Question

find a quadratic polynomial whose zeros are 3 upon 5 and minus one upon two ​

Answer 1

$a({x}^{2} - ( \alpha + \beta )x + \alpha \beta ) \\a( {x}^{2} - ( \frac{3}{5} - \frac{ 1}{2} )x + \frac{3}{5} \times \frac{ - 1}{2}) \\ a( {x}^{2} - ( \frac{1}{10})x - \frac{3}{10} ) \\ \frac{1}{10} a(10 {x}^{2} - x - 3) \\ it \: can \: also \: be \: written \: as \: \\ k(10 {x}^{2} - x - 3)$

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Answer 2

The quadratic polynomial whose zeros are $\dfrac{3}{5}$ and $-\dfrac{1}{2}$ is $x^{2} -\dfrac{1}{10}x-\dfrac{3}{10}.$

Step-by-step explanation:

Here, α = $\dfrac{3}{5}$ and β = $-\dfrac{1}{2}$

To find, the quadratic polynomial = ?

We know that,

The quadratic polynomial whose zeros are α and β

=$x^{2} -(\alpha+\beta)x+\alpha\beta$

$=x^{2} -(\dfrac{3}{5} -\dfrac{1}{2} )x+(\dfrac{3}{5})(-\dfrac{1}{2})$

$=x^{2} -(\dfrac{6-5}{10} )x-\dfrac{3}{10}$

$=x^{2} -\dfrac{1}{10}x-\dfrac{3}{10}$

Hence, the quadratic polynomial whose zeros are $\dfrac{3}{5}$ and $-\dfrac{1}{2}$is $x^{2} -\dfrac{1}{10}x-\dfrac{3}{10}$.