Question

How do you find the zeros of $3{x^2} - 10x + 3 = 0$?

Answer 1

Hint: This problem deals with solving a quadratic equation. Here, given a quadratic equation expression, we have to simplify the expression and make it into a standard form of quadratic equation. If the quadratic equation is in the form of $a{x^2} + bx + c = 0$, then we know that the roots of this quadratic equation are given by :
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step-by-step solution:
Given a quadratic equation which is $3{x^2} - 10x + 3 = 0$.
To solve a quadratic expression is the same as finding the roots of the quadratic equation.
Consider the given quadratic equation, as given below:
$ \Rightarrow 3{x^2} - 10x + 3 = 0$
The roots of the above quadratic equation are obtained by applying the formula of roots of the quadratic equation, as given below:
$ \Rightarrow x = \dfrac{{ - \left( { - 10} \right) \pm \sqrt {{{\left( {10} \right)}^2} - 4\left( 3 \right)\left( 3 \right)} }}{{2\left( 3 \right)}}$
$ \Rightarrow x = \dfrac{{10 \pm \sqrt {100 - 36} }}{6}$
Simplifying the above expression inside the under root, gives the following expression:
$ \Rightarrow x = \dfrac{{10 \pm \sqrt {64} }}{6}$
$ \Rightarrow x = \dfrac{{10 \pm 8}}{6}$
We know that the square root of 64 is 8, which is $\sqrt {64} = 8$, substituting this in the equation above.
$ \Rightarrow x = \dfrac{{10 + 8}}{6}$ and $x = \dfrac{{10 - 8}}{6}$
Simplifying the above roots as shown below:
$ \Rightarrow x = \dfrac{{18}}{6}$ and $x = \dfrac{2}{6}$
$\therefore x = 3$ and $x = \dfrac{1}{3}$
Hence the roots of the quadratic equation are distinct and are given by:
The values of $x$ are $3$ and $\dfrac{1}{3}$.

The solutions of the quadratic equation $3{x^2} - 10x + 3 = 0$ are $3$ and $\dfrac{1}{3}$.

Note: Note that in any quadratic equation $a{x^2} + bx + c = 0$, here in this ${b^2} - 4ac$ is called as the determinant. There are 3 general cases of the discriminant.
If ${b^2} - 4ac > 0$, then the quadratic equation has real and distinct roots.
If ${b^2} - 4ac = 0$, then the quadratic equation has real and equal roots.
If ${b^2} - 4ac < 0$, then the quadratic equation has no real roots but has complex conjugate roots.