Math, asked by nsvnarayanan, 10 months ago

please answer these three questions .

pls pls pls
fast
.......​

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Answers

Answered by crazyrock
0

answer:

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

(or)

10xsquare-7x+1

pls mark as brainliest.

Answered by hukam0685
0

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.

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