Question

22. Form a quadratic polynomial whose zeros are 3+ √2 and 3-√2.​

Answer 1

$\green{ \huge{\sf \underline{ \fbox{ \: Solution : \: \: }}}}$

Given ,

The zeores of quadratic equation are 3 + √2 and 3 - √2

So ,

$\star \: \sf Sum \: of \: zeroes = ( 3 + \sqrt{2} ) + (3 - \sqrt{2} ) \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf= 6 \\ \sf \star \: </p><p>Product \: of \: zeroes = ( 3 + \sqrt{2} ) (3 - \sqrt{2} ) \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf= {(3)}^{2} - {( \sqrt{2} )}^{2} \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf= 9 - 2 \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf = 7$

We know that , the quadratic equation is given by

$\mathtt{ \large \purple{ \fbox{ {x}^{2} - (sum \: of \: zeroes)x + (product \: of \: zeroes) = 0}}}$

Substitute the known values , we obtain

$\sf \hookrightarrow {x}^{2} - 6x + 7 = 0$

Hence , the required quadratic polynomial is x² - 6x + 7