Question

1.find the zeroes of the polynomial..
$(1) {x }^{2} - 2x - 3$
$(2) 49 {x}^{2} - 64$

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Answer 1

$(1). \: \: \: x = 3 \: \: \: and \: \: \: - 1 \\ (2). \: \: \: x = \frac{8}{7} \: \: \: and \: \: \: \frac{ - 8}{7}$

Step-by-step explanation:

$(1). \: \: \: {x}^{2} - 2x - 3 = 0 \\ {x}^{2} - 3x + x - 3 = 0 \\ x(x - 3) + 1(x - 3) = 0 \\ (x - 3)(x + 1) = 0 \\x - 3 = 0 \: \: \: and \: \: \: x + 1 = 0 \\ so \: \: \: x = 3 \: \: \: and \: \: \: - 1 \\ \\ (2). \: \: \: 49 {x}^{2} - 64 = 0 \\ 49 {x}^{2} = 64 \\ {x}^{2} = \frac{64}{49} \\ x = ±\sqrt{ \frac{64}{49} } = ± \frac{8}{7} \\ so \: \: \: x = \frac{8}{7} \: \: \: and \: \: \frac{ - 8}{7}$