Question

Find the values of x​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

Given -

p(x) = x² - (1 + √2)x + √2 = 0

To Find -

Value of x

In x² - (1 + √2)x + √2 = 0

here,

a = 1

b = -(1 + √2) = (-1 - √2)

c = √2

Using Quadratic formula -

• x = -b ± √b² - 4ac/2a

» -[(-1 - √2)] ± √(-1 - √2)² - 4×1×√2/2(1)

» (1 + √2) ± √1 + 2 + 2√2 - 4√2/2

» (1 + √2) ± (√3 - 2√2)/2

Zeroes are -

• x = (1 + √2) + (√3 - 2√2)/2

and

• x = (1 + √2) - (√3 - 2√2)/2

Verification -

Let α = 1 + √2 + √3 - 2√2/2 and β = 1 + √2 - √3 - 2√2/2

Now,

• α + β = -b/a

(1 + √2) + (√3 - 2√2)/2 + (1 + √2) - (√3 - 2√2)/2 = -(-1 - √2)/1

» (1 + √2) + (√3 - 2√2) + (1 + √2) - (√3 - 2√2)/2 = 1 + √2

» 1 + √2 + 1 + √2/2 = 1 + √2

» 2 + 2√2/2 = 1 + √2

» 2(1 + √2)/2 = 1 + √2

» 1 + √2 = 1 + √2

LHS = RHS

And

• αβ = c/a

(1 + √2) + (√3 - 2√2)/2 × (1 + √2) - (√3 - 2√2)/2 = √2/1

» (1 + √2)² - (√3 - 2√2)²/4 = √2

» 1 + 2 + 2√2 - (3 - 2√2)/4 = √2

» 3 + 2√2 - 3 + 2√2/4 = √2

» 4√2/4 = √2

» √2 = √2

LHS = RHS

Hence,

Verified..