Question

plz give me correct answer​

Answer 1

Answer:

Explanation:

ANSWER:-

We need to remember the formula ,

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

$\alpha\beta= \frac{c}{a}$

And also, $k(x^2-(\alpha+\beta)x+(\alpha\beta))$.

(1) Let alpha and Beta be the two zeroes.

$\alpha=-4\\ \beta = -6$

Placing the, we get

$k(x^2-(-4-6)x+(-4 \times -6))$

Required Quadratic Polynomial is $x^2+10x+24$, with coefficient of b as 10.

(2) We know that

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

Let alpha and Beta be the two zeroes.

• b here is 11
• a here is 6.

Placing them , we get sum of zeroes as $\frac{-11}{6}$.

(3) We know that :-

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

$\alpha\beta= \frac{c}{a}$.

Placing the sum first, we get

$\alpha+\beta = \frac{-(-10)}{1} \implies \frac{10}{1}$

$\alpha\beta = \frac{-39}{1}$

Placing them here as $k(x^2-(\alpha+\beta)x+(\alpha\beta))$, we get the required Quadratic Equation as

$x^2+10x-39$.

(4) Given,

$\alpha +\beta = \frac{15}{4}$ and $\alpha\beta = \frac{3}{1}$.

In the product, making the base same, we get

$\alpha\beta = \frac{12}{4}$,

so required Quadratic Equation is

$4x^2+15x+12$  ____________[ Formula used is $k(x^2-(\alpha+\beta)x+(\alpha\beta))$].