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# How do you solve ${x^2} + x - 12 = 0$using the quadratic formula?

Last updated date: 12th Jul 2024
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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.

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: $2$
$x = \dfrac{{ - 1 - 7}}{2} = \dfrac{{ - 8}}{2}$
$x = - 4$
$x = 3$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.