Question

6. Find the zeroes of the quadratic
polynomial f(x) = x2 – 2root(ax) +(a-b). Verify the
relationship between zeroes and coefficients.
$x ^{2} - 2 \sqrt{ax} + (a - b)$
​

Answer 1

Answer:

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Answer 2

The given question is we have to find Find the zeroes of the quadratic polynomial f(x) = x2 – 2root(ax) +(a-b). Verify the relationship between zeroes and coefficients.

The given expression is

$x ^{2} - 2 \sqrt{ax} + (a - b)</p><p>$

let f(x) be the above equation.

we can write

$a - b = ( { \sqrt{a} })^{2} - ({ \sqrt{b} )}^{2}$

substitute value of a-b in the equation we get

$x ^{2} - 2 \sqrt{ax} + ( { \sqrt{a} }^{2} - { \sqrt{b} }^{2} )$

expand the square values

${x}^{2} - 2 \sqrt{ax} +( \sqrt{a} + \sqrt{b} )( \sqrt{a} - \sqrt{b} ) = 0$

use splitting middle theorem

The steps will be explained in the picture

after solving f(x) we get x=√a+√b and x=√a-√b

The relationship between zeroes and coefficients.

sum of zeroes =

$\frac{ - co \: efficient \: of \: x}{coefficnt \: of \: {x}^{2} } \\ \: \sqrt{a} + \sqrt{b} + \sqrt{a} - \sqrt{b} = \frac{ - ( - 2 \sqrt{a}) }{1} \\ 2 \sqrt{a} = 2 \sqrt{a}$

cancelling the like terms on the LHS we get the above equations.

product of zeroes will be

$\frac{constant \: term}{co \: efficient \: of \: {x}^{2} } \\ ( \sqrt{a} + \sqrt{b} )( \sqrt{a} - \sqrt{b} ) = \frac{a - b}{1} \\ { \sqrt{a} }^{2} - { \sqrt{b} }^{2} = a - b \\ a - b = a - b$

Hence, verified the zeroes and the coefficients of the polynomial

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