Question

if alpha and beta are zeroes of the quadratic polynomial f(x) = x2+x-2 then find a polynomial whose zeroes are 2alpha + 1 and 2beta + 1​

Answer 1

Solution :-

f(x) = x² + x - 2

To find zeroes equate the given polynomial to 0

x² + x - 2 = 0

Splitting the middle term

x² + 2x - x - 2 = 0

x(x + 2) - 1(x + 2) = 0

(x + 2)(x - 1) = 0

x + 2 = 0 or x - 1 = 0

x = - 2 or x = 1

Zeroes of the polynomial x² + x - 2 are - 2 , 1

So

• α = - 2

• β = 1

Finding a polyinomial whose zeroes are (2α + 1) and (2β + 1)

2α + 1

= 2(-2) + 1

= - 4 + 1

(2α + 1)= - 3

2β + 1

= 2(1) + 1

= 2 + 1

(2β + 1) = 3

The new polynomial zeroes should be - 3 and 3

Here

• α = - 3

• β = 3

Sum of zeroes = α + β = - 3 + 3 = 0

Product of zeroes = αβ = - 3(3) = - 9

Quadratic polynomial ax² + bx + c = k{x² - x(α + β) + αβ}

(Where k ≠ 0)

= k{x² - x(0) + (-9)}

= k(x² - 0 - 9)

= k(x² - 9)

When k = 1

= 1(x² - 9)

= x² - 9

Therefore the polynomial is x² - 9.

Answer 2

Answer:

${\bold{\therefore Quadratic=x^{2}-9}}$

Step-by-step explantion:

${\huge{\underline{\mathfrak{SOLUTION}}}}$

• In the given question information given about a quadratic whose zeroes are alpha and beta.

• We have to fimd the quadratic form by new zeroes.

$\underline \bold{Given : } \\ \implies \alpha \: and \: \beta \:are \: zeroes \: of \: f(x) \\ \implies f(x) = {x}^{2} + x - 2 \\ \\ \underline \bold{To \: Find : } \\ \bold{Form \: quadratic \: by} \\ \implies( 2 \alpha + 1) \: and \: (2 \beta + 1)$

• According to given question :

$\bold{Middle \: term \: spliting} \\ \implies {x}^{2} + x - 2 = 0 \\ \implies {x}^{2} + 2x - x - 2 = 0 \\ \implies x(x + 2) - 1(x + 2) = 0 \\ \implies (x - 1)(x + 2) \\ \bold{\implies \alpha = 1 }\\ \bold{\implies \beta = - 2} \\ \\ \bold{zeroes \: of \: another \: quadratic : }\\ \bold{First \: zeroes} \\ \implies 2 \alpha + 1 \\ \implies 2 \times (1) + 1 \\ \implies \: 3 \\ \\ \bold{Second \: zeroes} \\ \implies 2 \beta + 1 \\ \implies 2 \times ( - 2) + 1 \\ \implies - 3 \\ \\ \bold{sum \: of \: zeroes} \\ \implies \alpha + \beta = 3 - 3 \\ \implies \alpha + \beta = 0 \\ \\ \bold{Product \: of \: zeroes} \\ \implies \alpha \beta = 3 \times - 3 \\ \implies \alpha \beta = - 9 \\ \\ \bold{quadratic \: from} \\ \implies {x}^{2} - (Sum \: of \: zeroes) x + (Product \: of \: zeroes) \\ \implies {x}^{2} - 0 \times x + ( - 9) \\ \bold{\implies {x}^{2} - 9 } \\ \\ \bold{\therefore Quadratic \: form = {x}^{2} - 9}$