Question

Find the roots of the equation
x⁴-(√5+√2)x²+√10=0

Answer 1

Answer:

$x = + - \sqrt[4]{5} \: or \: + - \sqrt[4]{2} .$

Step-by-step explanation:

$x = + - \sqrt[4]{5} \: or \: + - \sqrt[4]{2} \\ as \: \: {x}^{2} = \sqrt{5} \: or \: \sqrt{2} \\ \\ discr \: = ( \sqrt{5} + \sqrt{2} )^{2} - 4 \times \sqrt{10} \\ {( \sqrt{5} - \sqrt{2} }^{2} \\ just \: apply \: quadratic \: equatiom \: roots \: formula. \\ simple.$

x^2 = [ (sqrt 2 + sqrt 5) + - (sqrt 5 - sqrt2 ) ] / 2.

can verify by substituting the values as x^2 = sqrt5 or sqrt2. and x^4 = 5 or 2.

Answer 2

Answer :-

Roots of the equation are ± ∜5 and ± ∜2.

Explanation :-

x^4 - (√5 + √2)x² + √10 = 0

(x²)² - (√5 + √2)x² + √10 = 0

Let x² = y

⇒ y² - (√5 + √2)y + √10 = 0

Comparing the above equation with ay² + by + c = 0 we get,

• a = 1

• b = - (√5 + √2)

• c = √10

Discriminant (D) = b² - 4ac

= { - (√5 + √2) }² - 4( 1 )( √10 )

= (√5 + √2)² - 4√10

= ( √5 )² + ( √2 )² + 2( √5 )( √2 ) - 4( √5 )( √2 )

= ( √5 )² + ( √2 )² - 2( √5 )( √2 )

= (√5 - √2)²

Finding roots of the equation by Quadratic formula

y = ( - b ± √D ) / 2a

⇒ y = [ - { - (√5 + √2) } ± √(√5 - √2)² ] / 2( 1 )

⇒ y = {√5 + √2 ± (√5 - √2) } / 2

⇒ y = { √5 + √2 + (√5 - √2) } / 2 or y = { √5 + √2 - (√5 - √2) } / 2

⇒ y = (√5 + √2 + √5 - √2) / 2 or y = (√5 + √2 - √5 + √2) / 2

⇒ y = 2√5/2 or y = 2√2/2

⇒ y = √5 or y = √2

But y = x²

⇒ x² =√5 or x² = √2

Taking square root on both sides

⇒ √x² = ± √√5 or √x² = ± √√2

⇒ x = ± ∜5 or x = ± ∜2

∴ roots of the equation are ± ∜5 and ± ∜2.