Question

ANSWER the question please arugent.
section B​

Answer 1

Answer:

Step-by-step explanation:

$f(x)=ax^2-bx+c$$f(x)=ax^2+bx+c$

1)

   LCM of 13 and 17 : 221

simple ! When they give two prime numbers, product is the LCM.

HCF of 13 and 17: 1

because there is nothing that divides 13 and 17 except 1.

2)

Zeros  given are -2 and -5,

So, the standard form to write a quadratic equation is

$f(x)=ax^2+bx+c$

or,

Points to remember:

$\frac{3+\sqrt{5} }{5}$

$\frac{3+\sqrt{5} }{5}$

$f(x)=x^2-(sum of zeros)x+(product of zeros)$

$sum of zeros=-(\frac{b}{a})$

$product of zeros=(\frac{c}{a})$

so sum of zeros -2 and -5 = -7.

product of zeros -2 and -5 = 10.

so,

=$x^2-(-7)+10$

=$x^2+7x+10$

3)

Let $g(x)$ be a polynomial whose roots are

$\frac{3+\sqrt{5} }{5}$ and $\frac{3-\sqrt{5} }{5}$.

like above , sum of the zeros: $\frac{6}{5}$

product of zeros: $(\frac{3+\sqrt{5} }{5})(\frac{3-\sqrt{5} }{5})$

=$\frac{4}{25}$

so like above frame the equation!!!!!!

sry for i didn't have idea about the rest two.

Any inconveniences are regretted.

Answer 2

1st is correct....❤.....