Question

If α and β are the zeros of the quadratic polynomial f(x) = x² - 1 , find a quadratic polynomial whose zeros are 2α/β and 2β/α .

Answer 1

Step-by-step explanation:

We have,

We have,α + β = 24 …… E-1

We have,α + β = 24 …… E-1α – β = 8 …. E-2

We have,α + β = 24 …… E-1α – β = 8 …. E-2By solving the above two equations accordingly, we will get

We have,α + β = 24 …… E-1α – β = 8 …. E-2By solving the above two equations accordingly, we will get2α = 32 α = 16

We have,α + β = 24 …… E-1α – β = 8 …. E-2By solving the above two equations accordingly, we will get2α = 32 α = 16Substitute the value of α, in any of the equation. Let we substitute it in E-2, we will get

We have,α + β = 24 …… E-1α – β = 8 …. E-2By solving the above two equations accordingly, we will get2α = 32 α = 16Substitute the value of α, in any of the equation. Let we substitute it in E-2, we will getβ = 16 – 8 β = 8

We have,α + β = 24 …… E-1α – β = 8 …. E-2By solving the above two equations accordingly, we will get2α = 32 α = 16Substitute the value of α, in any of the equation. Let we substitute it in E-2, we will getβ = 16 – 8 β = 8Now,

We have,α + β = 24 …… E-1α – β = 8 …. E-2By solving the above two equations accordingly, we will get2α = 32 α = 16Substitute the value of α, in any of the equation. Let we substitute it in E-2, we will getβ = 16 – 8 β = 8Now,Sum of the zeroes of the new polynomial = α + β = 16 + 8 = 24

We have,α + β = 24 …… E-1α – β = 8 …. E-2By solving the above two equations accordingly, we will get2α = 32 α = 16Substitute the value of α, in any of the equation. Let we substitute it in E-2, we will getβ = 16 – 8 β = 8Now,Sum of the zeroes of the new polynomial = α + β = 16 + 8 = 24Product of the zeroes = αβ = 16 × 8 = 128

We have,α + β = 24 …… E-1α – β = 8 …. E-2By solving the above two equations accordingly, we will get2α = 32 α = 16Substitute the value of α, in any of the equation. Let we substitute it in E-2, we will getβ = 16 – 8 β = 8Now,Sum of the zeroes of the new polynomial = α + β = 16 + 8 = 24Product of the zeroes = αβ = 16 × 8 = 128Then, the quadratic polynomial is-K

We have,α + β = 24 …… E-1α – β = 8 …. E-2By solving the above two equations accordingly, we will get2α = 32 α = 16Substitute the value of α, in any of the equation. Let we substitute it in E-2, we will getβ = 16 – 8 β = 8Now,Sum of the zeroes of the new polynomial = α + β = 16 + 8 = 24Product of the zeroes = αβ = 16 × 8 = 128Then, the quadratic polynomial is-Kx2– (sum of the zeroes)x + (product of the zeroes) = x2 – 24x + 128

We have,α + β = 24 …… E-1α – β = 8 …. E-2By solving the above two equations accordingly, we will get2α = 32 α = 16Substitute the value of α, in any of the equation. Let we substitute it in E-2, we will getβ = 16 – 8 β = 8Now,Sum of the zeroes of the new polynomial = α + β = 16 + 8 = 24Product of the zeroes = αβ = 16 × 8 = 128Then, the quadratic polynomial is-Kx2– (sum of the zeroes)x + (product of the zeroes) = x2 – 24x + 128Hence, the required quadratic polynomial is f(x) = x2 + 24x + 128

Answer 2

Answer:

We have,α + β = 24 …… E-1

We have,α + β = 24 …… E-1α – β = 8 …. E-2

We have,α + β = 24 …… E-1α – β = 8 …. E-2By solving the above two equations accordingly, we will get

We have,α + β = 24 …… E-1α – β = 8 …. E-2By solving the above two equations accordingly, we will get2α = 32 α = 16

We have,α + β = 24 …… E-1α – β = 8 …. E-2By solving the above two equations accordingly, we will get2α = 32 α = 16Substitute the value of α, in any of the equation. Let we substitute it in E-2, we will get

We have,α + β = 24 …… E-1α – β = 8 …. E-2By solving the above two equations accordingly, we will get2α = 32 α = 16Substitute the value of α, in any of the equation. Let we substitute it in E-2, we will getβ = 16 – 8 β = 8

We have,α + β = 24 …… E-1α – β = 8 …. E-2By solving the above two equations accordingly, we will get2α = 32 α = 16Substitute the value of α, in any of the equation. Let we substitute it in E-2, we will getβ = 16 – 8 β = 8Now,

We have,α + β = 24 …… E-1α – β = 8 …. E-2By solving the above two equations accordingly, we will get2α = 32 α = 16Substitute the value of α, in any of the equation. Let we substitute it in E-2, we will getβ = 16 – 8 β = 8Now,Sum of the zeroes of the new polynomial = α + β = 16 + 8 = 24

We have,α + β = 24 …… E-1α – β = 8 …. E-2By solving the above two equations accordingly, we will get2α = 32 α = 16Substitute the value of α, in any of the equation. Let we substitute it in E-2, we will getβ = 16 – 8 β = 8Now,Sum of the zeroes of the new polynomial = α + β = 16 + 8 = 24Product of the zeroes = αβ = 16 × 8 = 128

We have,α + β = 24 …… E-1α – β = 8 …. E-2By solving the above two equations accordingly, we will get2α = 32 α = 16Substitute the value of α, in any of the equation. Let we substitute it in E-2, we will getβ = 16 – 8 β = 8Now,Sum of the zeroes of the new polynomial = α + β = 16 + 8 = 24Product of the zeroes = αβ = 16 × 8 = 128Then, the quadratic polynomial is-K

We have,α + β = 24 …… E-1α – β = 8 …. E-2By solving the above two equations accordingly, we will get2α = 32 α = 16Substitute the value of α, in any of the equation. Let we substitute it in E-2, we will getβ = 16 – 8 β = 8Now,Sum of the zeroes of the new polynomial = α + β = 16 + 8 = 24Product of the zeroes = αβ = 16 × 8 = 128Then, the quadratic polynomial is-Kx2– (sum of the zeroes)x + (product of the zeroes) = x2 – 24x + 128

We have,α + β = 24 …… E-1α – β = 8 …. E-2By solving the above two equations accordingly, we will get2α = 32 α = 16Substitute the value of α, in any of the equation. Let we substitute it in E-2, we will getβ = 16 – 8 β = 8Now,Sum of the zeroes of the new polynomial = α + β = 16 + 8 = 24Product of the zeroes = αβ = 16 × 8 = 128Then, the quadratic polynomial is-Kx2– (sum of the zeroes)x + (product of the zeroes) = x2 – 24x + 128Hence, the required quadratic polynomial is f(x) = x2 + 24x + 128

Step-by-step explanation: