Question

How do you solve $9{x^2} - 5x = 7$?

Answer 1

Hint: 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 quadratic formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step by step solution:
We have to solve the given equation $9{x^2} - 5x = 7$ using the quadratic formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.
We rearrange the given equation to put all the terms in the LHS so that we have $0$ in the RHS.
$
  9{x^2} - 5x = 7 \\
   \Rightarrow 9{x^2} - 5x - 7 = 0 \\
$
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 compare it with the general form of the quadratic equation given by $a{x^2} + bx + c = 0$
On comparing the above rearranged equation with the general form, we observe that
Coefficient $a$ of ${x^2}$ is $9$,
Coefficient $b$ of $x$ is $ - 5$,
and the constant term $c$ is $ - 7$.
Thus, $a = 9$, $b = - 5$and $c = - 7$.
Now we put the values of $a$, $b$ and $c$ in the quadratic formula to solve for value of $x$.
\[
  x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\
   \Rightarrow x = \dfrac{{ - ( - 5) \pm \sqrt {{{( - 5)}^2} - 4 \times 9 \times ( - 7)} }}{{2 \times 9}} \\
   \Rightarrow x = \dfrac{{ + 5 \pm \sqrt {25 - ( - 252)} }}{{18}} \\
   \Rightarrow x = \dfrac{{ + 5 \pm \sqrt {25 + 252} }}{{18}} \\
   \Rightarrow x = \dfrac{{ + 5 \pm \sqrt {277} }}{{18}} \\
\]
In simplified form,
\[x = \dfrac{{5 + \sqrt {277} }}{{18}}or\dfrac{{5 - \sqrt {277} }}{{18}}\]
Thus, the two values of $x$that we get on solving the given equation are \[\dfrac{{5 + \sqrt {277} }}{{18}}\] and \[\dfrac{{5 - \sqrt {277} }}{{18}}\].

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.