Question

find the zeroes of following quadratic polynomial and verify the relationship between
the zeroes and the cefficients
$y2 + 3 \div 2 \sqrt{5y } - 5$
​

Answer 1

Step-by-step explanation:

Given -

• p(y) = y² + 3√5y/2 - 5

To Find -

• Zeroes of quadratic polynomial
• Verify the relationship between the zeroes and the coefficient.

Now,

→ y² + 3√5y/2 - 5 = 0

→ 2y² + 3√5y - 10/2 = 0

→ 2y² + 3√5y - 10 = 0

here,

a = 2

b = 3√5

c = -10

By using quadratic formula :-

• x = -b ± √b² - 4ac/2a

→ -(3√5) ± √(3√5)² - 4×2×-10/2(2)

→ -3√5 ± √45 + 80/4

→ -3√5 ± √125/4

→ -3√5 ± 5√5/4

Zeroes are -

→ x = -3√5 + 5√5/4

→ 2√5/4

→ √5/2

And

→ x = -3√5 - 5√5/4

→ -8√5/4

→ -2√5

Hence,

The zeroes are √5/2 and -2√5.

Verification :-

• α + β = -b/a

→ -2√5 + √5/2 = -(3√5)/2

→ -4√5 + √5/2 = -3√5/2

→ -3√5/2 = -3√5/2

LHS = RHS

And

• αβ = c/a

→ -2√5 × √5/2 = -10/2

→ -5 = -5

LHS = RHS

Hence,

Verified..

It shows that our answer is absolutely correct.

Answer 2

${\huge{\underbrace{\overbrace{\red{Answer:-}}}}}$

$\implies$ √5/2

$\implies$ -4√5

$\large\underline\mathrm{Given:-}$

• p(y) = y² + 3√5y/2 - 5

$\large\underline\mathrm{To \: find}$

• zeroes of quadratic polynomial.
• verify the zeroes of the coefficient.

$\large\underline\mathrm{Solution}$

$\implies$ y² + 3√5y/2 - 5 =0

$\implies$ 2y² + 3√5y - 10/2 = 0

$\implies$ 2y² + 3√5y - 10 = 0

$\large\underline\mathrm{Thus,}$

$\implies$ a = 2

$\implies$ b = 3√5

$\implies$ c = -10

$\large\underline\mathrm{By \: using \: formula \: of \: the \: quadratic \: polynomial \: .}$

$\implies$ -b +,- √b² - 4ac/2a

$\implies$ -(3√5) +,- √(3√5)² - 4 × 2 × - 10/2 + 2)

$\implies$ -3√5 +,- √45 + 80/4

$\implies$ -3√5 +,- √125/4

$\implies$ -3√5 +,- 5√5/4

$\implies$ x = -3√5 + 5√5/4

$\implies$ 2√5/4

$\implies$ √5/2

$\implies$ x = -3√5 - 5√5/4

$\implies$ -8√5/4

$\implies$ -4√5

$\large\underline\mathrm{Hence,}$

$\large\underline\mathrm{The \: zeroes \: are \: √5/2 \: and -2√5 \: .}$

$\large\underline\mathrm{verification \: of \: quadratic \: polynomial \: :-}$

$\implies$ α + β = -b/a

$\implies$ -2√5 + √5/2 = -(3√5)/2

$\implies$ -4√5 + √5/2 = -(3√5)/2

$\implies$ -3√5/2 = -3√5/2

$\large\underline\mathrm{LHS \: = \: RHS}$

$\large\underline\mathrm{And}$

$\implies$ βα =c/a

$\implies$ -2√5 × √5/2 = 10/2

$\implies$ -5 = -5

$\large\underline\mathrm{LHS \: = \: RHS}$

$\large\underline\mathrm{Hope \: it \: helps \: you \: plz \: mark \: me \: brainlist}$