Question

if α and β are zeros of a quadratic polynomial x ^2- x - 30 in the form a quadratic polynomial whose zeroes are (2 -α )and(2 -β)

Answer 1

Hey mate!

Here's your answer!

We know that,
α+β = -b/a
αβ = c/a
Here, a=1; b=-1; c=-30
α+β = -(-1)/1 => 1
=> α = 1-β ...(i)
αβ = -30/1 => -30 ...(ii)
Substituting (i) in (ii),
(1-β)β = -30
=> β-β^2 = -30
=> β^2-β-30 = 0
=> (β+5)(β-6) = 0
=> β = -5,6
When β = -5,
α = 1-(-5)
=> α = 1+5
=> α = 6
When β = 6,
α = 1-6
=> α = -5
It is given that,
The zeroes of the second polynomial is (2-α) and (2-β). Therefore,
2-α+2-β = -b/a
=> 2-6+2-(-5) = -b/a
=> -b/a = 3
Also,
(2-α)(2-β) = c/a
=> (2-6)(2-(-5)) = c/a
=> c/a = -4×10
=> c/a = -40
We know that,
Equation of a polynomial = x^2 - (sum of the zeroes) + product of the zeroes
Therefore, the equation of the polynomial is x^2 - 3x + (-40) = 0
=> x^2 - 3x - 40 = 0

That's the answer!

Hope it helps :)

Answer 2

ɪғ α ᴀɴᴅ β ᴀʀᴇ ᴢᴇʀᴏs ᴏғ ᴀ ϙᴜᴀᴅʀᴀᴛɪᴄ ᴘᴏʟʏɴᴏᴍɪᴀʟ x ^2⃣- x - 3⃣0⃣ ɪɴ ᴛʜᴇ ғᴏʀᴍ ᴀ ϙᴜᴀᴅʀᴀᴛɪᴄ ᴘᴏʟʏɴᴏᴍɪᴀʟ ᴡʜᴏsᴇ ᴢᴇʀᴏᴇs ᴀʀᴇ (2⃣ -α )ᴀɴᴅ(2⃣ -β)
ʜᴇʏ ᴍᴀᴛᴇ!

ʜᴇʀᴇ's ʏᴏᴜʀ ᴀɴsᴡᴇʀ!

ᴡᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,
α+β = -ʙ/ᴀ
αβ = ᴄ/ᴀ
ʜᴇʀᴇ, ᴀ=1⃣; ʙ=-1⃣; ᴄ=-3⃣0⃣
α+β = -(-1⃣)/1⃣ => 1⃣
=> α = 1⃣-β ...(ɪ)
αβ = -3⃣0⃣/1⃣ => -3⃣0⃣ ...(ɪɪ)
sᴜʙsᴛɪᴛᴜᴛɪɴɢ (ɪ) ɪɴ (ɪɪ),
(1⃣-β)β = -3⃣0⃣
=> β-β^2⃣ = -3⃣0⃣
=> β^2⃣-β-3⃣0⃣ = 0⃣
=> (β+5⃣)(β-6⃣) = 0⃣
=> β = -5⃣,6⃣ 
ᴡʜᴇɴ β = -5⃣,
α = 1⃣-(-5⃣)
=> α = 1⃣+5⃣
=> α = 6⃣
ᴡʜᴇɴ β = 6⃣,
α = 1⃣-6⃣
=> α = -5⃣
ɪᴛ ɪs ɢɪᴠᴇɴ ᴛʜᴀᴛ, 
ᴛʜᴇ ᴢᴇʀᴏᴇs ᴏғ ᴛʜᴇ sᴇᴄᴏɴᴅ ᴘᴏʟʏɴᴏᴍɪᴀʟ ɪs (2⃣-α) ᴀɴᴅ (2⃣-β). ᴛʜᴇʀᴇғᴏʀᴇ, 
2⃣-α+2⃣-β = -ʙ/ᴀ
=> 2⃣-6⃣+2⃣-(-5⃣) = -ʙ/ᴀ
=> -ʙ/ᴀ = 3⃣
ᴀʟsᴏ, 
(2⃣-α)(2⃣-β) = ᴄ/ᴀ
=> (2⃣-6⃣)(2⃣-(-5⃣)) = ᴄ/ᴀ
=> ᴄ/ᴀ = -4⃣×1⃣0⃣
=> ᴄ/ᴀ = -4⃣0⃣
ᴡᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ, 
ᴇϙᴜᴀᴛɪᴏɴ ᴏғ ᴀ ᴘᴏʟʏɴᴏᴍɪᴀʟ = x^2⃣ - (sᴜᴍ ᴏғ ᴛʜᴇ ᴢᴇʀᴏᴇs) + ᴘʀᴏᴅᴜᴄᴛ ᴏғ ᴛʜᴇ ᴢᴇʀᴏᴇs 
ᴛʜᴇʀᴇғᴏʀᴇ, ᴛʜᴇ ᴇϙᴜᴀᴛɪᴏɴ ᴏғ ᴛʜᴇ ᴘᴏʟʏɴᴏᴍɪᴀʟ ɪs x^2⃣ - 3⃣x + (-4⃣0⃣) = 0⃣
=> x^2⃣ - 3⃣x - 4⃣0⃣ = 0⃣

ᴛʜᴀᴛ's ᴛʜᴇ ᴀɴsᴡᴇʀ!

ʜᴏᴘᴇ ɪᴛ ʜᴇʟᴘs :)