Question

How do you solve $ - {x^2} - 3x + 5 = 0$ using the quadratic formula?

Answer 1

Hint: The given problem requires us to solve a quadratic equation $ - {x^2} - 3x + 5 = 0$ using the quadratic formula. There are various methods that can be employed to solve a quadratic equation like completing the square method, using quadratic formula and by splitting the middle term. Using the quadratic formula gives us the roots of the equation directly with ease. So, we will first compare the given equation with the standard form of the quadratic equation and put the values known quantities of in the formula.

Complete step by step answer:
In the given question, we are required to solve the equation $ - {x^2} - 3x + 5 = 0$ with the help of a quadratic formula. The quadratic formula can be employed for solving an equation only if we compare the given equation with the standard form of quadratic equation.Comparing with standard quadratic equation $a{x^2} + bx + c = 0$
Here,$a = - 1$, $b = - 3$ and $c = 5$.
Now, Using the quadratic formula, we get the roots of the equation as:
$x = \dfrac{{( - b) \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Substituting the values of a, b, and c in the quadratic formula, we get,
$x = \dfrac{{ - \left( { - 3} \right) \pm \sqrt {{{( - 3)}^2} - 4 \times \left( { - 1} \right) \times \left( 5 \right)} }}{{2 \times \left( { - 1} \right)}}$
Simplifying the calculations, we get,
$ \Rightarrow x = \dfrac{{3 \pm \sqrt {9 + 20} }}{{ - 2}}$
Adding like terms under the square root, we get,
$ \Rightarrow x = \dfrac{{3 \pm \sqrt {29} }}{{ - 2}}$
Separating the denominators, we get,
$ \Rightarrow x = - \dfrac{3}{2} \pm \dfrac{{\sqrt {29} }}{2}$
So, $x = - \dfrac{3}{2} + \dfrac{{\sqrt {29} }}{2}$ and $x = - \dfrac{3}{2} - \dfrac{{\sqrt {29} }}{2}$ are the roots of the equation $ - {x^2} - 3x + 5 = 0$.

So, the roots of the equation $ - {x^2} - 3x + 5 = 0$ are: $x = - \dfrac{3}{2} + \dfrac{{\sqrt {29} }}{2}$ and $x = - \dfrac{3}{2} - \dfrac{{\sqrt {29} }}{2}$.

Note: Quadratic equations are the polynomial equations with degree of the variable or unknown as 2. Quadratic equations may also be solved by a hit and trial method if the roots of the equation are easy to find. If one root of a quadratic equation with rational coefficients is irrational, then the other root will also be the conjugate pair of the first irrational root. For example: If one root is $a + \sqrt b $, then the root will be $a - \sqrt b $.