Question

Q23. Find the quadratic polynomial whose zeros are 5-3√2 and 5+ 3√2?​

Answer 1

Given:-

• $\sf \alpha = 5 - 3 \sqrt{2}$
• $\sf \beta = 5 + 3 \sqrt{2}$

To find:-

• The quadratic polynomial whose zeroes are $\alpha$ and $\beta$

Answer:-

We know that, a quadratic polynomial p(x) can be expressed as,

$\sf \: p(x) = {x}^{2} - (sum \: of \: zeroes)x + (product \: of \: zeroes)$

▪Finding sum of zeroes:-

$\sf \alpha + \beta = 5 - 3 \sqrt{2} + 5 + 3 \sqrt{2}$

$\sf \longrightarrow \alpha + \beta = 5 \: \: \cancel{- 3 \sqrt{2}} + 5 \: \: \cancel{ + 3 \sqrt{2} }$

$\sf \longrightarrow \alpha + \beta = 10 \quad \quad \dots(1)$

▪Finding product or zeroes:-

$\sf \alpha \times \beta = (5 - 3 \sqrt{2} )(5 + 3 \sqrt{2} )$

• Using (a - b)(a + b) = a² - b²,

$\sf \longrightarrow \alpha \beta = {5}^{2} - (3 \sqrt{2} ) {}^{2}$

$\sf \longrightarrow \alpha \beta = 25 - 18$

$\sf \longrightarrow \alpha \beta = 7 \quad \quad \dots(2)$

▪Finding the polynomial:-

$\sf \: p(x) = {x}^{2} - (sum \: of \: zeroes)x + (product \: of \: zeroes)$

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

From (1) and (2),

$\longrightarrow \sf p(x) = {x}^{2} - 10x + 7$

Hence the answer is,

$\longrightarrow \underline{ \underline{ \sf{ \green{p(x) = {x}^{2} - 10x + 7}}}}$