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How do you solve using the quadratic formula \[3{x^2} - 2x - 5 = 0?\]

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Last updated date: 29th Feb 2024
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IVSAT 2024
Answer
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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: \[2\]
\[
x = \dfrac{{2 - 8}}{6} \\
x = \dfrac{{ - 6}}{6} \\
x = - 1 \\
\]
So, the final answer is
\[x = - 1\]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 \[x\].
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