Question

How do you solve \[\left( x-5 \right)\left( x+6 \right)=12\] ?

Answer 1

Hint: In order to solve this question, we will first multiply both the terms by expanding the brackets and solve the equation as a quadratic equation and we will simplify the equation by finding the roots and by applying quadratic equation formula that is, $\left( {{x}_{1}},{{x}_{2}} \right)=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, where $\left( {{x}_{1}},{{x}_{2}} \right)$ are the roots of the equation and $a,b,c$ are the coefficients of the expression.

Complete step by step solution:
We have the given equation:
\[\left( x-5 \right)\left( x+6 \right)=12\]
First, we have to convert our equation in quadratic equation format.
Therefore, on expanding we get:
\[x\left( x+6 \right)-5\left( x+6 \right)=12\]
On simplifying, we get:
\[{{x}^{2}}+6x-5x-30=12\]
Now, on taking 12 at left hand side and simplifying, we get:
\[{{x}^{2}}+x-42=0\]
So, now we will solve with a quadratic formula.
Now for a quadratic equation in the form of $a{{x}^{2}}+bx+c=0$ , we will use the following quadratic equation formula: $\left( {{x}_{1}},{{x}_{2}} \right)=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Therefore from the above equation and on comparing, we get \[a=1,b=1,c=-42\]
Now, substituting the values in formula we get: $\left( {{x}_{1}},{{x}_{2}} \right)=\dfrac{-\left( 1 \right)\pm \sqrt{{{\left( 1 \right)}^{2}}-4\times 1\times \left( -42 \right)}}{2\times 1}$
Now we will first simplify, root part in our equation:
$\Rightarrow \sqrt{{{1}^{2}}-4\times 1\times -42}$
Now, we will apply rule: $-\left( -a \right)=a$
So, we get our equation as:
$\Rightarrow \sqrt{{{1}^{2}}+4\times 1\times 42}$
Now, multiply and add the numbers. $\sqrt{{{1}^{2}}+168}$
So, we get:
$\Rightarrow \sqrt{168}$
Now, we will factor the number: $169={{13}^{2}}$
 $\Rightarrow \sqrt{{{13}^{2}}}$
Now, we will apply radical rule: $\sqrt[n]{{{a}^{n}}}=a$
So, we get:
$\Rightarrow 13$
Now coming to our equation and substituting what we equate, we get:
$\left( {{x}_{1}},{{x}_{2}} \right)=\dfrac{-1\pm 13}{2}$
Now, on separating the above equation, we get:
\[{{x}_{1}}=\dfrac{-1+13}{2},{{x}_{2}}=\dfrac{-1-13}{2}\]
On simplifying the solutions, we get:
\[{{x}_{1}}=\dfrac{12}{2},{{x}_{2}}=\dfrac{-14}{2}\]
On dividing the numbers, we get our solution as:
\[{{x}_{1}}=6,{{x}_{2}}=-7\]
Therefore, the solution to the quadratic equation are:
\[{{x}_{1}}=6\] and \[{{x}_{2}}=-7\]

Note: There are three forms of quadratic equation, Standard form:$y=a{{x}^{2}}+bx+c$ where $a,b,c$ are just numbers. Factored form:\[~y\text{ }=\text{ }\left( ax\text{ }+\text{ }c \right)\left( bx\text{ }+\text{ }d \right)\] again the $a,b,c,d$ are just numbers. Vertex form: \[y\text{ }=\text{ }a{{\left( x\text{ }+\text{ }b \right)}^{2}}\text{ }+\text{ }c\] again the $a,b,c$ are just numbers.
We have to identify by looking at the equation which form we have to use, then only we can solve the equation.
As we have a standard form of quadratic equation, we can verify the answer by substituting the answer in the original equation.