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# How do you solve using the quadratic formula $3{x^2} - 2x - 5 = 0?$

Last updated date: 29th Feb 2024
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Hint:This question involves the operation of addition/ subtraction/ multiplication/ division. Also, we need to know the basic form of a quadratic equation and the formula to find the value of $x$ in a quadratic equation. We need to know the square root value of basic numbers. We have ${x^2}$ terms in the given equation so, we would find two answers $x$ by solving the given equation.

Complete step by step solution:
The given question is shown below,
$3{x^2} - 2x - 5 = 0 \to \left( 1 \right)$
We know that the basic form of a quadratic equation is,
$a{x^2} + bx + c = 0 \to \left( 2 \right)$
The formula for finding the value $x$ from the above equation is shown below,
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \to \left( 3 \right)$
By comparing the equation $\left( 1 \right)$and$\left( 2 \right)$, we get the value
of $a,b$and$c$ as follows,
$\left( 1 \right) \to 3{x^2} - 2x - 5 = 0$
$\left( 2 \right) \to a{x^2} + bx + c = 0$
So, we get the value of $a$ is $3$, the value of $b$is$- 2$, and the value of $c$is$- 5$. Let’s substitute these values in the equation $\left( 3 \right)$, we get
$\left( 3 \right) \to x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
$x = \dfrac{{ - \left( { - 2} \right) \pm \sqrt {{{\left( { - 2} \right)}^2} - 4 \times 3 \times - 5} }}{{2 \times 3}}$
So, we get
$x = \dfrac{{2 \pm \sqrt {4 + 60} }}{6} \\ x = \dfrac{{2 \pm \sqrt {64} }}{6} \\$
We know that, ${8^2} = 64$. So, the above equation can also be written as,
$x = \dfrac{{2 \pm \sqrt {{8^2}} }}{6} \\ x = \dfrac{{2 \pm 8}}{6} \\$
So, the above equation can be solved by using two cases.
Case: $1$
$x = \dfrac{{2 + 8}}{6} \\ x = \dfrac{{10}}{6} \\ x = \dfrac{5}{3} \\$
Case: $2$
$x = \dfrac{{2 - 8}}{6} \\ x = \dfrac{{ - 6}}{6} \\ x = - 1 \\$
$x = - 1$and$x = \dfrac{5}{3}$.
Note:This type of question involves the operation of addition/ subtraction/ multiplication/ division. Note that the denominator would not be equal to zero. When ${n^2}$ is placed inside the square root we can cancel the square root and square with each other. Note that if $\pm$is present in the equation we would find two values for $x$. Also, remember the basic form of a quadratic equation
and the formula for finding the value $x$.