Question

How do you solve $2{{x}^{2}}-3x+4=0$ using the quadratic formula?

Answer 1

We have been given a quadratic equation of x as $2{{x}^{2}}-3x+4=0$. We use the quadratic formula to solve the value of the x. we have the solution in the form of $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ for general equation of $a{{x}^{2}}+bx+c=0$. We put the values and find the solution.

Complete answer:
We know for a general equation of quadratic $a{{x}^{2}}+bx+c=0$, the value of the roots of x will be $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. This is the quadratic equation solving method. The root part $\sqrt{{{b}^{2}}-4ac}$ of $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ is called the discriminant of the equation.
In the given equation we have $2{{x}^{2}}-3x+4=0$. The values of a, b, c are $2,-3,4$ respectively.
We put the values and get x as $x=\dfrac{-\left( -3 \right)\pm \sqrt{{{\left( -3 \right)}^{2}}-4\times 4\times 2}}{2\times 2}=\dfrac{3\pm \sqrt{-23}}{4}=\dfrac{3\pm i\sqrt{23}}{4}$.
The roots of the equation are imaginary numbers.
The discriminant value being negative, we get the imaginary numbers root values.
In this case the value of $D=\sqrt{{{b}^{2}}-4ac}$ is non-square. ${{b}^{2}}-4ac={{\left( -3 \right)}^{2}}-4\times 4\times 2=-23$.
This is a negative value. That’s why the roots are imaginary.

Note: We find the value of x for which the function $f\left( x \right)={{x}^{2}}-4x+1$. We can see $f\left( 2+\sqrt{3} \right)={{\left( 2+\sqrt{3} \right)}^{2}}-4\left( 2+\sqrt{3} \right)+1=4+3+4\sqrt{3}-8-4\sqrt{3}+1=0$. So, the root of the $f\left( x \right)={{x}^{2}}-4x+1$ will be the $2+\sqrt{3}$. This means for $x=a$, if $f\left( a \right)=0$ then $\left( x-a \right)$ is a root of $f\left( x \right)$. We can also do the same process for $2-\sqrt{3}$.
We can also solve using the square form.
We have $2{{x}^{2}}-3x+4=2\left[ {{\left( x-\dfrac{3}{4} \right)}^{2}}+\dfrac{23}{16} \right]$.
We get $2\left[ {{\left( x-\dfrac{3}{4} \right)}^{2}}+\dfrac{23}{16} \right]=0$. Taking solution, we get

& {{\left( x-\dfrac{3}{4} \right)}^{2}}+\dfrac{23}{16}=0 \\\ & \Rightarrow {{\left( x-\dfrac{3}{4} \right)}^{2}}=-\dfrac{23}{16} \\\ & \Rightarrow \left( x-\dfrac{3}{4} \right)=\dfrac{\pm i\sqrt{23}}{4} \\\ & \Rightarrow x=\dfrac{3\pm i\sqrt{23}}{4} \\\ \end{aligned}$$. Thus, verified the solution of the equation $2{{x}^{2}}-3x+4=0$ is $$x=\dfrac{3\pm i\sqrt{23}}{4}$$.