How do you factor and solve $3{x^2} - 12x = 0$?
Answer
591.9k+ views
Hint: To solve this equation, first using a quadratic formula finds out the coefficients correctly by using the standard form equation. Then substitute those values in the formula and evaluate the exponent, multiply and add the numbers, evaluate the square root. To solve for the unknown variable, separate into two equations: one with a plus and the other with a minus. Then rearrange and isolate the variable to find the solution.
Complete step-by-step solution:
The given equation is $3{x^2} - 12x = 0$
To solve this equation you can use quadratic formula
That is
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Once in standard form, identify a, b and c from the original equation and plug them into the quadratic formula.
Therefore $a = 3,b = - 12$ and $c = 0$
Substituting this value in the formula, we get
$ \Rightarrow x = \dfrac{{ - \left( { - 12} \right) \pm \sqrt {{{\left( { - 12} \right)}^2} - 4\left( 3 \right)\left( 0 \right)} }}{{2\left( 3 \right)}}$
When we evaluate the component
$ \Rightarrow x = \dfrac{{12 \pm \sqrt {144 - 0} }}{6}$
On further reducing the component we get
$ \Rightarrow x = \dfrac{{12 \pm \sqrt {144} }}{6}$
The removing the square root we get,
$ \Rightarrow x = \dfrac{{12 \pm 12}}{6}$
Now find the solutions separately by
Subtracting the components first,
$i)x = \dfrac{{12 - 12}}{6} = \dfrac{0}{6}$
$\therefore x = 0$
Now add the components,
We get
$ii)\,x = \dfrac{{12 + 12}}{6} = \dfrac{{24}}{6}$
$\therefore x = 4$
Therefore on factoring and solving the equation we get solution as
$x = 0$ and $x = 4$
Note: Alternative method to solve this equation $3{x^2} - 12x = 0$
Take $3x$ as common from the equation, we get
$ \Rightarrow 3x\left( {x - 4} \right) = 0$
Now equate the components separately to zero
Then we get
$ \Rightarrow 3x = 0$
$\therefore x = 0$
Next, $x - 4 = 0$
Bringing $4$ to the RHS, We get
$x = 4$
Hence we get the required answer.
Complete step-by-step solution:
The given equation is $3{x^2} - 12x = 0$
To solve this equation you can use quadratic formula
That is
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Once in standard form, identify a, b and c from the original equation and plug them into the quadratic formula.
Therefore $a = 3,b = - 12$ and $c = 0$
Substituting this value in the formula, we get
$ \Rightarrow x = \dfrac{{ - \left( { - 12} \right) \pm \sqrt {{{\left( { - 12} \right)}^2} - 4\left( 3 \right)\left( 0 \right)} }}{{2\left( 3 \right)}}$
When we evaluate the component
$ \Rightarrow x = \dfrac{{12 \pm \sqrt {144 - 0} }}{6}$
On further reducing the component we get
$ \Rightarrow x = \dfrac{{12 \pm \sqrt {144} }}{6}$
The removing the square root we get,
$ \Rightarrow x = \dfrac{{12 \pm 12}}{6}$
Now find the solutions separately by
Subtracting the components first,
$i)x = \dfrac{{12 - 12}}{6} = \dfrac{0}{6}$
$\therefore x = 0$
Now add the components,
We get
$ii)\,x = \dfrac{{12 + 12}}{6} = \dfrac{{24}}{6}$
$\therefore x = 4$
Therefore on factoring and solving the equation we get solution as
$x = 0$ and $x = 4$
Note: Alternative method to solve this equation $3{x^2} - 12x = 0$
Take $3x$ as common from the equation, we get
$ \Rightarrow 3x\left( {x - 4} \right) = 0$
Now equate the components separately to zero
Then we get
$ \Rightarrow 3x = 0$
$\therefore x = 0$
Next, $x - 4 = 0$
Bringing $4$ to the RHS, We get
$x = 4$
Hence we get the required answer.
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