Question

Please solve the above question fast it's urgent ​

Answer 1

Answer:

given : two zeros of quadratic polynomials are

3+√2 and 3-√2.

quadratic polynomial given its two zeros can be formed as below

p (x) = x² - x (sum of zeros) + product of zeros

=> p(x) = x² - x (3+√2+3-√2) + (3+√2)(3-√2)

=> x² - x (6 ) + (9 - 2)

=> x² - 6x + 7

Answer 2

✪AnSwEr

${\tt{\red{\underline{\large{Given}}}}}$

• Zeroes of quadratic polynomial
• $\tt{3}\pm\sqrt{2}$

${\sf{\green{\underline{\large{To\:Find}}}}}$

• Quadratic polynomial

${\sf{\pink{\underline{\Large{Explanation}}}}}$

Let the zeroes of the polynomial be$\tt\alpha{and}\beta$

$\tt\alpha{=3+\sqrt{2}}\\{\beta=3-\sqrt{2}}$

Then

$\tt\alpha{+}\beta\frac{-b}{a}$

&

$\tt\alpha{\times}\beta{=}\frac{c}{a}$

Here

☞

__________________________

$\tt\implies\alpha{+}\beta{=}{3-\sqrt{2}}+3+\sqrt{2}$

=>6

&

$\tt\implies\alpha{\times}\beta{=}{3-\sqrt{2}}{\times}3+\sqrt{2}$

=>(3-√2)(3+√2) (breaking in form of a²-b²)

=>3²-√2²

=>9-2

=>7

___________________________

From this we conclude that

• a=1
• -b=6
• c=7

Creating polynomial

ax²+bx+c

=>x²-6x+7

Additional Information

$\rightarrow\tt\alpha{+}\beta{=}\dfrac{6}{1}$

$\rightarrow\tt\alpha{+}\beta{=}\dfrac{Cofficient\:of\:X}{Cofficient\:of\:x^2}=$

&

$\rightarrow\tt\alpha{\times}\beta{=}\dfrac{7}{1}$

$\rightarrow\tt{\large\alpha{\times}\beta{=}\dfrac{Constant\:term}{Cofficient\:of\:x^2}}$

Hence,relation verified