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# How do you solve $6 - 7x = 3{x^2}$?

Last updated date: 28th Feb 2024
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Hint: First of all we will take all the terms on one side of the equation. Remember when you move any term from one side to another then the sign of the term also changes. Then will compare the simplified equation with the standard quadratic equation or roots can be found by splitting the middle term and then will find the resultant required value for “x”.

Complete step-by-step solution:
Take the given expression: $6 - 7x = 3{x^2}$
The above equation can be re-written as: $3{x^2} = 6 - 7x$
Move all the terms from the right hand side of the equations to the right hand side of the equation. Remember when you move any term from one side of the equation to the other side of the equation then the sign of the term also changes. Positive terms become negative and vice versa.
$3{x^2} + 7x - 6 = 0$
Split the middle term in such a way that its product is equal to the product of the first and the last term.
$+ 7 = 9 - 2 \\ - 18 = 9( - 2) \\$
Now, frame the equation accordingly
$3{x^2} + 9x - 2x - 6 = 0$
Now make the pair of the first two and the last two terms.
$\underline {3{x^2} + 9x} - \underline {2x - 6} = 0$
Find the common factors from the paired terms.
$\Rightarrow 3x(x + 3) - 2(x + 3) = 0$
Take out the common factor common
$\Rightarrow (x + 3)(3x - 2) = 0$
$\Rightarrow x + 3 = 0$or $\Rightarrow 3x - 2 = 0$
Make the subject “x”. term multiplicative on one side if moved to the opposite side then it goes to the denominator.
$\Rightarrow x = 3$or $x = \dfrac{2}{3}$
This is the required solution.

Note: Be careful about the sign convention while splitting the middle term. Here we were able to find the roots by splitting the middle term but in case roots are not real we won’t be able to do so, for that remember imaginary roots formula which can be expressed as $x = \dfrac{{ - b \pm \sqrt \Delta }}{{2a}}$ and considering the general form of the quadratic equation $a{x^2} + bx + c = 0$ . Be careful about the sign convention and simplification of the terms in the equation.