=-5$ or $x=4$
Hence,After solving the equation $\left( x-1 \right)\left( x+2 \right)=18$ we get the factor $-5$ and $4.$
Additional Information:
Factoring in a polynomial has writing it as a product of two or more polynomials. In the method of factorisation we reduce in the simplest form. The factor can be an integer. A variable or algebraic itself in any type of equation. There are various methods of factorisation; they are factoring out GCF, the grouping method. The difference of square pattern etc. We have several examples of factoring. However for this you should take common factors using distributive property.
As an example: $6{{x}^{2}}+4x=2x\left( 3x+2 \right)$
Note: In the factorisation problem a common mistake is done when we have to expand a binomial. Which is raised to power but students will distribute the exponents. As an example we take ${{\left( x+3 \right)}^{2}}={{x}^{2}}+{{3}^{2}}={{x}^{2}}+9.$ Which is incorrect. Hence you will understand it by the given example. Another root the students will distribute the root but the question is asked to simplify the expression within radical also a square root.
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