Question

quadratic equation with zeros 2/ 3, - 2 is​

Answer 1

Answer:

Given:-

$\alpha + \beta = 2 |3 \: \: \: \: \alpha \beta = - 2$

We know that the ideal equation of quadratic:-

=

$ax ^{2} + \: bx \: + \: c\ \\ = x^{2} - ( \alpha + \beta )x + \alpha \beta = 0$

Put the values in given equation we get:-

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

l hope it's help you ♥️

Answer 2

Answer:

x= 3y + 4z - 5

therefore

y - 1 = z

sin(x-y)= sinxcosy- cosxsiny

Step-by-step explanation:

x= 3y + 4z - 5

therefore

y - 1 = z

sin(x-y)= sinxcosy- cosxsiny