Question

please answer these three questions .

pls pls pls
fast
.......​

Answer 1

answer:

1. -5
2. 60
3. xsquare-7/10x+1/10

(or)

10xsquare-7x+1

pls mark as brainliest.

Answer 2

Answer:

Answer 6: -5

Answer 7: 60

Answer 8:

$10 {x}^{2} - 7x + 1 = 0$

Step-by-step explanation:

Q6.We know that sum of zeros(roots) of Quadratic equation= -b/a

In this question

${x}^{2} + kx + 6 = 0 \\ \\ a = 1 \\ b = - k \\ c = 6 \\ \\ let \: roots \: are \: \alpha \: and \: \beta \\ \alpha + \beta = 5 = \frac{ - k}{1} \\ \\ so \\ \\ k = - 5$

Q7.

$3 {x}^{2} + 23x + k = 0 \\ \\ let \: roots \: are \: \alpha \: and \: \beta \\ \alpha \beta = \frac{c}{a} \\ \\ \alpha \beta = \frac{k}{3} \\ \\ ATQ \\ \\ \alpha \beta = 20 \\ \\ 20 = \frac{k}{3} \\ \\ k = 20 \times 3 \\ \\ k = 60$

Q8

${x}^{2} - 7x + 10 = 0 \\ \\ we \: know \: that \\ \alpha + \beta = \frac{ - b}{a} = - \frac{( - 7)}{1} \\ \\ \alpha + \beta = 7...eq1 \\ \\ \alpha \beta = \frac{c}{a} \\ \\ \alpha \beta = 10...eq2 \\ \\ eq1 \div eq2 \\ \\ \frac{ \alpha + \beta }{ \alpha \beta } = \frac{7}{10} \\ \\ \frac{ \alpha }{ \alpha \beta } + \frac{ \beta }{ \alpha \beta } = \frac{7}{10} \\ \\ \frac{1}{ \alpha } + \frac{1}{ \beta } = \frac{7}{10} ...eq3$

product of roots

$\frac{1}{ \alpha \beta } = \frac{1}{10}$

Quadratic equation having roots

$\frac{1}{ \alpha } \: and \: \frac{1}{ \beta } \\ \\ {x}^{2} - \bigg( \frac{1}{ \alpha } + \frac{1}{ \beta }\bigg )x + \frac{1}{ \alpha \beta } = 0 \\ \\ {x}^{2} - \bigg( \frac{7}{10} \bigg )x + \frac{1}{10} = 0 \\ \\ 10 {x}^{2} - 7x + 1 = 0$

Hope it helps you.