Question

find the zeroes of 8x²-4 and verify the relation between the zero and the cofficient .
(the question is not wrong)​

Answer 1

Answer:

8x2-4

8x2=4

X2=4-8

X2=-4

x=✓-4

x=-2

Answer 2

Answer:

$\sf Let \: f(x) = {8x}^{2} – 4 \\ \\ \sf \implies 4 ((√2x)2 – (1)2) \\ \\ \sf \implies 4(√2x + 1)(√2x – 1)$

• To find the zeroes, set f(x) = 0

$\sf \: (√2x + 1)(√2x – 1) = 0 \\ \\ \sf \: (√2x + 1) = 0 \: or \: (√2x – 1) = 0 \\ \\ \sf \: x = \frac{ - 1}{ \sqrt{2} } \: or \: x = \frac{1}{ \sqrt{2} }$

So, the zeroes of f(x) are (-1)/√2 and x = 1/√2

Again,

Sum of zeroes

$\sf \implies \frac{ - 1}{ \sqrt{2} } + \frac{1}{ \sqrt{2} } \\ \\ \sf \implies \frac{ - 1 + 1}{ \sqrt{2} } = 0 \\ \\ \sf \implies \frac{ - b}{a} \\ \\ \sf \implies \frac{(-Coefficient of x)}{(Cofficient of x2)}$

Product of zeroes

$\sf \implies \frac{ - 1}{ \sqrt{2} } \times \frac{1}{ \sqrt{2} } \\ \\ \sf \implies \frac{ - 1}{2} \\ \\\sf \implies \frac{ - 4}{8} \\ \\ \sf \implies \frac{c}{a} \\ \\\sf \implies \frac{Constant \: term}{Coefficient \: of \: {x}^{2}}$

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