Question

How do you solve using the quadratic formula \[2{{x}^{2}}-2x=1\]?

Answer 1

Hint: This type of problem is based on the concept of quadratic equations. First, we have to consider the given function and subtract 1 from both the sides of the equation. Then, use the quadratic formula, that is, \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]. Here, a=2, b=-2 and c=-1. Substituting these values in the quadratic formula, we get the value of x. On further simplification, that is, by cancelling out the common terms and taking square roots, we get the value of x which is the required answer.

Complete step by step answer:
According to the question, we are asked to find the value of x using the quadratic formula of the equation \[2{{x}^{2}}-2x=1\].
We have been given the quadratic equation is \[2{{x}^{2}}-2x=1\].
Let us first subtract 1 from both the sides of the given equation. We get
\[2{{x}^{2}}-2x-1=1-1\]
On further simplification, we get
\[2{{x}^{2}}-2x-1=0\]
We know that, for a quadratic equation \[a{{x}^{2}}+bx+c=0\], the values of x are \[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].
Comparing with standard quadratic equation, we get
a=2, b=-2 and c=-1.
Substituting these values in the quadratic formula, we get
\[x=\dfrac{-(-2)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( 2 \right)\left( -1 \right)}}{2\left( 2 \right)}\].
On further simplifications, we get
\[x=\dfrac{2\pm \sqrt{4-4\left( -2 \right)}}{4}\]
\[\Rightarrow x=\dfrac{2\pm \sqrt{4+8}}{4}\]
\[\Rightarrow x=\dfrac{2\pm \sqrt{12}}{4}\]
We know that, \[12=3\times 4\]. We get,
\[\Rightarrow x=\dfrac{2\pm \sqrt{3\times 4}}{4}\]
Using the property \[\sqrt{a\times b}=\sqrt{a}\times \sqrt{b}\], we get
\[x=\dfrac{2\pm \sqrt{3}\times \sqrt{4}}{4}\]
We know that \[\sqrt{4}=2\].
Therefore, we get
\[x=\dfrac{2\pm 2\sqrt{3}}{4}\]
Let us now cancel out the common terms.
Here, the common term is 2.
\[x=\dfrac{2\left( 1\pm \sqrt{3} \right)}{2\times 2}\]
\[\Rightarrow x=\dfrac{\left( 1\pm \sqrt{3} \right)}{2}\]
Therefore, the values of x are \[\dfrac{1+\sqrt{3}}{2}\] and \[\dfrac{1-\sqrt{3}}{2}\].
Hence, the values of x for the equation \[2{{x}^{2}}-2x=1\] are \[\dfrac{1+\sqrt{3}}{2}\] and \[\dfrac{1-\sqrt{3}}{2}\].

Note: Whenever we get such types of problems, we should make necessary calculations to make the right-hand side of the equation to be 0 and then solve the problem. Here, we can also divide the whole equation by 2 and then use the quadratic formula to solve the question.