Question

How do you solve $ {{x}^{2}}-2x-4=0 $ by factoring?

Answer 1

Hint:
 For answering this question we will use factorization. Factorization is the process of deriving factors of a number that divides the given number evenly. Factorization writes a number as the product of smaller numbers. Factorization is the process of reducing the bracket of a quadratic equation, instead of expanding the bracket and converting the equation to a product of factors that cannot be reduced further. There are many methods for the factorization process. Now, we will do the given question by the method of splitting the constant and doing the sum-product pattern.

Complete step by step answer:
Now considering from the question we have the expression $ {{x}^{2}}-2x-4=0 $ which is in the form of $ a{{x}^{2}}+bx+c=0 $ the zeroes of this quadratic equation is given by the formula: $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $
 $ \delta ={{b}^{2}}-4ac $ is the value of the discriminant which specifies the nature of the zeroes.
If $ \delta >0 $ and a perfect square then there are two real rational solutions.
If $ \delta >0 $ and is not a perfect square then there are two real irrational solutions.
If $ \delta <0 $ then there are two complex conjugate solutions that are unreal.
If $ \delta =0 $ then there is one real solution equal to $ \dfrac{-b}{2a} $ .
By comparing the coefficients, we get
 $ \begin{align}
  & a=1 \\
 & b=-2 \\
 & c=-4 \\
\end{align} $
Now,
 $ \Rightarrow \delta ={{b}^{2}}-4ac $
 $ \Rightarrow \delta ={{\left( -2 \right)}^{2}}-4\left( 1 \right)\left( -4 \right) $
 $ \Rightarrow \delta =4+16 $
 $ \Rightarrow \delta =20 $
Now we can say that the roots are real.
Now, by using above result we can get the factors of the given equation
 $ {{x}_{1}}=\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a} $
 $ {{x}_{1}}=\dfrac{2+\sqrt{20}}{2} $
 $ {{x}_{1}}=\dfrac{2+\sqrt{4\times 5}}{2} $
 $ {{x}_{1}}=\dfrac{2+2\sqrt{5}}{2} $
By taking $ 2 $ as common in the above equation we get,
 $ {{x}_{1}}=1+\sqrt{5} $
Therefore $ {{x}_{1}}=1+\sqrt{5} $ is one of the factor of the given equation.
 $ {{x}_{2}}=\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a} $
 $ {{x}_{2}}=\dfrac{2-\sqrt{20}}{2} $
 $ {{x}_{2}}=\dfrac{2-\sqrt{4\times 5}}{2} $
 $ {{x}_{2}}=\dfrac{2-2\sqrt{5}}{2} $
By taking $ 2 $ as common in the above equation we get,
 $ {{x}_{2}}=1-\sqrt{5} $
Therefore $ {{x}_{2}}=1-\sqrt{5} $ is another factor for the given equation.
Therefore we can conclude that $ {{x}_{1}}=1+\sqrt{5},\text{ }{{x}_{2}}=1-\sqrt{5} $ are the factors for the given equation.


Note:
During answering questions of this type we should be sure with the calculations and concept. For any quadratic equation in the form of $ a{{x}^{2}}+bx+c=0 $ the zeroes of this quadratic equation is given by the formula: $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $ . $ \delta ={{b}^{2}}-4ac $ is the value of the discriminant which specifies the nature of the zeroes.If $ \delta >0 $ and a perfect square then there are two real rational solutions. If $ \delta >0 $ and is not a perfect square then there are two real irrational solutions. If $ \delta <0 $ then there are two complex conjugate solutions that are unreal. If $ \delta =0 $ then there is one real solution equal to $ \dfrac{-b}{2a} $ . Let us consider an example if the given expression is $ {{x}^{2}}-2x-4=0 $ then the zeroes of this equation will be given as $ \dfrac{2\pm \sqrt{4-4\left( -4 \right)}}{2}=\dfrac{2\pm \sqrt{20}}{2}=1\pm \sqrt{5} $ and for this expression the discriminant $ \delta =4-4\left( -4 \right)=20>0 $ we had observed that here we have two distinct irrational roots as the discriminant value is not a prefect square.