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How do you solve \[{x^2} + x - 12 = 0\]using the quadratic formula?

seo-qna
Last updated date: 12th Jul 2024
Total views: 380.7k
Views today: 4.80k
Answer
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380.7k+ views
Hint: This question involves the arithmetic operations like 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 values of basic numbers. We have the term \[{x^2}\] in the question, so we would find two values \[x\] by solving the given equation.

Complete step-by-step answer:
The given equation is shown below,
\[{x^2} + x - 12 = 0 \to \left( 1 \right)\]
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 given 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\[c\].
\[\left( 1 \right) \to {x^2} + x - 12 = 0\]
\[\left( 2 \right) \to a{x^2} + bx + c = 0\]
So, we get the value of \[a\] is \[1\], the value of \[b\] is \[1\] , and the value of \[c\] is \[ - 12\]. Let’s substitute these values in the equation\[\left( 3 \right)\], we get
\[\left( 3 \right) \to x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
\[x = \dfrac{{ - 1 \pm \sqrt {{{\left( 1 \right)}^2} - 4 \times 1 \times - 12} }}{{2 \times 1}}\]
\[
  x = \dfrac{{ - 1 \pm \sqrt {1 + 48} }}{2} \\
  x = \dfrac{{ - 1 \pm \sqrt {49} }}{2} \\
 \]
We know that \[{7^2} = 49\]. So, the above equation can also be written as,
\[x = \dfrac{{ - 1 \pm 7}}{2}\]
Case: \[1\]
\[x = \dfrac{{ - 1 + 7}}{2} = \dfrac{6}{2}\]
\[x = 3\]
Case: \[2\]
\[x = \dfrac{{ - 1 - 7}}{2} = \dfrac{{ - 8}}{2}\]
\[x = - 4\]
So, the final answer is,
\[x = 3\]and \[x = - 4\]

Note: This type of questions involves the arithmetic operation like addition/ subtraction/ multiplication/ division. Note that the denominator value would not be equal to zero. When\[{n^2}\]is placed inside the square root we can cancel the square and square root of each other. If \[ \pm \]is present in the calculation we would find two values\[x\]. Also, note that if \[{x^2}\]is present in the given equation in the question it must have two factors for the equation. Also, note that if \[ - \]is present inside the root we would put\[j\]it with the answer.