Question

How do you solve $- 4{x^2} + x + 9 = 0$ using the quadratic formula?

Answer 1

The given equation is a quadratic equation in one variable $x$ . The general form of a quadratic equation is given by $a{x^2} + bx + c = 0$ . Solving this equation gives two values of the variable $x$ as the result. We can solve this equation by using the quadratic formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ .

Complete step by step solution:
We have to solve the given equation $- 4{x^2} + x + 9 = 0$ using the quadratic formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
To find the value of $x$ , we have to put the values of $a$ , $b$ and $c$ in the quadratic formula. To get the values of $a$ , $b$ and $c$ from the given equation, we rearrange the equation and compare it with the general form of the quadratic equation.
General form of quadratic equation is written in the form of $a{x^2} + bx + c = 0$ , where $a$ is the coefficient of ${x^2}$ , $b$ is the coefficient of $x$ and $c$ is the constant term. The RHS is $0$ .
On comparing the above rearranged equation with the general form, we observe that
Co-efficient $a$ of ${x^2}$ is $- 4$ ,
Co-efficient $b$ of $x$ is $1$ ,
and the constant term $c$ is $9$ .
Thus, $a = - 4$ , $b = 1$ and $c = 9$ .
Now we put the values of $a$ , $b$ and $c$ in the quadratic formula to solve for value of $x$ .

$$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\\ \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {{{(1)}^2} - 4 \times ( - 4) \times 9} }}{{2 \times ( - 4)}} \\\ \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {1 - ( - 144)} }}{{ - 8}} \\\ \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {1 + 144} }}{{ - 8}} \\\ \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {145} }}{{ - 8}} \\\$$

Multiplying $- 1$ in the numerator and the denominator, we get:

$$\Rightarrow x = \dfrac{{ - 1 \pm \sqrt {145} }}{{ - 8}} \times \dfrac{{ - 1}}{{ - 1}} \\\ \Rightarrow x = \dfrac{{1 \pm \sqrt {145} }}{8} \\\$$

In simplified form,
$x = \dfrac{{1 + \sqrt {145} }}{8}or\dfrac{{1 - \sqrt {145} }}{8}$
$x = \dfrac{{1 + \sqrt {145} }}{8}$
Thus, the two values of $x$ that we get on solving the given equation are $\dfrac{{1 + \sqrt {145} }}{8}$ and $\dfrac{{1 - \sqrt {145} }}{8}$.

Note: Another method to solve for $x$ in the quadratic equation is by factorization. Using quadratic formula is simpler than factorization as it involves direct calculation using values of $a$ , $b$ and $c$ . We get two values of $x$ while solving the quadratic equation. We can check the answer by putting the result in the given equation to satisfy LHS = RHS.