Form A Polynomial With The Given Zeros
Let zeros of a quadratic polynomial be α and β.
x = β, x = β
x – α = 0, x – β = 0
The obviously the quadratic polynomial is
(x – α) (x – β)
i.e., x2 – (α + β) x + αβ
x2 – (Sum of the zeros)x + Product of the zeros
Form A Polynomial With The Given Zeros Example Problems With Solutions
Example 1: Form the quadratic polynomial whose zeros are 4 and 6.
Sol. Sum of the zeros = 4 + 6 = 10
Product of the zeros = 4 × 6 = 24
Hence the polynomial formed
= x2 – (sum of zeros) x + Product of zeros
= x2 – 10x + 24
Example 2: Form the quadratic polynomial whose zeros are –3, 5.
Sol. Here, zeros are – 3 and 5.
Sum of the zeros = – 3 + 5 = 2
Product of the zeros = (–3) × 5 = – 15
Hence the polynomial formed
= x2 – (sum of zeros) x + Product of zeros
= x2 – 2x – 15
Example 3: Find a quadratic polynomial whose sum of zeros and product of zeros are respectively \(\frac { 1 }{ 2 }\), – 1
Sol. Let the polynomial be ax2 + bx + c and its zeros be α and β.
(i) Here, α + β = \(\frac { 1 }{ 4 }\) and α.β = – 1
Thus the polynomial formed
= x2 – (Sum of zeros) x + Product of zeros
\(={{\text{x}}^{\text{2}}}-\left( \frac{1}{4} \right)\text{x}-1={{\text{x}}^{\text{2}}}-\frac{\text{x}}{\text{4}}-1\)
The other polynomial are \(\text{k}\left( {{\text{x}}^{\text{2}}}\text{-}\frac{\text{x}}{\text{4}}\text{-1} \right)\)
If k = 4, then the polynomial is 4x2 – x – 4.
Example 4: Find a quadratic polynomial whose sum of zeros and product of zeros are respectively \(\sqrt { 2 }\), \(\frac { 1 }{ 3 }\)
Sol. Here, α + β =\(\sqrt { 2 }\), αβ = \(\frac { 1 }{ 3 }\)
Thus the polynomial formed
= x2 – (Sum of zeroes) x + Product of zeroes
= x2 – \(\sqrt { 2 }\) x + \(\frac { 1 }{ 3 }\)
Other polynomial are \(\text{k}\left( {{\text{x}}^{\text{2}}}\text{-}\frac{\text{x}}{\text{3}}\text{-1} \right)\)
If k = 3, then the polynomial is
3x2 – \(3\sqrt { 2 }x\) + 1
Example 5: Find a quadratic polynomial whose sum of zeros and product of zeros are respectively 0, √5
Sol. Here, α + β = 0, αβ = √5
Thus the polynomial formed
= x2 – (Sum of zeroes) x + Product of zeroes
= x2 – (0) x + √5 = x2 + √5
Example 6: Find a cubic polynomial with the sum of its zeroes, sum of the products of its zeroes taken two at a time, and product of its zeroes as 2, – 7 and –14, respectively.
Sol. Let the cubic polynomial be ax3 + bx2 + cx + d
⇒ x3 + \(\frac { b }{ a }\)x2 + \(\frac { c }{ a }\)x + \(\frac { d }{ a }\) …(1)
and its zeroes are α, β and γ then
α + β + γ = 2 = \(\frac { -b }{ a }\)
αβ + βγ + γα = – 7 = \(\frac { c }{ a }\)
αβγ = – 14 = \(\frac { -d }{ a }\)
Putting the values of \(\frac { b }{ a }\), \(\frac { c }{ a }\), and \(\frac { d }{ a }\) in (1), we get
x3 + (–2) x2 + (–7)x + 14
⇒ x3 – 2x2 – 7x + 14
Example 7: Find the cubic polynomial with the sum, sum of the product of its zeroes taken two at a time and product of its zeroes as 0, –7 and –6 respectively.
Sol. Let the cubic polynomial be ax3 + bx2 + cx + d
⇒ x3 + \(\frac { b }{ a }\)x2 + \(\frac { c }{ a }\)x + \(\frac { d }{ a }\) …(1)
and its zeroes are α, β and γ then
α + β + γ = 0 = \(\frac { -b }{ a }\)
αβ + βγ + γα = – 7 = \(\frac { c }{ a }\)
αβγ = – 6 = \(\frac { -d }{ a }\)
Putting the values of \(\frac { b }{ a }\), \(\frac { c }{ a }\), and \(\frac { d }{ a }\) in (1), we get
x3 – (0) x2 + (–7)x + (–6)
⇒ x3 – 7x + 6
Example 8: If α and β are the zeroes of the polynomials ax2 + bx + c then form the polynomial whose zeroes are \(\frac { 1 }{ \alpha } \quad and\quad \frac { 1 }{ \beta } \)
Since α and β are the zeroes of ax2 + bx + c
So α + β = \(\frac { -b }{ a }\) , α β = \(\frac { c }{ a }\)
Sum of the zeroes = \(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta } =\frac { \alpha +\beta }{ \alpha \beta } \)
\(=\frac{\frac{-b}{c}}{\frac{c}{a}}=\frac{-b}{c}\)
Product of the zeroes
\(=\frac{1}{\alpha }.\frac{1}{\beta }=\frac{1}{\frac{c}{a}}=\frac{a}{c}\)
But required polynomial is
x2 – (sum of zeroes) x + Product of zeroes
\(\Rightarrow {{\text{x}}^{2}}-\left( \frac{-b}{c} \right)\text{x}+\left( \frac{a}{c} \right)\)
\(\Rightarrow {{\text{x}}^{2}}+\frac{b}{c}\text{x}+\frac{a}{c}\)
\(\Rightarrow c\left( {{\text{x}}^{2}}+\frac{b}{c}\text{x}+\frac{a}{c} \right)\)
⇒ cx2 + bx + a