Question

How do you solve $9{x^2} = 4 + 7x$ using the quadratic formula?

Answer 1

Hint: The given problem requires us to solve a quadratic equation using 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 $9{x^2} = 4 + 7x$ with the help of a quadratic formula. So, first taking all the terms to left side of equation, we get,
$ \Rightarrow 9{x^2} - 7x - 4 = 0$
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,$a = 9$, $b = - 7$ and$c = - 4$.

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,
$x = \dfrac{{ - \left( { - 7} \right) \pm \sqrt {{{( - 7)}^2} - 4 \times 9 \times \left( { - 4} \right)} }}{{2 \times 9}}$
Computing the square of the term, we get,
$ \Rightarrow x = \dfrac{{7 \pm \sqrt {49 + 144} }}{{2 \times 9}}$
$ \Rightarrow x = \dfrac{{7 \pm \sqrt {193} }}{{2 \times 9}}$
Simplifying the expression, we get,
$ \Rightarrow x = \dfrac{{7 \pm \sqrt {193} }}{{18}}$
Separating the denominators for both the terms, we get,
$ \therefore x = \dfrac{7}{{18}} \pm \dfrac{{\sqrt {193} }}{{18}}$
So, $x = \dfrac{7}{{18}} + \dfrac{{\sqrt {193} }}{{18}}$ and $x = \dfrac{7}{{18}} - \dfrac{{\sqrt {193} }}{{18}}$ are the roots of the equation $9{x^2} = 4 + 7x$.

So, the roots of the equation $9{x^2} = 4 + 7x$ are: $x = \dfrac{7}{{18}} + \dfrac{{\sqrt {193} }}{{18}}$ and $x = \dfrac{7}{{18}} - \dfrac{{\sqrt {193} }}{{18}}$.

Note: Quadratic equations are the polynomial equations with degree of the variable or unknown as 2. Complex roots always occur in pairs. So, a quadratic equation can have either two real roots or two complex roots. If a quadratic equation has one complex root as $\left( {a + ib} \right)$, then it also has $\left( {a - ib} \right)$ as a root.