Question

if one root of the equation (with rational coefficients) x2+px+q is 2+√3, then the value of p and q is
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Answer 1

Given ,

one root of quadratic equation x² - px + q = 0 is 2 + √3

then, other root must be (2 - √3)

Now, sum of roots = -coefficient of x/coefficient of x²

(2 + √3) + (2 - √3) = -(-p)/2

⇒ 4 = p/2

⇒p = 8

product of roots = constant/Coefficient of x²

(2 + √3)(2 - √3) = q/2

⇒(2² - √3²) = q/2

⇒ 4 - 3 = q/2

⇒ 1 = q/2

⇒q = 2

∴ pq = 8 × 2 = 16 , your options are wrong . Answer should be 16

Answer 2

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

• p(x) = x² + px + q
• one root = 2+√3

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

• value of p and q

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

If one root is 2+√3

then, the other root be 2 - √3

• p(x) = x² + px + q

This equation is in the form of :-

• ax² + bx + c

• p(x) = x² +px + q

• a = 1
• b= p
• c = q

Sum of zeroes = - Coefficient of x/coefficients of x²

⠀⠀⠀⠀⠀➝ (2 + √3)+(2-√3) = - (p)/1

⠀⠀⠀⠀⠀➝ 2 + 2 = - p/1

⠀⠀⠀⠀⠀➝ 4 = -p

⠀⠀⠀⠀⠀➝ -4 = p

⠀⠀⠀⠀⠀➝ p = -4

Product of zeroes = constant term/coefficient of x²

⠀⠀⠀⠀⠀➝ (2+√3)(2-√3) = q/1

⠀⠀⠀⠀⠀➝ 4 - 3 = q/1

⠀⠀⠀⠀⠀➝ 1= q

⠀⠀⠀⠀⠀➝ q = 1

Hence,

• p = -4
• q = 1