Question

How do you solve by factoring \[2{x^2} + 7x + 3 = 0\] ?

Answer 1

Hint: Here we have to find the factors of the given equation. So we will use the method in which we will make the coefficient of first term one and then we will find a pair of factors such that their product is third term of the equation and their sum is second term. Then we will write the equation in that form of two numbers and proceed with the similar step of our regular equation.

Complete step-by-step answer:
Given the equation
 \[2{x^2} + 7x + 3 = 0\]
Now to make the coefficient of first term 1 we will divide both sides by 2.
 \[{x^2} + \dfrac{7}{2}x + \dfrac{3}{2} = 0\]
This is our equation.
Now we will find a pair of numbers such that their product is third term of the equation and their sum is second term.
If we split \[\dfrac{3}{2}\] as \[3\& \dfrac{1}{2}\] we can say that their sum is also \[\dfrac{7}{2}\] . So we have found the pair now. We will solve for factors.
 \[{x^2} + \dfrac{1}{2}x + 3x + \dfrac{3}{2} = 0\]
Now taking \[x\] common from first two terms and \[3\] common from last two terms we get,
 \[x\left( {x + 3} \right) + \dfrac{1}{2}\left( {x + 3} \right) = 0\]
Now taking the brackets separately,
 \[\left( {x + 3} \right)\left( {x + \dfrac{1}{2}} \right) = 0\]
Equating them one by one to zero,
 \[\left( {x + 3} \right) = 0\& \left( {x + \dfrac{1}{2}} \right) = 0\]
Taking constants on other side,
 \[x = - 3 \; and \; x = - \dfrac{1}{2}\]
These are the factors of the given equation.
So, the correct answer is “ \[x = - 3\ \; and \; x = - \dfrac{1}{2}\] ”.

Note: Note that we can solve the same problem using quadratic formula \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] . In that we need not to divide the equation by 2 because we will just compare the given equation with the general equation and put the values in the quadratic equation formula. On solving it we will get the factors or roots of the equation.