Question

How do you solve $6{x^2} + 3x = 0$?

Answer 1

Factoring reduces the higher degree equation into its reduced equation. In the above given equation we need to reduce the quadratic equation either solving this by factorization or using algebraic identity where the square of addition of terms will give the factors from the equation.

Complete step by step solution:
The above equation is a quadratic equation which includes at least one monomial that has the highest power of 2. But we have only two terms one with power of 2 and other with power of 1, the third term is not present.

We can generate the third term by using the algebraic identity which is given as:
${(a + b)^2} = {a^2} + 2ab + {b^2}$

So, we need to convert and bring the equation in the above form. First thing we need to do is get the terms a and b.

So, we will divide the above equation with 6 we will get,
${x^2} + \dfrac{1}{2}x = 0$

We will generate a new term which on getting cancelled will give the above equation.
${\left( {{x^{}} + \dfrac{1}{4}} \right)^2} - \dfrac{1}{{16}} = 0$

On expanding the terms
${x^2} + 2 \times \dfrac{1}{4} \times x + \dfrac{1}{{16}} - \dfrac{1}{{16}}$

The last two terms get cancelled. We get the equation which was divided by 6. So further on finding the factor we get,
${\left( {x + \dfrac{1}{4}} \right)^2} = \dfrac{1}{{16}}$

Now taking square root on both sides,
$\left( {x + \dfrac{1}{4}} \right) = \pm \dfrac{1}{4}$

Considering positive value, we get
$x = \dfrac{1}{4} - \dfrac{1}{4} = 0$

And value is negative we get,
$x = - \dfrac{1}{4} - \dfrac{1}{4} = - \dfrac{1}{2}$

Therefore, the values we get are $0$or $- \dfrac{1}{2}$.

Note: An alternative method to solve this equation can be pulling out common factors from the equation and then equating the individual terms with zero so that we can get the factors. By writing $3x \times (2x + 1) = 0$we get $3x = 0$ and$2x = - 1$.

Therefore, we get values as $0$ and $- \dfrac{1}{2}$.