Question

How do you solve $9{x^2} = 25$ using the quadratic formula?

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. The quadratic formula to solve for such an equation is given by, $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.

Complete solution step by step:
We have to solve the given equation $9{x^2} = 25$ 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.
$
9{x^2} = 25 \\
\Rightarrow 9{x^2} - 25 = 0 \\
$
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
Coefficient $a$ of ${x^2}$ is $9$,
Co-efficient $b$ of $x$ is $0$,
and the constant term $c$ is $ - 25$.
Thus, $a = 9$, $b = 0$and $c = - 25$.
Now we put the values of $a$, $b$ and $c$ in the quadratic formula to solve for the value of $x$.
\[
x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\
\Rightarrow x = \dfrac{{ - 0 \pm \sqrt {{0^2} - 4 \times 9 \times ( - 25)} }}{{2 \times 9}} \\
\Rightarrow x = \dfrac{{ \pm \sqrt { - 36 \times ( - 25)} }}{{18}} \\
\Rightarrow x = \dfrac{{ \pm \sqrt {900} }}{{18}} \\
\]
Since square root of $900$ is $30$, we write the above equation as:
\[ \Rightarrow x = \dfrac{{ \pm 30}}{{18}}\]
In simplified form,
$x = \dfrac{{30}}{{18}}or\dfrac{{ - 30}}{{18}}$
Dividing the numerator and denominator by common factor $6$,
$x = \dfrac{5}{3}or\dfrac{{ - 5}}{3}$.
Thus, the two values that we get on solving the given equation by quadratic formula are $\dfrac{5}{3}$ and $\dfrac{{ - 5}}{3}$.
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.