Question

$\huge\sf\underline{QuEsTiOn:–}$

Find the zeroes of the quadratic polynomial (8x² ˗ 4) and verify the relation between the zeroes and the coefficients.
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Answer 1

Step-by-step explanation:

f(x) = 8x2 – 4

= 4 ((√2x)2 – (1)2)

= 4(√2x + 1)(√2x – 1)

for zeroes, f(x) = 0

(√2x + 1)(√2x – 1) = 0

(√2x + 1) = 0 or (√2x – 1) = 0

x = (-1)/√2 or x = 1/√2

So,

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

now,

Sum of zeroes = -1/√2 + 1/√2 = (-1+1)/√2 = 0 = -b/a

Product of zeroes = -1/√2 x 1/√2 = -1/2 = c/a

Answer 2

Answer:

$Let \: f(x)=8x²−4 \\ = 4((√2x)²−(1)²) \\=4(√2x+1)(√2x−1) \\ to \: find \: the \: zeroes \: , Let f(x)=0 \\ (√2x+1)(√2x−1)=0 \\ (√2x+1)=0 or (√2x−1)=0 \\ x=(−1)/√2 or \: x=1/√2$

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

Again

Sum of zeroes = −1/√2+1/√2=(−1+1)/√2=0=−b/a=(-coefficient of x)/(−coefficient of x²)Product of zeroes = −1/√2×1/√2=2−1=8−4=ac=coefficient of x2Constant term