Question

How do you solve $2{x^2} + 4x - 5 = 0$

Answer 1

Hint: Find the discriminant value and use the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$. The first step will be to find the value of discriminant of this equation using the formula \[{b^2} - 4ac\]. Since, the value of \[{b^2} - 4ac\] is not a perfect square, we know that the roots are real but irrational. Therefore, moving ahead we will use the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ to get the roots or solution of the question.

Complete step-by-step solution:
The given equation is $2{x^2} + 4x - 5 = 0$
Now, we know that the roots or solution of any equation of the form $a{x^2} + bx + c = 0$ where the value of discriminant \[{b^2} - 4ac\] is not a perfect square is mathematically given by:-
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Calculating the value of discriminant \[{b^2} - 4ac\] we will get
\[
  {b^2} - 4ac = \left( {16} \right) - \left( {4 \times 2 \times - 5} \right) \\
   = 56 \\
\]
Here, a is the coefficient of ${x^2}$, b is the coefficient of $x$ and $c$ is the constant.
Since, 56 is not a perfect square we will have to use the given formula to find the solution
Comparing the given general solution form with our question we will find that:-
$a=2, b=4, c=-5$.
Therefore, putting the respective values of $a, b$ and $c$ in the equation we will get:-
\[
  x = \dfrac{{ - 4 \pm \sqrt {{{(4)}^2} - (4 \times 2 \times ( - 5))} }}{{2 \times 2}} \\
   = \dfrac{{ - 4 \pm \sqrt {16 + 40} }}{4} \\
   = \dfrac{{ - 4 \pm \sqrt {56} }}{4} \\
\]
We can write $\sqrt {56} = 2 \times \sqrt {14} $ and then dividing the whole equation by 2 we will get
 \[
  \dfrac{{ - 4 \pm 2\sqrt {14} }}{4} = \dfrac{{2\left( { - 2 \pm \sqrt {14} } \right)}}{4} \\
   = \dfrac{{ - 2 \pm \sqrt {14} }}{2} \\
\]
Which means our first solution is $\dfrac{{ - 2 + \sqrt {14} }}{2}$ and second solution is $\dfrac{{ - 2 - \sqrt {14} }}{2}$

Note: The above process to find the roots of the equation is done only when the discriminant value of the equation \[{b^2} - 4ac\] is not a perfect square. Only when this value is not a perfect square, we can say that our equation has real and irrational roots. Here, \[{b^2} - 4ac\] was equal to 56, which is not a perfect square, hence the method. In other cases, if \[{b^2} - 4ac\] is less than 0 then roots are imaginary.