Question

How do you solve \[2{x^2} + 6x = 0\] ?

Answer 1

Hint: The given equation in the form of quadratic equation, you have 2 different methods to solve this equation. Firstly, by taking \[2x\] common in the given equation, you can also solve this by using the formula \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] . By further simplification, you get the required result.

Complete step-by-step answer:
Method:1
Consider given equation
 \[ \Rightarrow \,\,\,2{x^2} + 6x = 0\]
Take out the greatest common factor GCF, then
 \[ \Rightarrow \,\,\,2x\left( {x + 3} \right) = 0\]
Equating each term of the above equation to zero
 \[ \Rightarrow \,\,\,2x = 0\] or \[\left( {x + 3} \right) = 0\]
On simplification we get
 \[ \Rightarrow \,\,\,x = \dfrac{0}{2}\] or \[x = - 3\]
The roots are
 \[\therefore \,\,\,\,\,x = 0\] or \[x = - 3\]

Method: 2
Also, solve this problem by using the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] . The standard quadratic equation is in the form \[a{x^2} + bx + c = 0\]
Consider the given quadratic equation \[2{x^2} + 6x = 0\] . When we compare the given equation to the standard form of the quadratic equation
Here a=2, b=6 and c=0 then substituting the values to the formula it is written as
 \[ \Rightarrow \,\,\,\,x = \,\,\dfrac{{ - 6 \pm \sqrt {{6^2} - 4\left( 2 \right)\left( 0 \right)} }}{{2\times 2}}\]
On simplification we have
 \[ \Rightarrow \,\,\,\,x = \,\,\dfrac{{ - 6 \pm \sqrt {36 - 0} }}{4}\]
On simplifying we get
 \[ \Rightarrow \,\,\,\,x = \,\,\dfrac{{ - 6 \pm \sqrt {36} }}{4}\]
The number 36 is a perfect square. As we know 36 is the square of 6 i.e., \[{6^2} = 36\] , then
 \[ \Rightarrow \,\,\,\,x = \dfrac{{ - 6 \pm \sqrt {{6^2}} }}{4}\]
The square and square root will cancels and we get
 \[ \Rightarrow \,\,\,\,x = \dfrac{{ - 6 \pm 6}}{4}\]
Let we write two terms separately, so we have
 \[ \Rightarrow \,\,\,\,x = \dfrac{{ - 6 + 6}}{4}\] or \[x = \dfrac{{ - 6 - 6}}{4}\]
On simplifying we get
 \[ \Rightarrow \,\,\,\,x = \dfrac{0}{4}\] or \[x = \dfrac{{ - 12}}{4}\]
 \[\therefore \,\,\,\,\,x = 0\] or \[x = - 3\]
So, the correct answer is “ \[x = 0\] OR \[x = - 3\] ”.

Note: The equation is a quadratic equation. The general form of quadratic equation is \[a{x^2} + bx + c\] We can solve the equation by using the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] . Sometimes the factorisation method is not applicable to solve the quadratic equation. Because sometimes the root values will be imaginary forms.