Question

How do you solve \[2{{x}^{2}}-4x-5=0\] using the quadratic formula?

Answer 1

Hint: Suppose that we have a quadratic equation in the form \[a{{x}^{2}}+bx+c=0\] where a, b and c is any number and a is non-zero. Using quadratic formula \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\], we can calculate the roots of the given equation. The behaviour of any quadratic equation \[a{{x}^{2}}+bx+c=0\] can be predicted by its discriminant given by \[{{b}^{2}}-4ac\].

Complete step by step answer:
We know that the sum of roots of this quadratic equation, that is \[{{x}_{1}}+{{x}_{2}}\] equals to negative of the ratio of x- coefficient to \[{{x}^{2}}\] - coefficient \[\Rightarrow \dfrac{-b}{a}\], which we write as \[{{x}_{1}}+{{x}_{2}}=\dfrac{-b}{a}\]. Also, the product of the roots of this quadratic equation, that is \[{{x}_{1}}\times {{x}_{2}}\] equals to the ratio of constant term to \[{{x}^{2}}\] - coefficient \[\Rightarrow \dfrac{c}{a}\].
That is, we can write it as \[{{x}_{1}}\times {{x}_{2}}=\dfrac{c}{a}\].
Using algebraic formula \[{{({{x}_{1}}-{{x}_{2}})}^{2}}={{({{x}_{1}}+{{x}_{2}})}^{2}}-4({{x}_{1}}\times {{x}_{2}})\], we get \[{{x}_{1}}-{{x}_{2}}=\dfrac{\sqrt{({{b}^{2}}-4ac)}}{a}\].
From the above equation, we have the quadratic formula which gives the information about the roots of the quadratic equation: \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].
In the given question, we are given the quadratic equation \[2{{x}^{2}}-4x-5=0\]. Here 2, -4 and -5 are the \[{{x}^{2}}\]- coefficient, x - coefficient and constant term respectively.
Comparing with the general quadratic equation \[a{{x}^{2}}+bx+c=0\], we can write a = 2,
b = -4 and c = -5.
Then, \[{{x}_{1}}+{{x}_{2}}=\dfrac{-(-4)}{2}\] which is simply 2 and \[{{x}_{1}}\times {{x}_{2}}=\dfrac{-5}{2}\].
On plugging the values of coefficient of \[{{x}^{2}}\] (a), coefficient of x (b) and the constant term (c) into the quadratic formula: \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\], we get
\[\begin{align}
  & \Rightarrow x=\dfrac{-(-4)\underline{+}\sqrt{({{(-4)}^{2}}-4(2)(-5))}}{2(2)} \\
 & \Rightarrow x=\dfrac{4\underline{+}\sqrt{(16-(-40))}}{4} \\
 & \Rightarrow x=\dfrac{4\underline{+}\sqrt{(16+40)}}{4} \\
\end{align}\]
\[\Rightarrow x=\dfrac{4\underline{+}\sqrt{56}}{4}\]
Here, we have 56 under square root. We can write it as \[\sqrt{4}\times \sqrt{14}\to 2\sqrt{14}\]. Then,
\[\Rightarrow x=\dfrac{4\underline{+}2\sqrt{14}}{4}=\dfrac{2\underline{+}\sqrt{14}}{2}\]
\[\therefore x=\dfrac{2+\sqrt{14}}{2}\] and \[x=\dfrac{2-\sqrt{14}}{2}\] are roots of the given quadratic equation \[2{{x}^{2}}-4x-5=0\].

Note:
We can also solve quadratic equations by factoring methods in which we try to obtain a common factor by splitting up the x term. We can understand the behaviour of roots of any quadratic equation from the value of discriminant \[D={{b}^{2}}-4ac\] where a, b and c are \[{{x}^{2}}\] - coefficient, x - coefficient and constant term respectively.