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 \(x^2 -4x +1.\)