Question

Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. (i)$\dfrac{1}{4} , 1$ (ii) $\sqrt 2 , \dfrac{1}{3}$ (iii) $0, \sqrt5$ (iv) $1, 1$ (v) $-\dfrac{1}{4} ,\dfrac{1}{4}$(vi) $4, 1$

Answer 1

(i) $\dfrac{1}{4} ,1$
Let the polynomial be $ax^2 + bx + c$and its zeroes are α and ẞ.
$α + β = \dfrac{1}{3} = -\dfrac{b}{a}$
$αβ=-1 = -\dfrac{4}{4} =\dfrac{c}{a}$
If $a =4,$ then $b= -1, c = -4$

Therefore, the quadratic polynomial is $4x^2 - x - 4.$

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(ii)√2,1/3
Let the polynomial be $ax^2 + bx + c$ and its zeroes are α and ẞ.
$α + β = √2= \dfrac{3√2}{3} = -\dfrac{b}{a}$
$αβ = \dfrac{1}{3} = \dfrac{c}{a}$

If $a= 3,$ then $b=-3√2, c=1$
Therefore, the quadratic polynomial is $3x^2 - 3√2x + 1.$

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(iii) 0,√5
Let the polynomial be $ax^2 + bx + c$ and its zeroes be $α$and $β.$
$α + ẞ = 1 = -\dfrac{(-1)}{1} =\dfrac{ -b}{a}$
$αβ = √5 = \dfrac{√5}{1} = \dfrac{c}{a}$

If $a=1$, then $b=0, c=√5$

Therefore, the quadratic polynomial is $x^2+√5.$

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(iv)1, 1
Let the polynomial be $ax^2 + bx + c$ and its zeroes be $α$ and $β.$.
$α + ẞ = 1 = -\dfrac{(-1)}{1} =\dfrac{ -b}{a}$
$αβ = 1 = \dfrac{1}{1 }= \dfrac{c}{a}$

If a=1, then $b=-1, c=√5$

Therefore, the quadratic polynomial is $x^2 -x +1.$

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**(v) **$-\dfrac{1}{4} ,\dfrac{1}{4}$
Let the polynomial be $ax^2 + bx + c$and its zeroes are $α$ and $β.$

$α+β$ =$$ -\dfrac{1}{4} $=-\dfrac{b}{a}$
$αβ = \dfrac{1}{4} = \dfrac{c}{a}$

If a=4, then $b=1, c=1$

Therefore, the quadratic polynomial is$4x^2 +x +1.$

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(vi)4, 1
Let the polynomial be $ax^2 + bx + c$ and its zeroes be 𝛼 and 𝛽.
$α + ẞ = 4 = -\dfrac{(-4)}{1}= \dfrac{ -b}{a}$
$αβ = 1 =\dfrac{1}{1}= \dfrac{c}{a}$

If $a=1$a=1 then $b=-4, c=1$

Therefore, the quadratic polynomial is $x^2 -4x +1.$