Question

Equation of the following The sum of the seeds of 3 .A. 2x^2 -3x 6 = 0 B. -x^2 + 3x - 3 = 0 C.√2x^2-3x/√2 + 1 = 0D. 3x^2 - 3x + 3 = 0`​

Answer 1

Step-by-step explanation:

At first, rearrange these equations in the formula,

ax²+bx+c = 0

Then, we all know that, if this equation has 2 seeds, α and β, then the sum of the seeds is:

$\alpha + \beta = - \frac{b}{a}$

for example,

$3x^{2} - 3x + 3 = 0$

let's rearrange this equation using ax²+bx+c=0 formula.

$(3) {x}^{2} + ( - 3)x + (3) = 0$

now, the sum of two seeds is :

$\alpha + \beta = - \frac{ (- 3)}{(3)} = 1$

Now, solve the next problems easily using this method.