Question

find the zeros of a polynomial x square - 18 x + 12 and verify the relationship between zeros and coefficients of variables in the polynomial.

Answer 1

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Answer 2

Answer:

Given: Quadratic Polynomial is x² - 18x + 12

To find: Zeroes and then verify the relationship between zeroes and coefficient.

To find zero we put polynomial equal to 0

x² - 18x + 12 = 0

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-(-18)\pm\sqrt{(-18)^2-4(12)}}{2}$

$x=\frac{18\pm\sqrt{324-48}}{2}$

$x=\frac{18\pm2\sqrt{69}}{2}$

x = 9 + √69    and x = 9 - √69

let, α = 9 + √69 and β = 9 - √69

Now, $sum\:of\:zeroes=\frac{-coefficient\:of\:x}{coefficient\:of\:x^2}$

LHS = α + β = 9 + √69 + 9 - √69 = 18

RHS = -(-18)/1 = 18

LHS = RHS

$product\:of\:zeroes=\frac{constant\:term}{coefficient\:of\:x^2}$

LHS = α × β = ( 9 + √69 ) ( 9 - √69 ) = 81 - 69 = 12

RHS = 12/1 = 12

LHS = RHS

Hence Verified.