Question

If ∝ and β are the roots of the quadratic equation x² – x – 2 = 0 then ∝² + β² is equal to​

Answer 1

${\huge{\fbox{\colorbox{black}{\color{hotpink}{\textbf{\textsf{Answer:-}}}}}}}$

To Solve:

• If ∝ and β are the roots of the quadratic equation x² – x – 2 = 0 then ∝² + β² is equal to

Solⁿ:

• Factorize: x² - x - 2 = 0
• Factors will be the ∝ and β

${x}^{2} - x - 2 = 0 \\ \\ x {}^{2} - 2x + 1x - 2 = 0 \\ \\ x(x - 2) - 1(x - 2) = 0 \\ \\ (x - 2)(x - 1) = 0 \\ \\ x = 2 \: \: \: \: \: \: \: \: \: x = 1 \\ \\\alpha = 2 \: \: \: \: \: \: \: \: \: \beta = 1 \\ \\ { \alpha}^{2} + { \beta}^{2} \\ \\ {(2) }^{2} + {(1)}^{2} \\ \\ 4 + 1 \: \: \: \: { \boxed{ \pink{= 5}}}$

ㅤ

HOPE IT HELPS

MARK BRAINLIEST PLS :)

Answer 2

Given that, α and β are the roots of the quadratic equation x² – x – 2 = 0.

We'll have to find the value of α² + β².

When we compare x² – x – 2 = 0 to ax² + bx + c = 0, we get the values of a, b and c as :

⇒ a = 1

⇒ b = – 1

⇒ c = – 2

So, now the sum and product of the zeroes of the equation are :

⋆ Sum of the zeroes = – b/a

⇒ α + β = – (– 1)/1

⇒ α + β = 1/1

⇒ α + β = 1 . . . . . (1)

⋆ Product of zeroes = c/a

⇒ αβ = c/a

⇒ αβ = – 2/1

⇒ αβ = – 2 . . . . . (2)

We know that, α² + β² can be written as :

⇒ α² + β² = (α + β)² – 2αβ

Now, substituting the values of (1) and (2) in the above formula :

⇒ α² + β² = (α + β)² – 2αβ

⇒ α² + β² = (1)² – 2 (– 2)

⇒ α² + β² = 1 + 4

⇒ α² + β² = 5

⋆ Therefore, α² + β² is 5 when α and β are the roots of the quadratic equation x² – x – 2 = 0.