Question

Find a quadratic polynomial with zeroes as


root2,-3 by root2

?​

Answer 1

Answer:

$x^{2} + \frac{1}{ \sqrt{2} } x - 3$

Step-by-step explanation:

Let the zeroes of the quadratic polynomial be

α = √2

β = $\frac{ - 3}{ \sqrt{2} }$

Then, α + β = $\sqrt{2} + ( \frac{ - 3}{ \sqrt{2} } ) = \sqrt{2 } - \frac{3}{ \sqrt{2} } = \frac{2 - 3}{ \sqrt{2} } = \frac{ - 1}{ \sqrt{2} }$

αβ = $\sqrt{2 } \times \frac{ - 3}{ \sqrt{2} } = - 3$

Sum of zeroes = α + β = $\frac{ - 1}{ \sqrt{2} }$

Product of zeroes = αβ = - 3

Then, the quadratic polynomial:

= x² - ( sum of zeroes) x + product of zeroes

$= x^{2} - ( \frac{ - 1}{ \sqrt{2} } )x + ( - 3)$

$= {x}^{2} + \frac{1}{ \sqrt{2} } - 3$

Hope this helps : )

Mark as brainliest if you are satisfied.

Answer 2

Step-by-step explanation:

$◆ Answer ◆$

= 1 + 2 + 3 + 4 + 5

= 3 + 3 + 4 + 5

= 6 + 4 + 5

= 10 + 5

= 15

Hope, it's help you.

✌️✌️