Question

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Answer 1

Answer:

Let α and β are zeros of required quadratic polynomial.

⇒ α+β=4 ---- ( 1 ) [ given ]

⇒ αβ=1 ----- ( 2 ) [ Given ]

The required polynomial,

x

2

−(α+β)x+(αβ)=0

Now, using ( 1 ) and ( 2 )

⇒ x

2

−4x+1=0

Answer 2

Given

• Sum of quadratic polynomial is 4
• Product of quadratic polynomial is 1

To Find

• Quadratic polynomial

Concept used

• Required polynomial=k[x²-(S)x+P]

Solution

We have sum and product of quadratic polynomial and we need to find the polynomial

So generally we use

Required polynomial

k[x²-(S)x+P

Here

• Sum(S)=4
• Product(P)=1

Putting values

$\rm \: k \left[ {x}^{2} - (s) x + p \right] \\ \\ \implies \rm \: k\left[ {x}^{2} - (4)x + 1 \right] \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies \rm \: k\left[ {x}^{2} - 4x + 1 \right] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \boxed{ \bf \: polynomial = {x}^{2} - 4x + 1 } \\ \rm \:where \: k = 1$

So the required polynomial is x²-4x+1

Now

Verification

We know

$\rm \: \alpha + \beta = \dfrac{ - b}{a} \\ \\ \bf \: putting \: values \\ \\ \implies \: 4 = \frac{ 4}{1} \\ \\ \boxed{ \bf \: LHS = RHS} \\ \rm \: hence \: proved$

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