Question

How does \[{t^2} - 2t - 1 = 0\] become \[t = 1 \pm \sqrt 2 \]?

Answer 1

Hint: Here, we will find the process at which the given quadratic equation is converted to find the roots of the variable. We will use the Quadratic roots formula and by substituting the coefficients of the variables, we will find the roots of the variable. Thus, the process of conversion of the given quadratic equation into the roots is the required answer.

Formula Used:
Quadratic roots is given by the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]

Complete step by step solution:
We are given with a Quadratic equation \[{t^2} - 2t - 1 = 0\].
On comparing with the quadratic equation \[a{x^2} + bx + c = 0\], we get
\[\begin{array}{l}a = 1\\b = - 2\\c = - 1\end{array}\]
Now, by substituting 1 for the coefficient of \[{x^2}\], \[ - 2\] as coefficient of \[x\] and \[ - 1\] as the constant term \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\], we get
\[t = \dfrac{{ - \left( { - 2} \right) \pm \sqrt {{{\left( { - 2} \right)}^2} - 4\left( 1 \right)\left( { - 1} \right)} }}{{2\left( 1 \right)}}\]
Now, by simplifying the equation, we get
\[ \Rightarrow t = \dfrac{{2 \pm \sqrt {4 + 4} }}{2}\]
Now, by adding the terms, we get
\[ \Rightarrow t = \dfrac{{2 \pm \sqrt 8 }}{2}\]
Now, by simplifying the equation, we get
\[ \Rightarrow t = \dfrac{{2 \pm \sqrt {2 \times 4} }}{2}\]
Now, by simplifying the equation, we get
\[ \Rightarrow t = \dfrac{{2 \pm 2\sqrt 2 }}{2}\]
Now, by taking out the common factors, we get
\[ \Rightarrow t = \dfrac{{2\left( {1 \pm \sqrt 2 } \right)}}{2}\]
Now, by dividing, we get
\[ \Rightarrow t = \left( {1 \pm \sqrt 2 } \right)\]

Therefore, \[{t^2} - 2t - 1 = 0\] becomes \[t = 1 \pm \sqrt 2 \] by the process of factorization using the quadratic roots formula.

Additional Information:
We know that the Quadratic equation is an equation with the highest degree 2. Factorization is a process of rewriting the expression in terms of the product of the factors. The process of converting the higher degree polynomial as the product of factors of its lower degree polynomial which cannot be factored further is called Factorization.

Note:
We know that some quadratic equations cannot be solved by using the factorization method and square root method. But whatever be the quadratic equation, it is quite easy to solve by using the method of quadratic formula. We should be careful that the quadratic equation should be arranged in the right form. We should also notice that we have both the positive and negative signs in the formula, so the solutions for the equations would be according to the signs. The values of the variable satisfying the given equation are called the roots of a quadratic equation.