Answer
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Hint- Here, we will proceed with the help of factorization as well as discriminant formula to solve for the two roots of the given quadratic equation. For discriminant method we will apply the general formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ for any quadratic equation $a{x^2} + bx + c = 0$.
Complete step-by-step answer:
Given quadratic equation in variable x is ${x^2} + 7x + 12 = 0{\text{ }} \to {\text{(1)}}$
Here, we can solve this quadratic equation with the help of factorization method. The given quadratic equation can be written as
$ \Rightarrow {x^2} + 7x + 12 = 0 \Rightarrow {x^2} + 3x + 4x + 12 = 0 \Rightarrow x\left( {x + 3} \right) + 4\left( {x + 3} \right) = 0 \Rightarrow \left( {x + 3} \right)\left( {x + 4} \right) = 0$
Either $
\left( {x + 3} \right) = 0 \\
\Rightarrow x = - 3 \\
$ or $
\left( {x + 4} \right) = 0 \\
\Rightarrow x = - 4 \\
$
Hence, the two roots of the given quadratic equation are -3 and -4.
We can also solve the given quadratic equation by using the discriminant method.
For any general quadratic equation $a{x^2} + bx + c = 0{\text{ }} \to {\text{(2)}}$
According to discriminant method, the roots of this quadratic equation is given by
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}{\text{ }} \to {\text{(3)}}$
By comparing equations (1) and (2), we get
a=1, b=7 and c=12
Using the formula given by equation (3), the roots of the given quadratic equation are given by
\[x = \dfrac{{ - 7 \pm \sqrt {{{\left( 7 \right)}^2} - 4 \times 1 \times 12} }}{{2 \times 1}} = \dfrac{{ - 7 \pm \sqrt {49 - 48} }}{2} = \dfrac{{ - 7 \pm \sqrt 1 }}{2} = \dfrac{{ - 7 \pm 1}}{2}\]
Either \[x = \dfrac{{ - 7 + 1}}{2} = \dfrac{{ - 6}}{2} = - 3\] or \[x = \dfrac{{ - 7 - 1}}{2} = \dfrac{{ - 8}}{2} = - 4\]
So, the two roots of the given quadratic equation are -3 and -4.
Clearly, we are getting the same results from both factorization method and discriminant method.
Note- In these types of problems, we can use either factorization method or discriminant method to obtain the roots of the given quadratic equation. But discriminant method is usually adopted because it is an easier method as compared to factorization method for finding the roots of the given quadratic equation.
Complete step-by-step answer:
Given quadratic equation in variable x is ${x^2} + 7x + 12 = 0{\text{ }} \to {\text{(1)}}$
Here, we can solve this quadratic equation with the help of factorization method. The given quadratic equation can be written as
$ \Rightarrow {x^2} + 7x + 12 = 0 \Rightarrow {x^2} + 3x + 4x + 12 = 0 \Rightarrow x\left( {x + 3} \right) + 4\left( {x + 3} \right) = 0 \Rightarrow \left( {x + 3} \right)\left( {x + 4} \right) = 0$
Either $
\left( {x + 3} \right) = 0 \\
\Rightarrow x = - 3 \\
$ or $
\left( {x + 4} \right) = 0 \\
\Rightarrow x = - 4 \\
$
Hence, the two roots of the given quadratic equation are -3 and -4.
We can also solve the given quadratic equation by using the discriminant method.
For any general quadratic equation $a{x^2} + bx + c = 0{\text{ }} \to {\text{(2)}}$
According to discriminant method, the roots of this quadratic equation is given by
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}{\text{ }} \to {\text{(3)}}$
By comparing equations (1) and (2), we get
a=1, b=7 and c=12
Using the formula given by equation (3), the roots of the given quadratic equation are given by
\[x = \dfrac{{ - 7 \pm \sqrt {{{\left( 7 \right)}^2} - 4 \times 1 \times 12} }}{{2 \times 1}} = \dfrac{{ - 7 \pm \sqrt {49 - 48} }}{2} = \dfrac{{ - 7 \pm \sqrt 1 }}{2} = \dfrac{{ - 7 \pm 1}}{2}\]
Either \[x = \dfrac{{ - 7 + 1}}{2} = \dfrac{{ - 6}}{2} = - 3\] or \[x = \dfrac{{ - 7 - 1}}{2} = \dfrac{{ - 8}}{2} = - 4\]
So, the two roots of the given quadratic equation are -3 and -4.
Clearly, we are getting the same results from both factorization method and discriminant method.
Note- In these types of problems, we can use either factorization method or discriminant method to obtain the roots of the given quadratic equation. But discriminant method is usually adopted because it is an easier method as compared to factorization method for finding the roots of the given quadratic equation.
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