Question

Find the roots of $3{x^2} + 13x - 10 = 0$.

Answer 1

Hint: Compare the given equation with the general quadratic equation $a{x^2} + bx + c = 0$ to obtain the values of a, b, and c. Substitute these values in the quadratic equation formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ to compute the required roots.

Complete step by step answer:
We are given a quadratic equation $3{x^2} + 13x - 10 = 0$
How do we know this is a quadratic equation? Well, any equation of the form $a{x^2} + bx + c = 0$ where a, b, and c are real numbers, and a is a non-zero constant is called a quadratic equation.

Now, the values of x which will satisfy the quadratic equation will be called the roots of the quadratic equation. Also, a quadratic equation will always have two roots. The roots may be the same or different.
Thus, we are going to locate the two roots of the given quadratic equation $3{x^2} + 13x - 10 = 0$.
To do this, we may use the quadratic equation formula.
The formula for finding the two roots of the quadratic equation $a{x^2} + bx + c = 0$ is as follows:
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Comparing the given equation $3{x^2} + 13x - 10 = 0$ with the general quadratic equation, we have
a = 3, b = 13, and c = -10.
Now, substitute these values in the quadratic equation formula.
This will give us
$
  x = \dfrac{{ - 13 \pm \sqrt {{{13}^2} - 4 \times 3 \times - 10} }}{{2 \times 3}} \\
   \Rightarrow x = \dfrac{{ - 13 \pm \sqrt {169 + 120} }}{6} \\
   \Rightarrow x = \dfrac{{ - 13 \pm \sqrt {289} }}{6} \\
   \Rightarrow x = \dfrac{{ - 13 \pm 17}}{6} \\
 $
Thus, we have $x = \dfrac{{ - 13 + 17}}{6}$ or \[x = \dfrac{{ - 13 - 17}}{6}\]
$ \Rightarrow x = \dfrac{4}{6}$ or \[x = \dfrac{{ - 30}}{6}\]
Reducing the fractions in their lowest form, we get $x = \dfrac{2}{3}$ or \[x = - 5\].
Therefore, $\dfrac{2}{3}$ and \[ - 5\] are the roots of the equation $3{x^2} + 13x - 10 = 0$
We can also verify this by substituting the values for x in $3{x^2} + 13x - 10 = 0$
Let us verify and see.
Consider the left hand side of the equation $3{x^2} + 13x - 10 = 0$
Put $x = \dfrac{2}{3}$, then we have
\[
  3{(\dfrac{2}{3})^2} + 13(\dfrac{2}{3}) - 10 \\
   = 3 \times \dfrac{4}{9} + \dfrac{{26}}{3} - 10 \\
   = \dfrac{4}{3} + \dfrac{{26}}{3} - 10 \\
   = \dfrac{{30}}{3} - 10 \\
   = 10 - 10 \\
   = 0 \\
 \]
Similarly, we can verify for \[x = - 5\]
\[
= 3{( - 5)^2} + 13( - 5) - 10 \\
= 3 \times 25 - 65 - 10 \\
= 75 - 75 \\
= 0 \\
 \]

Note: Alternate method to compute the roots of $3{x^2} + 13x - 10 = 0$
Here, we will use the factorization method
Consider the product of 3 (the power of ${x^2}$) and -10 which is -30.
We need to find two numbers a and b such that their product = ab = -30 and their sum = a + b = 13
The various factorizations of 30 are as follows:
\[
  2 \times 15 \\
  3 \times 10 \\
  5 \times 6 \\
 \]
We can see that the product of 15 and -2 is -30 and their sum is 13.
So, we use these numbers to split the middle term of $3{x^2} + 13x - 10 = 0$ as follows:
\[
  3{x^2} + 13x - 10 = 0 \\
   \Rightarrow 3{x^2} + 15x - 2x - 10 = 0 \\
   \Rightarrow 3x(x + 5) - 2(x + 5) = 0 \\
   \Rightarrow (x + 5)(3x - 2) = 0 \\
 \]
Therefore, we have \[(x + 5) = 0\] or \[(3x - 2) = 0\]
Simplifying further, we get the roots as $\dfrac{2}{3}$ and \[ - 5\].