Question

Solve the given quadratic equation $2{{x}^{2}}-6x+3$.

Answer 1

To solve the above quadratic equation and find the roots of the equation we will be using the quadratic formula to find the roots of the quadratic equation by substituting the coefficients of ${{x}^{2}}$and $x$ in the formula which will result in the solution of the quadratic equation.

Complete step by step solution:
The equation that has been given to us to solve is $2{{x}^{2}}-6x+3$
Now to solve this quadratic equation we will need to find the roots of the equation and to find the roots of the equation we will use the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
In this above equation to find the values of a, b, c we will be comparing it with the general quadratic equation which is $a{{x}^{2}}+bx+c$ now when we compare it with the given equation we can see that the value of a is 2 value of b is -6 and value of c is 3 now we will be substituting all these values in the formula mentioned to find the roots of the equation and we get

& \Rightarrow x=\dfrac{-\left( -6 \right)\pm \sqrt{{{\left( -6 \right)}^{2}}-4\left( 2 \right)\left( 3 \right)}}{2\left( 2 \right)} \\\ & \Rightarrow x=\dfrac{6\pm \sqrt{36-24}}{4} \\\ & \Rightarrow x=\dfrac{6\pm \sqrt{12}}{4} \\\ & \Rightarrow x=\dfrac{6\pm 2\sqrt{3}}{4} \\\ \end{aligned}$$ Now as we can see that there will be two roots of the equation as the maximum power of x is 2 which indicates that there will be two roots of the equation. From the above answer for x we can see that there are two answers which are $$x=\dfrac{6+2\sqrt{3}}{4}$$ and $$x=\dfrac{6-2\sqrt{3}}{4}$$ which on further solving will give $$x=\dfrac{3+\sqrt{3}}{2}$$ and $$x=\dfrac{3-\sqrt{3}}{2}$$ . To find whether this is our answer we will substitute the values we got in the given equation and we will see that the right hand side and the left hand side produce the same value which is zero(0). **So the solution of the given quadratic equation is $$x=\dfrac{3+\sqrt{3}}{2}$$ and $$x=\dfrac{3-\sqrt{3}}{2}$$.** **Note:** The common mistakes that happen in while solving the quadratic equation is that what if we get the value of $${{b}^{2}}-4ac$$ as negative then there will be complex roots of the equation which will result in the solution that there are no real roots to the equation.