Question

If alpha and beta are the zeroes of the quadratic polynomial 2x2+11x+5 find the value of 1/alpha + 1/beta -2 ×alpha ×beta Please send whole solution ASAP

Answer 1

Gɪᴠᴇɴ :-

• ɑ & β are the zeroes of the quadratic polynomial 2x²+11x+5.

Tᴏ Fɪɴᴅ :-

• (1/ɑ) + (1/β) - 2 * ɑ * β

Fᴏʀᴍᴜʟᴀ ᴜsᴇᴅ :-

The sum of the roots of the Equation ax² + bx + c = 0 , is given by = (-b/a)

and ,

→ Product of roots of the Equation is given by = c/a.

Sᴏʟᴜᴛɪᴏɴ :-

Comparing the given quadratic polynomial 2x²+11x+5 with ax² + bx + c we get,

→ a = 2

→ b = 11

→ c = 5

So,

→ sum = (ɑ + β) = (-b/a) = (-11/2)

→ Product = ɑ * β = c/a = (5/2)

Therefore,

→ (1/ɑ) + (1/β) - 2 * ɑ * β

→ (ɑ² + β² - 2(ɑ*β)² / (ɑ*β)

→ [{(ɑ + β)² - 2ɑβ} - 2(ɑ*β)²] / (ɑ*β)

Putting values Now, we get,

→ [{(-11/2)² - 2*(5/2)} - 2(5/2)²] / (5/2)

→ [{(121/4) - 5} - (25/2)] / (5/2)

→ [ (121 - 20/4) - (25/2)] / (5/2)

→ [ (101/4) - (25/2) ] / (5/2)

→ [ (101 - 50)/4 ] / (5/2)

→ (51/4) / (5/2)

→ (51/4) * (2/5)

→ (51/10) (Ans.)

Answer 2

$\maltese$ ${\red{\bold{\underline{ANSWER : }}}}$ $\maltese$

51/10 .

$\maltese$ ${\red{\bold{\underline{SOLUTION : }}}}$ $\maltese$

=>On comparing the given quadratic polynomial 2x²+11x+5 with ax² + bx + c and we get,

• a = 2

•b = 11

• c = 5

Now,

•sum of zeroes = (ɑ + β) = (-b/a) = (-11/2)

• Product of zeroes = ɑ × β = c/a = (5/2)

Since,

=> (1/ɑ) + (1/β) - 2 × ɑ × β

=> (ɑ² + β² - 2(ɑ×β)² / (ɑ×β)

=> [{(ɑ + β)² - 2ɑβ} - 2(ɑ×β)²] / (ɑ×β)-----(1)

[Putting the value of α + β and αβ in equ. 1 .]

=> (-11/2)² - 2×(5/2) - 2(5/2)² / (5/2)

=> (121/4) - 5 - (25/2)/ (5/2)

=>(121 - 20/4) - (25/2) / (5/2)

=>(101/4) - (25/2) / (5/2)

=>(101 - 50)/4 / (5/2)

=>(51/4) / (5/2)

=> (51/4) × (2/5)

=> (51 / 2 × 5)

=> (51/10) .