Question

How do you solve quadratic $3{x^2} - 2x = 2x + 7$ using any method?

Answer 1

Hint: This problem deals with solving a quadratic equation. Here, given a quadratic equation expression, we have to simplify the expression and make it into a standard form of quadratic equation. If the quadratic equation is in the form of $a{x^2} + bx + c = 0$, then we know that the roots of this quadratic equation are given by :
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step-by-step solution:
The given expression is a quadratic expression but not a quadratic equation. We have to convert the expression into an equation and then solve the quadratic equation.
The given expression is $3{x^2} - 2x = 2x + 7$, consider it as given below:
$ \Rightarrow 3{x^2} - 2x = 2x + 7$
Bringing all the terms on to the left hand side of the equation, as given below:
$ \Rightarrow 3{x^2} - 2x - 2x - 7 = 0$
$ \Rightarrow 3{x^2} - 4x - 7 = 0$
Now this is in the standard form of a quadratic equation, now comparing the coefficients $a,b$ and $c$:
$ \Rightarrow a = 3,b = - 4$ and $c = - 7$
Now applying the formula to find the value of the roots of $x$, as given below:
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Substituting the values of $a,b$ and $c$ in the above formula:
$ \Rightarrow x = \dfrac{{ - \left( { - 4} \right) \pm \sqrt {{{\left( { - 4} \right)}^2} - 4\left( 3 \right)\left( { - 7} \right)} }}{{2\left( 3 \right)}}$
$ \Rightarrow x = \dfrac{{4 \pm \sqrt {16 + 84} }}{6}$
$ \Rightarrow x = \dfrac{{4 \pm \sqrt {100} }}{6}$
We know that the square root of 100 is 10, $\sqrt {100} = 10$
$ \Rightarrow x = \dfrac{{4 \pm 10}}{6}$
Now considering the two cases, with plus and minus, as shown:
$ \Rightarrow x = \dfrac{{4 + 10}}{6};x = \dfrac{{4 - 10}}{6}$
$ \Rightarrow x = \dfrac{{14}}{6};x = \dfrac{{ - 6}}{6}$
Hence the value of the roots are equal to :
$\therefore x = \dfrac{7}{3};x = - 1$

The values of $x$ are $\dfrac{7}{3}$ and $-1$ respectively.

Note: Please note that for any given quadratic equation $a{x^2} + bx + c = 0$, here ${b^2} - 4ac$ is called the discriminant of the quadratic equation. Usually there are 3 general cases of the discriminant.
If the discriminant is greater than zero, ${b^2} - 4ac > 0$, then the quadratic equation has real and distinct roots.
If the discriminant is equal to zero, ${b^2} - 4ac = 0$, then the quadratic equation has real and equal roots.
If the discriminant is less than zero, ${b^2} - 4ac < 0$, then the quadratic equation has no real roots but has complex conjugate roots.