Question

How do you solve $3{{x}^{2}}+7x+4=0$ ?

Answer 1

Hint: The equation given in a quadratic equation is a quadratic equation. We can solve it by method of factorization. In quadratic equation $a{{x}^{2}}+bx+c$ while performing factorization we write bx as $dx+ex$ such that $d+e=b$ and product of d and e is equal to product of a and c, $de=ac$

Complete step by step answer:
The given equation is $3{{x}^{2}}+7x+4=0$
We can see the above equation is a quadratic equation. We can solve it by factoring the equation.
When we solve the equation $a{{x}^{2}}+bx+c$ by factorization we split box into 2 terms such that their sum is bx and product of their coefficient is equal to $ac$
In our equation value of a is 3 , value of b is 7 and value of c is 4
So $ac$ = 12
Now we will list all the pairs whose product is 12 and take the one whose sum is 7.
The pairs whose product is 12 are
(1, 12), (2, 6) , (3,4)
We can say the pair (3,4) has a sum equal to 7.
Now we can write $7x=3x+4x$ so replacing it the quadratic equation
$\Rightarrow 3{{x}^{2}}+7x+4=3{{x}^{2}}+3x+4x+4$
Now we can take 3x common in first half of the equation and take 4 common in second half of the equation
$\Rightarrow 3{{x}^{2}}+3x+4x+4=3x\left( x+1 \right)+4\left( x+1 \right)$
Now we can take x+1 common in the equation
$\Rightarrow 3x\left( x+1 \right)+4\left( x+1 \right)=\left( x+1 \right)\left( 3x+4 \right)$
We can write $\left( x+1 \right)\left( 3x+4 \right)=0$
So the value of x can be -1 or $-\dfrac{4}{3}$

Note:
Sometimes it is difficult to find the pair whose sum is b because roots of the equation may be irrational or fraction in that case it is very difficult to find roots by factorization. We can find the roots by direct formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .