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How do you solve $7{x^2} - 3x = 2$ using the quadratic formula?

seo-qna
Last updated date: 20th Jun 2024
Total views: 386.1k
Views today: 4.86k
Answer
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Hint: We will first write the given equation in the form of a general quadratic equation and then write the formula of finding the roots of a quadratic and use it.

Complete step-by-step answer:
We are given that we need to solve the equation $7{x^2} - 3x = 2$ using the quadratic formula.
The general quadratic equation is of the form: $a{x^2} + bx + c = 0$ and its roots are given by the formulas: $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ where D is the discriminant and is given by $D = {b^2} - 4ac$.
We can write the given equation $7{x^2} - 3x = 2$ as following:-
$ \Rightarrow 7{x^2} - 3x - 2 = 0$
Comparing it to the general quadratic equation, we will get: a = 7, b = -3 and c = -2.
Now, to get the roots, let us find the roots by putting the above mentioned values in the formulas mentioned above. So, we will get the discriminant as $D = {\left( { - 3} \right)^2} - 4 \times 7 \times \left( { - 2} \right)$.
Simplifying the calculation by opening the required square on the right hand side to obtain the following:-
$ \Rightarrow D = 9 - 4 \times 7 \times \left( { - 2} \right)$
Simplifying the calculations further to obtain the following expression:-
$ \Rightarrow $D = 9 + 56
Simplifying the adding the numbers in the right hand side to obtain:-
$ \Rightarrow $D = 65
Now, let us put this in the formula: $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$, we will then get:-
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {65} }}{{2a}}$
Now putting a = 7, b = -3 and c = -2 in the above expression, we will then obtain expression:-
$ \Rightarrow x = \dfrac{{ - \left( { - 3} \right) \pm \sqrt {65} }}{{2\left( 7 \right)}}$
Simplifying the calculations by taking – ( - a ) = a in the above expression, we will then obtain:-
$ \Rightarrow x = \dfrac{{3 \pm \sqrt {65} }}{{14}}$

Hence, the roots are $x = \dfrac{{3 + \sqrt {65} }}{{14}}$ and $x = \dfrac{{3 - \sqrt {65} }}{{14}}$.

Note:
The students must commit to the memory that:
The general quadratic equation is of the form: $a{x^2} + bx + c = 0$ and its roots are given by the formulas: $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ where D is the discriminant and is given by $D = {b^2} - 4ac$.
The roots of the equation depends on the value of D as well. If D > 0, we have real and distinct roots, if D = 0, we have real and equal roots and if D < 0 we get imaginary roots which exist in conjugate pairs.