Question

How do you solve $2{x^2} + 3x + 7 = 0$?

Answer 1

Hint:In order to determine the solution to the above quadratic equation having variable $x$,first we compare it with the standard equation i.e. $a{x^2} + bx + c$ to obtain the value of the coefficients. Then find the determinant to find out the nature of roots ,in our case it comes to be $D < 0$ which means roots are distinct but imaginary and now apply the quadratic formula $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ to get your required solutions.

Complete step by step solution:
Given a quadratic equation, $2{x^2} + 3x + 7 = 0$ let it be $f(x)$ and the roots to the equation be
${x_1},{x_2}$.
$f(x) = 2{x^2} + 3x + 7 = 0$
Comparing the equation with the standard Quadratic equation $a{x^2} + bx + c$
a becomes 2
b becomes 3
And c becomes 7
Let’s now find the determinant value of this quadratic equation, as by looking to the value of
determinant we can find out the nature of both the roots.
$D = {b^2} - 4ac$
Putting the values, we get
$
D = {\left( 3 \right)^2} - 4(2)(7) \\
= 9 - 56 \\
= - 47 \\
$
As we can see $D < 0$,so we can conclude that the both roots of the equation will be distinct but
imaginary
Now using Quadratic formula to find out the actual value of the roots
x=$\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Since,${b^2} - 4ac = D$
$
x = \dfrac{{ - b \pm \sqrt D }}{{2a}} \\
\Rightarrow {x_1} = \dfrac{{ - b + \sqrt D }}{{2a}},{x_2} = \dfrac{{ - b - \sqrt D }}{{2a}} \\
\\
$Now putting the value of $D,b,a$,we get
$
\Rightarrow {x_1} = \dfrac{{ - (3) + \sqrt { - 47} }}{{2(2)}},{x_2} = \dfrac{{ - (3) - \sqrt { - 47}
}}{{2(2)}} \\
\Rightarrow {x_1} = \dfrac{{ - 3 + \sqrt { - 47} }}{4},{x_2} = \dfrac{{ - 3 - \sqrt { - 47} }}{4} \\
\Rightarrow {x_1} = \dfrac{{ - 3 + \sqrt {( - 1)47} }}{4},{x_2} = \dfrac{{ - 3 - \sqrt {( - 1)47} }}{4}
\\
$
As we know that the square of iota $i$ is equal to $ - 1$ i.e. ${i^2} = - 1$
So replacing $ - 1$with ${i^2}$,we get
$
\Rightarrow {x_1} = \dfrac{{ - 3 + \sqrt {{i^2}(47)} }}{4},{x_2} = \dfrac{{ - 3 - \sqrt {{i^2}(47)} }}{4} \\
\Rightarrow {x_1} = \dfrac{{ - 3 + i\sqrt {47} }}{4},{x_2} = \dfrac{{ - 3 - i\sqrt {47} }}{4} \\
$
Therefore, the solution to the quadratic equation $2{x^2} + 3x + 7 = 0 $having roots ${x_1},{x_2}$ is${x_1} = \dfrac{{ - 3 + i\sqrt {47} }}{4},{x_2} = \dfrac{{ - 3 - i\sqrt {47} }}{4}$,where $i$ is the imaginary number.

Additional Information:
Quadratic Equation: A quadratic equation is a equation which can be represented in the form of $a{x^2} + bx + c$ where $x$ is the unknown variable and a,b,c are the numbers known where $a \ne 0$.If $a = 0$then the equation will become linear equation and will no more quadratic .
The degree of the quadratic equation is of the order 2.
Every Quadratic equation has 2 roots.
Discriminant: $D = {b^2} - 4ac$
Using Discriminant, we can find out the nature of the roots
If D is equal to zero, then both of the roots will be the same and real.
If D is a positive number then, both of the roots are real solutions.
If D is a negative number, then the root are the pair of complex solutions

Note:
1. One must be careful while calculating the answer as calculation error may come.
2.Don’t forget to compare the given quadratic equation with the standard one every time.
3. Cube of iota $i$is equal to $ - i$or ${i^3} = {i^2}.i = - i$.