Question

find a quadratic polynomial having zeros 2+7√3​

Answer 1

$\large\bf\underline{Given:-}$

• Zeroes of required polynomial = 2 +7√3

$\large\bf\underline {To \: find:-}$

• quadratic polynomial

$\huge\bf\underline{Solution:-}$

One zero of required polynomial = 2+7√3

Then the other zero is 2 -7√3

Let α and β are the zeroes of required polynomial.

➝ α + β = 2 + 7√3 + 2 -7√3

➝ α + β = 4

➝ αβ = (2+7√3) × (2 -7√3)

➝ αβ = 4 - 147

➝ αβ = -143

Formula for quadratic polynomial:-

$\bf \: {x}^{2} - ( \alpha + \beta )x + \alpha \beta$

➝ x² -(4)x - 143

➝ x² - 4x - 143

So, the quadratic polynomial is x² - 4x - 143

$\underline{ \bf \: verification : - }$

• p(x) = x² - 4x - 143
• a = 1
• b = -4
• c = -143

Sum of zeroes = -b/a

➝ 4 = -(-4)/1

➝ 4 = 4

Product of zeroes = c/a

➝ -143 = -143/1

➝ -143 = -143

LHS = RHS

Hence Verified

Answer 2

Solution ↓

1 root of required polynomial. = 2+7√3

Other root = 2-7√3

So,

$\alpha \: and \: \beta$

↑Are the root of the polynomial.↑

Sum of the zeroes↓

$\rule{300}2$

$\alpha+\beta=2+7\sqrt3+2-7\sqrt3 \\ \alpha+\beta=4$

product of the zeroes↓

$\alpha\beta=( 2+7\sqrt3)(2-7\sqrt3)\\ \alpha\beta=143$

Therefore,

required polynomial is ↓

$x^2-(4)x+143\\ \leadsto x^2-4x+143$

$\rule{150}2$

Hence,

$\leadsto x^2-4x+143$ (is the required polynomial..)