Question

Find the zeros of the quadratic polynomial x^2 − 6x + 8 and verify the relation between the zeros and coefficients​

Answer 1

Answer:

$x {}^{2} - 6x + 8 = 0 \\ x {}^{2} - 4x - 2x + 8 = 0 \\ x(x - 4) - 2(x - 4) = 0 \\ (x - 4)(x - 2) = 0 \\ \\ x = 4 = \alpha \\ x = 2 = \beta \\ here \: \: \: a = 1 \: \: \: \: b = - 6 \: \: \: \: c = 8 \\ \\ sum \: of \: zeroes \: = \frac{ - b}{a} \\ = \: \: \alpha + \beta = \frac{6}{1} \\ = \: \: 4 + 2 = 6 \\ = \: \: 6 = 6 \\ = \: \: lhs = rhs \\ \\ product \: of \: zeroes \: = \frac{c}{a} \\ = \: \: \alpha \beta = \frac{8}{1} \\ = \: \: 4 \times 2 = 8 \\ = \: \: 8 = 8 \\ = \: \: lhs = rhs \\ \\ hence \: verified$

Step-by-step explanation:

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