Math, asked by kassandragvndr2961, 1 year ago

Find the quadratic polynomial with zero 5and-2

Answers

Answered by Lalablackmama
0

Ah excuse me

Quadratic equations can only have 2 roots but u have mentioned 3 roots which is impossible.

Theorem : A quadratic equation cannot have more than two roots.

Proof : Let us consider α,β and γ are the three roots of the given quadratic equation ax2 + bx + c = 0, where a,b,c ϵ R and a \ne 0. Then each α,β and γ will satisfy this quadratic equation.

∴ aα2 + bα + c = 0 ------ (1)

aβ2 + bβ + c = 0 ------(2)

aγ2 + bγ + c = 0 -------(3)

Subtract equation (2) from (1) we get

aα2 + bα + c - ( aβ2 + bβ + c) = 0

⇒ a(α2−β2) + b(α−β ) = 0

a(α−β )(α+β ) + b(α−β )= 0

(α−β )(a (α+β ) + b)= 0

a(α+β) + b = 0 ---------- (4) [α−β≠ 0]

Subtract equation (3) from (2) we get

aβ2 + bβ + c - ( aγ2 + bγ + c) = 0

⇒ a(β2−γ2) + b(β−γ ) = 0

a(β−γ )(β+γ ) + b(β−γ )= 0

(β−γ )(a (β+γ ) + b)= 0

a(β+γ) + b = 0 ---------- (5) [α−γ≠ 0]

Subtracting equation (5) from (4) , we get

a(α−γ )= 0

⇒ α=γ

But this is not possible, because α and γ are distinct and a≠ 0. So their product can not be zero.

Thus our assumption that quadratic equation has three distinct real roots is wrong.

Hence, a quadratic equation cannot have more than two roots.


misbhavk0009: Let alpha equal to 5 and beta equal to minus 2 to form quadratic equation the equation is x squared minus (alpha+beta)x
misbhavk0009: Sorry
misbhavk0009: The formula is x square-(alpha+betha)x+(alpha×betha). therefore applying the values of alpha and beta we get:x^-(5+(-2))x+(5(-2)). => X^-(3)x+(3). => X^-3x+3 ...... that's the quadratic equation for the zeroes given
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