Question

How do you solve $3{x^2} - 16x + 5 = 0$?

Answer 1

Hint:The general form of the quadratic equation is $a{x^2} + bx + c = 0$ and according to the quadratic formula the value of the variable can be given by $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ . Compare the given equation in the question with the general form and find the values for constants ‘a’, ‘b’, and ‘c’. Now use the quadratic formula to find the solution.

Complete step by step answer:
Here in this problem, we are given a Quadratic equation, i.e. $3{x^2} - 16x + 5 = 0$ and we need to solve this equation to find its root.
Before starting with the solution to the problem, we must understand a few concepts regarding Quadratic equations. Quadratics or quadratic equations can be defined as a polynomial equation of a second degree, which implies that it comprises a minimum of one term that is squared. The solutions to the quadratic equation are the values of the unknown variable $x$, which satisfy the equation. These solutions are called roots or zeros of quadratic equations.
It is in form $a{x^2} + bx + c = 0$ , where $'a',{\text{ }}'b'{\text{ and }}'c'$ are constant in the equation. These coefficients ‘a’ and ‘b’ play an important role in the solution of an equation.
According to the Quadratic formula, the equation of the form $a{x^2} + bx + c = 0$ can be solved as:
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ , which represents the two roots of the equation or possible values of ‘x’ that satisfies the equation.
Let’s compare the given equation in this problem, i.e. $3{x^2} - 16x + 5 = 0$ with the general form of the equation, i.e. $a{x^2} + bx + c = 0$.
$ \Rightarrow {\text{We get: }}a = 3,b = - 16,c = 5$
These values of the constants can be substituted in the Quadratic formula as:
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} = \dfrac{{ - \left( { - 16} \right) \pm \sqrt {{{\left( { - 16} \right)}^2} - 4 \times 3 \times 5} }}{{2 \times 3}}$
Now the above expression can be solved to get the roots of the equation as:
 $ \Rightarrow x = \dfrac{{ - \left( { - 16} \right) \pm \sqrt {{{\left( { - 16} \right)}^2} - 4 \times 3 \times 5} }}{{2 \times 3}} = \dfrac{{16 \pm \sqrt {256 - 60} }}{6}$
We can simplify the radicand first as:
$ \Rightarrow x = \dfrac{{16 \pm \sqrt {256 - 60} }}{6} = \dfrac{{16 \pm \sqrt {196} }}{6} = \dfrac{{16 \pm 14}}{6}$
So we can get the two roots of the value of $'x'$ by using the plus and minus separately
$ \Rightarrow x = \dfrac{{16 + 14}}{6}{\text{ = }}\dfrac{{30}}{6}{\text{ and }}x = \dfrac{{16 - 14}}{6} = \dfrac{2}{6}$
Therefore, we get the values as:
$ \Rightarrow x = 5{\text{ or }}x = \dfrac{1}{3}$

Thus, the given equation $3{x^2} - 16x + 5 = 0$ has the solution $x = 5{\text{ or }}x = \dfrac{1}{3}$.

Note: In this question, the use of the quadratic formula is a crucial part of the solution. We obtained two values for the variable $'x'$ . The answer can be cross-checked by substituting the value of the variable again in the LHS, i.e. ${\text{ For }}x = 5,{\text{ }}3 \times {\left( 5 \right)^2} - 16 \times 5 + 5 = 75 - 80 + 5 = 0 = RHS$ and for $x = \dfrac{1}{3}$ , $3{\left( {\dfrac{1}{3}} \right)^2} - 16 \times \dfrac{1}{3} + 5 = \dfrac{1}{3} - \dfrac{{16}}{3} + 5 = - 5 + 5 = 0 = RHS$ . So our solution is thus verified.