Question

Please solve this work sheet
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Answer 1

Answer:

1.

$(k-1)x^2+kx+1$

Given: one of the zero is -3

Let the other zero be $\beta$

sum of zeros:

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

$-3+\beta=\frac{-k}{k-1}$

$\beta=\frac{-k}{k-1}+3$

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

$\beta=\frac{-k+3k-3}{k-1}$

$\beta=\frac{2k-3}{k-1}$......(1)

product of zeros:

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

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

$\beta=\frac{1}{-3(k-1)}$......(2)

from (1) and (2)

$\frac{1}{-3(k-1)}=\frac{2k-3}{k-1}$

$\frac{1}{-3}=\frac{2k-3}{1}$

$1=-6k+9$

$-8=-6k$

$4=3k$

$k=\frac{4}{3}$

2.

Given:

$\alpha\:$and$\beta$ are zeros of $x^2-5x+6$

sum of zeros:

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

$\alpha+\beta=\frac{-(-5)}{1}$

$\alpha+\beta=5$

product of zeros:

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

$\alpha.\beta=\frac{6}{1}$

$\alpha\beta=6$

Now,

$\alpha+\beta-3\alpha\beta$

$=5-3(6)$

$=5-18$

$=-13$