Question

Solve the equation by using formula: ${{x}^{2}}-2x-2=0$.

Answer 1

Hint: Use formula, $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ where quadratic equation is $a{{x}^{2}}+bx+c=0$. Thus, putting values in formulas and getting answers.

Complete step-by-step answer:

We are given a quadratic equation ${{x}^{2}}-2x-2=0$ and we have to solve it to find the value of x.

${{x}^{2}}-2x-2=0$ is considered a quadratic equation as a quadratic equation is any equation that can be rearranged in standard form as $a{{x}^{2}}+bx+c=0$ .

Here x represents unknown a, b, c are known numbers where $a\ne 0$ otherwise it becomes linear as no ${{x}^{2}}$ term is there. The numbers a, b, c are coefficient of equation and may be distinguished by calling them respectively. The quadratic coefficient, the linear coefficient, the linear coefficient and the constant or free term.

The values of x that satisfy the equation are called solutions of the equation and roots or zeros of the expression on the left hand side. A quadratic equation has at most two solutions. If there is no real solution, then there are two complex solutions. If there is only one solution, one says that it is a double root. A quadratic equation always has two roots, if complex roots are included and a double root is counted for two. A quadratic equation of form $a{{x}^{2}}+bx+c=0$ can be factored as $a\left( x-\alpha \right)\left( x-\beta \right)=0$ where $\alpha $ and $\beta $ are solutions of x.

Because, the quadratic equation involves only one known, it is called univariate. The quadratic equation only contains powers of x that are non-negative integers and therefore, it is a polynomial equation. In particular it is a second degree polynomial equation.

We are given equation ${{x}^{2}}-2x-2=0$ and we will solve by using formula

$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

Where quadratic equation is $a{{x}^{2}}+bx+c=0$

So, by comparing we can say that $a=1,b=-2,c=-2$ so the value of x will be $\dfrac{-\left( -2

\right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\times 1\times \left( -2 \right)}}{2\times 1}$

So, on simplifying we get,

$x=\dfrac{2\pm \sqrt{4+8}}{2}=\dfrac{2\pm 2\sqrt{3}}{2}$

Hence, x is $1+\sqrt{3}$ or $1-\sqrt{3}$ .


Note: We can also solve by the complete square method which is, ${{x}^{2}}-2x-2=0$ as ${{x}^{2}}-2x=2$. Now adding 1 to both sides, ${{x}^{2}}-2x+1=3$ and write it as ${{\left( x-1 \right)}^{2}}=3$. Thus, taking square roots and doing necessary calculations.