Question

two roots of quadratic equation are √2+√3 and √2-√3 , frame the quadratic equation
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Answer 1

GIVEN :-

• Roots of quadratic equation are √2 + √3 and √2 - √3.

TO FIND :-

• A quadratic polynomial.

SOLUTION :-

As we know that the quadratic polynomial is given by,

$\\ : \implies \displaystyle \sf \:polynomial \: = x ^{2} - ( \alpha + \beta )x + ( \alpha \beta )$

• α = √2 + √3
• β = √2 - √3

$\\ \\ : \implies \displaystyle \sf \:polynomial \: = x ^{2} - \big( \sqrt{2} + \sqrt{3} + \sqrt{2} - \sqrt{3} \big)x + \big( \sqrt{2} + \sqrt{3} \big) \big( \sqrt{2} - \sqrt{3} \big)$

$: \implies \displaystyle \sf \:polynomial \: = x ^{2} -2 \sqrt{2} x + \big( \sqrt{2} \big) ^{2} - \big( \sqrt{3} \big) ^{2}$

$: \implies \displaystyle \sf \:polynomial \: = x ^{2} -2 \sqrt{2} x + 2 - 3$

$: \implies \underline{ \boxed{\displaystyle \sf \bold{ \:polynomial \: = x ^{2} -2 \sqrt{2} x - 1}}}$

Hence the required quadratic polynomial is x² - 2√2 x - 1.

Answer 2

Given

• Two roots of quadratic equation are √2+√3 and √2-√3

We Find

• A quadratic equation

We know

We used the quadratic polynomial is given by,

$\begin{gathered}\\ \implies \displaystyle \sf \:polynomial \: = x ^{2} - ( \alpha + \beta )x + ( \alpha \beta ) \\ \\\end{gathered}$

So, As Given question

α = √2 + √3

β = √2 - √3

According to the question

$\begin{gathered}\\ \\ \implies \displaystyle \sf \:polynomial \: = x ^{2} - \big( \sqrt{2} + \sqrt{3} + \sqrt{2} - \sqrt{3} \big)x + \big( \sqrt{2} + \sqrt{3} \big) \big( \sqrt{2} - \sqrt{3} \big) \\ \\ \\\end{gathered}$

$\begin{gathered} \implies \displaystyle \sf \:polynomial \: = x ^{2} -2 \sqrt{2} x + \big( \sqrt{2} \big) ^{2} - \big( \sqrt{3} \big) ^{2} \\ \\ \\\end{gathered}$

$\begin{gathered} \implies \displaystyle \sf \:polynomial \: = x ^{2} -2 \sqrt{2} x + 2 - 3 \\ \\ \\\end{gathered}$

$\begin{gathered} \implies\displaystyle \sf \bold{ \:polynomial \: = x ^{2} -2 \sqrt{2} x - 1} \\ \\\end{gathered} \\ \\ </p><p></p><p></p><p>$

Hence, quadratic polynomial is x² - 2√2 x - 1.