When solving a quadratic equation: x2 + 2x - 8 = 0
We know that sum -2 and + 4 equal to middle part and their product is equal to -8. We can safely write the factors as (x + a)(x + b) =0 ie (x-2)(x+4)=0
But, if the equation is 3x^2 + 5x -12 =0 , we come to know that a and b as +9 and -4. But we can't write it directly (x + 9)(x -4)=0 because of the quotient 3. We then need to divide 9 and -4 by 3. SO, we get the equation: (x+3)(x-4/3)=0.
Just wanted to confirm if this is a general rule (to divide the roots by co-ef of x^2) and works all the time.
We know that sum -2 and + 4 equal to middle part and their product is equal to -8. We can safely write the factors as (x + a)(x + b) =0 ie (x-2)(x+4)=0
But, if the equation is 3x^2 + 5x -12 =0 , we come to know that a and b as +9 and -4. But we can't write it directly (x + 9)(x -4)=0 because of the quotient 3. We then need to divide 9 and -4 by 3. SO, we get the equation: (x+3)(x-4/3)=0.
Just wanted to confirm if this is a general rule (to divide the roots by co-ef of x^2) and works all the time.
- VK
I will (Learn. Recognize. Apply)
I will (Learn. Recognize. Apply)
