Question

if alpha ,beta are the roots of quadratic equation 2x square minus 3x plus 4 equals to 0 then find. 1. 1 divided by alpha square plus 1 divided by beta square​

Answer 1

$\large\bf\underline {To \: find:-}$

• we need to find the value of 1/α² + 1/β²

$\huge\bf\underline{Solution:-}$

$\bf\underline{\red{Given:-}}$

❥ α and β are the roots of the given equation 2x² - 3x + 4 .

● a = 2

● b = -3

● c = 4

☘️ we know that :-

→ α + β = -b/a

→ α + β = -(-3)/2

→ α + β = 3/2 .....1)

→ αβ = c/a

→ αβ = 4/2

→ αβ = 2 .......2)

▶Now , finding Value of 1/α² + 1/β² :--

→ 1/α² + 1/β²

→ (β² + α²)/(αβ)² ......3)

• we know that,

→ a² + b² = (a + b)² - 2ab .....4)

● Now , from 1) , 2) , 3) and 4)

→ 1/α² + 1/β² = [( α + β)² - 2αβ]/(αβ)²

→ 1/α² + 1/β² = [(3/2)² - 2 × 2]/2²

→ 1/α² + 1/β² = [ 9/4 - 4 ]/4

→ 1/α² + 1/β² = [ (9 - 16)/4]/4

→ 1/α² + 1/β² = -7/4/4

→ 1/α² + 1/β² = -7/4 × 1/4

→ 1/α² + 1/β² = -7/16

Hence,

༒ Value of 1/α² + 1/β² is -7/16

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Answer 2

we need to find the value of 1/α² + 1/β²

$\huge\pink{solution}$

❥ α and β are the roots of the given equation 2x² - 3x + 4 .

● a = 2

● b = -3

● c = 4

☘️ we know that :-

→ α + β = -b/a

→ α + β = -(-3)/2

→ α + β = 3/2 .....1)

→ αβ = c/a

→ αβ = 4/2

→ αβ = 2 .......2)

▶Now , finding Value of 1/α² + 1/β² :--

→ 1/α² + 1/β²

→ (β² + α²)/(αβ)² ......3)

we know that,

→ a² + b² = (a + b)² - 2ab .....4)

● Now , from 1) , 2) , 3) and 4)

→ 1/α² + 1/β² = [(a + b)² - 2ab]/(αβ)²

→ 1/α² + 1/β² = [(3/2)² - 2 × 2]/2²

→ 1/α² + 1/β² = [ 9/4 - 4 ]/4

→ 1/α² + 1/β² = [ (9 - 16)/4]/4

→ 1/α² + 1/β² = -7/4/4

→ 1/α² + 1/β² = -7/4 × 1/4

→ 1/α² + 1/β² = -7/16

Hence,

༒ Value of 1/α² + 1/β² is -7/16

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