Question

How do you factor and solve ${{x}^{2}}-6x-5=0$ ?

Answer 1

In order to solve this question, we will solve the equation as a quadratic equation and we will simplify the equation by finding the roots and by identifying which formula is to use.

Complete step-by-step solution:
We have the given equation:
${{x}^{2}}-6x-5=0$
Now for a quadratic equation in the form of $a{{x}^{2}}+bx+c=0$ , we will use the following 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 $a=1,b=-6,c=-5$
Now, substituting the values in formula we get: $\left( {{x}_{1}},{{x}_{2}} \right)=\dfrac{-\left( -6 \right)\pm \sqrt{{{\left( -6 \right)}^{2}}-4\times 1\times \left( -5 \right)}}{2\times 1}$
Now we will first simplify, root part in our equation:
$\Rightarrow \sqrt{36+20}$
$\Rightarrow \sqrt{56}$
Now by doing prime factorization of $56$, we get:
$\Rightarrow \sqrt{{{2}^{3}}\times 7}$
Now by applying exponent rule that is ${{a}^{b+c}}={{a}^{b}}.{{a}^{c}}$ we get:
$\Rightarrow \sqrt{{{2}^{2}}\times 2\times 7}$
Now by applying radical rule that is $\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}$ we get:
$\Rightarrow \sqrt{{{2}^{2}}}\sqrt{2\times 7}$
Now again by applying radical rule that is $\sqrt[n]{{{a}^{n}}}=a$ we get:
$\sqrt{{{2}^{2}}}=2$
$\Rightarrow 2\sqrt{2\times 7}$
Therefore, on simplifying we get:
$\Rightarrow 2\sqrt{14}$
Now coming to our equation and substituting what we equate, we get:
$$$$ $\left( {{x}_{1}},{{x}_{2}} \right)=\dfrac{6\pm 2\sqrt{14}}{2}$
Now as we can see that there are common multiple of $2$, so we will use factorization method.
Therefore, on factorization, we get:
$\left( {{x}_{1}},{{x}_{2}} \right)=\dfrac{2\left( 3\pm \sqrt{14} \right)}{2}$
Now, on dividing and equating, we get:
$\left( {{x}_{1}},{{x}_{2}} \right)=3\pm \sqrt{14}$
Now on equating the above equation we get the value of x, i.e.:
${{x}_{1}}=3+\sqrt{14}$ and ${{x}_{2}}=3-\sqrt{14}$
Therefore, the solution to the given quadratic equation are:
${{x}_{1}}=3+\sqrt{14}$ and ${{x}_{2}}=3-\sqrt{14}$

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.
If we have a standard form of a quadratic equation, we can verify the answer by substituting the answer in the original equation.