Question

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

Answer 1

Hint: We can solve the given equation in the question by factorization. In quadratic equation $a{{x}^{2}}+bx+c$ while performing factorization we write bx as sum of 2 terms such that the product of their coefficient is equal to product of a and c that is $ac$

Complete step by step answer:
The given equation is $-3{{x}^{2}}-x+2=0$
We can write $3{{x}^{2}}+x-2=0$ by multiplying -1 in both LHS and RHS
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 1 and value of c is -2
So $ac$ = -6
Now we will list all the pairs whose product is -6 and take the one whose sum is 1.
The pairs whose product is -6 are
(1, -6), (2, -3) , (3, -2) , (6,-1)
We can say the pair (3,-2) has a sum equal to 1.
Now we can write $x=3x-2x$ so replacing it the quadratic equation
$\Rightarrow 3{{x}^{2}}+x-2=3{{x}^{2}}+3x-2x-2$
Now we can take 3x common in first half of the equation and take -2 common in second half of the equation
$\Rightarrow 3{{x}^{2}}+3x-2x-2=3x\left( x+1 \right)-2\left( x+1 \right)$
Now we can take x+1 common in the equation
$\Rightarrow 3x\left( x+1 \right)-2\left( x+1 \right)=\left( 3x-2 \right)\left( x+1 \right)$
We can write $\left( 3x-2 \right)\left( x+1 \right)=0$
So the value of x can be -1 or $\dfrac{2}{3}$

Note:
We can verify whether our answer is correct or not by putting the roots in the equation. Our roots are -1 and $\dfrac{2}{3}$ so replacing x in $-3{{x}^{2}}-x+2$ we get
$-3{{\left( -1 \right)}^{2}}-\left( -1 \right)+2$
The value of above term is $-3+1+2=0$ so -1 is correct
Putting $\dfrac{2}{3}$
$-3{{\left( \dfrac{2}{3} \right)}^{2}}-\dfrac{2}{3}+2$
The value of above term is
$-\dfrac{6}{3}+2=0$
So the answer $\dfrac{2}{3}$ is also correct
If the root of a quadratic equation is irrational or fraction then it is very difficult to solve it by factorization. In that case we can find the roots by direct formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .