Question

If α and β are two zeroes of the polynomial 25x2

-15x+2, find a

quadratic polynomial whose zeroes are 1/2α and1/2β.
Pls help !!!!!​

Answer 1

Given:

$\alpha \: \ \: and \: \: \beta \: are \: \: the \: \: zeroes \: \: of \: \\ \: the \: \: polynomial \: \: 25 {x}^{2} - 15x + 2$

To find :

$\star \: the \: quadratic \: polynomial \: whose \: zeroes \: are \\ \: \frac{1}{2 \alpha } \: \:and \: \: \frac{1}{2 \beta } .$

Solution :

$\star \: p(x) = 25 {x}^{2} - 15x + 2$

Here ,

• a = 25 , b = -15 and c = 2

$\star \: \alpha + \beta = \frac{ - b}{a} = \frac{ - ( - 15)}{25} = \frac{3}{5} \\ \\ \star \: \alpha \beta = \frac{c}{a} = \frac{2}{25}$

Now , to find the polynomial whose zeroes are

$\frac{1}{2 \alpha } \:a nd \: \frac{1}{2 \beta }$

• sum of Zeroes :

$\star \: \frac{1}{2 \alpha } + \frac{1}{2 \beta } \\ = > \frac{2( \alpha + \beta )}{4 \alpha \beta } \\ = > \frac{ \alpha + \beta }{2 \alpha \beta } \\ = > \frac{ \frac{3}{5} }{2 (\frac{2}{25}) } \\ = > \frac{3}{5} \times \frac{25}{4} = \frac{15}{4}$

•Product of zeroes :

$\star \: \frac{1}{2 \alpha } \frac{1}{2 \beta } = \frac{1}{4 \alpha \beta } = \frac{1}{4 (\frac{2}{25}) } = \frac{25}{8}$

Now, Polynomial :

• x² - ( sum of Zeroes ) x + product of zeroes .

$\implies \: {x}^{2} - ( \frac{15}{4} )x + \frac{25}{8} = 0 \\ \\ \implies \: \frac{8 {x}^{2} - 30x + 25}{8} = 0 \\ \\ \implies \: 8{x}^{2} - 30x + 25 = 0$

The required quadratic equation is 8x² - 30 x +25.

Answer 2

Step-by-step explanation:

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