Question

How do you solve ${x^2} + x + 10 = 0$ using quadratic formula?

Answer 1

Hint: Here, we find the values of variables from the given equation and then simply substitute them into the quadratic formula to find the roots of the equation.

Complete step-by-step solution:
Firstly, we will identify the variables $a,b$ and $c$ from the given equation by writing it into the standard quadratic equation.
Standard quadratic equation is given by: $a{x^2} + bx + c = 0$.
We can see that the given equation ${x^2} + x + 10 = 0$ is of the form $a{x^2} + bx + c = 0$.
Remember that $a,b$ are the coefficients of ${x^2},x$ and $c$ is a constant.
Now, according to this we have values of variables as $a = 1,b = 1,c = 10$.
In the quadratic formula, we have a discriminant $\Delta $ which helps in determining the nature of the roots of the quadratic equation.
$\Delta = {b^2} - 4ac$ …. $\left( 1 \right)$
Now, substituting the values of $a,b$ and $c$ in the above equation to get the value of discriminant, $\Delta $ .
$\Delta = {1^2} - 4 \times 1 \times 10$
Simplifying the equation,
$\Delta = 1 - 40 = - 39$
Since this is negative, that means this quadratic equation has no real roots. The roots are distinct and imaginary.
Now using quadratic formula: $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Using equation (1),
$x = \dfrac{{ - b \pm \sqrt \Delta }}{{2a}}$
Substituting the values of $a,b$ and $\Delta $ in the above equation, we get,
$x = \dfrac{{ - 1 \pm \sqrt { - 39} }}{2}$
We cannot find the square root of a negative number then in such cases it has complex conjugate pair of roots where $\sqrt { - 1} $ is represented as $'i'$ .
So, we get: $\Delta = i\sqrt {39} $
$x = \dfrac{{ - 1 \pm i\sqrt {39} }}{2}$
To solve for the unknown variable $x$ , separate them into two equations, one with plus and other with minus.
Hence, these are the values of $x$.

Note: Remember not to forget to calculate the discriminant to know the nature of the roots.
If ${b^2} - 4ac > 0$ , the roots are real and distinct.
If ${b^2} - 4ac = 0$ , the roots are real and equal.
And if ${b^2} - 4ac < 0$ , the roots are distinct but not real, they are complex.