Question

6y^2- 7y+ 2 find one by alpha and one by beta​

Answer 1

Answer:

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Answer 2

Answer:

The roots are: $\alpha =\frac{1}{2}, \beta =\frac{2}{3}$

Step-by-step explanation:

The quadratic equation is: $6y^{2}-7y+2=0$

Use factorization method to determine the roots:

$6y^{2}-7y+2=0\\6y^{2}-3y-4y+2=0\\3y(2y-1)-2(2y-1)=0\\(2y-1)(3y-2)=0$

• If (2y - 1) = 0 then y is $\frac{1}{2}$.
• If (3y - 2) = 0, then y is $\frac{2}{3}$ .

Thus, the roots are: $\alpha =\frac{1}{2}, \beta =\frac{2}{3}$