Question

How do you solve \[3{{q}^{2}}-16q=-5\]?

Answer 1

Hint: The degree of an equation is the highest power to which the variable is raised. We can decide if the equation is linear, quadratic, cubic, etc. using the degree of the equation. For a quadratic equation \[a{{x}^{2}}+bx+c=0\], using the formula method we can find the roots of the equation as \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].

Complete step-by-step answer:
The given equation is \[3{{q}^{2}}-16q=-5\]. The highest power of the variable in the equation is 2, so the degree of the equation is also 2. It means that the equation is quadratic. To solve the equation, we first have to express it in its standard form, we can do it as, adding 5 to both sides of the given equation, it can be written as \[3{{q}^{2}}-16q+5=0\]. Comparing with the general equation of the quadratic \[a{{x}^{2}}+bx+c=0\], we get \[a=3,b=-16\And c=5\].
We know that using the formula method, we can find the roots of the quadratic as \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]. As the variable for this equation is \[q\], we need to use the formula in terms of \[q\]. Substituting the value of the coefficient in this formula, we get
\[\begin{align}
  & \Rightarrow q=\dfrac{-\left( -16 \right)\pm \sqrt{{{\left( -16 \right)}^{2}}-4\left( 3 \right)\left( 5 \right)}}{2\left( 3 \right)} \\
 & \Rightarrow q=\dfrac{16\pm \sqrt{196}}{6} \\
 & \Rightarrow q=\dfrac{16\pm 14}{6} \\
\end{align}\]
\[\Rightarrow q=\dfrac{16+14}{6}\] or \[q=\dfrac{16-14}{6}\]
\[\Rightarrow q=\dfrac{30}{6}=5\] or \[q=\dfrac{2}{6}=\dfrac{1}{3}\]
Hence, the roots of the equation are \[q=5\] or \[q=\dfrac{1}{3}\].

Note: There are many methods to solve a quadratic equation, as the factorization method, completing the square method, and formula method. We can use any of them to solve a quadratic equation. The formula method should be preferred because it gives two roots of the equation, whether they are real or imaginary.