Question

Find the zeroes of polynomial x2-4x+3 and verify relationship between zeroes and coefficient
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Answer 1

${x}^{2} - 4x + 3 = 0 \\ \\ {x}^{2} - x - 3x + 3 = 0 \\ \\ x(x - 1) - 3(x - 1) = 0 \\ \\ (x - 1) \: (x - 3) = 0 \\ \\ x = 1 \: \: \: and \: \: x = 3$

Sum of roots :-

$\alpha + \beta = 1 + 3 \\ \\ = 4 = \frac{ - ( - 4)}{1} = \frac{ - b}{a}$

Product of roots :-

$\alpha \times \beta = 1 \times 3 \\ = 3 = \frac{3}{1} = \frac{c}{a}$

Mark it brainliest.

Answer 2

ANSWER:

• Zeros are 3 and 1.

GIVEN:

• P(x) = x²-4x+3

TO FIND:

• Zeros and to verify the relationship between zeros and coefficients.

SOLUTION:

=> x²-4x+3 = 0

=> x²-3x-x+3 = 0

=> (x²-3x)+(-x+3) = 0

=> x(x-3)-1(x-3) = 0

=> (x-3)(x-1) = 0

Either (x-3) = 0

=> x = 3

Either (x-1) = 0

=> x = 1

Formula:

=> Sum of zeroes (α+β) = -(Coefficient of x)/Coefficient of x²

=> Product of zeroes (αβ) = Constant term/ Coefficient of x²

Sum of zeros:

= 3+1

=> -(-4)/1 = -(Coefficient of x)/Coefficient of x²

Product of zeros:

= 3(1)

=> 3/1 = Constant term/ Coefficient of x²

Verified