Question

If alpha and beta are zeros of x²-x-2, find a polynomial whose zeros are 3alpha+1 and 3beta+1

Answer 1

P(x) = x² - x -2

0 = x² - x - 2

Here a = 1 , b = -1 and c = -2

D = b² - 4ac

D = (-1)² - 4 × 1 × -2

D = 1 + 8

D = 9

$\boxed{\sf x = \frac{ - b \: \ \pm \sqrt{discriminant} }{2a} }$

$\alpha = \frac{ - ( - 1) + \sqrt{9} }{2 \times 1}$

$\implies\alpha = \frac{1 + 3}{2}$

$\implies \: \alpha = \frac{4}{2}$

$\implies \: \boxed{ \sf\alpha = 2}$

$\beta = \frac{ - ( - 1) - \sqrt{9} }{2 \times 1}$

$\implies \: \beta = \frac{1 - 3}{2}$

$\implies \: \beta = \frac{ - 2}{2}$

$\implies \: \boxed{ \sf{ \beta = - 1}}$

$3 \alpha + 1 = 3 \times 2 + 1 \\ = 6 + 1 \\ = 7$

$3 \beta + 1 = 3( - 1) + 1 \\ = - 3 + 1 \\ = - 2$

So , the roots of the new equation should be 7 and -2 .

We know that for a quadratic equations ,

P(x) = x² - (sum of the roots)x + product of roots

So , our new quadratic equation is

P(x) = x² - {7 + (-2)}x + 7× -2

P(x) = x² - (7- 2)x - 14

P(x) = x² - 5x -14

So , it can be written as the type

P(x) = k (x² - 5x -14)

Where k is a constant .

Answer 2

Given

$\sf \alpha\ and \beta\ are \ the \ zeroes \ of \ polynomial \\ \\ \longrightarrow\sf x^2-x-2$

To find :-

we have to find a polynomial whose zeroes of

$\sf (3 \alpha+1 )\ and \ (3\beta+1)$

• Here $\sf \alpha\ and \ \beta$ are the zeroes of given polynomial !

• Now first find the zeroes of given polynomial .

$\underline{\bigstar{\sf\ \ \ \ \ \ Solution :-\ \ \ \ \ }}$

$\dashrightarrow\sf x^2-x-2=0\\ \\ \dashrightarrow\sf x^2-(2-1)x-2=0\\ \\ \dashrightarrow\sf x^2-2x+x-2=0\\ \\ \dashrightarrow\sf x(x-2)+1(x-2)=0\\ \\ \dashrightarrow\sf (x-2)(x+1)\\ \\ \sf \ zeroes\ are \ (2) \ and \ (-1)\\ \\ \sf\ So ,\ \alpha= 2 \ ; \beta= (-1)$

• Now find the value of given zeroes

$\dashrightarrow\sf 3\alpha+1= 3\times 2+1 \ \ \ (\alpha=2)\\ \\ \dashrightarrow\sf 3\alpha+1 =6+1= 7 \\ \\ \sf \ \ Now \ the \ second\ zero \\ \\ \dashrightarrow\sf 3\beta+1= 3\times (-1)+1 \ \ \ [\beta=(-1)]\\ \\ \dashrightarrow\sf 3\beta= -3+1 = (-2)$

• Here we can say that

• $\sf \alpha= 7 \ \ and \ \beta=(-2)$

Hence required polynomial !

$\sf f(x)= k(x^2-(sum\ of \ zeroes) +(product\ of \ zeroes))$

• Where k is constant
• and f(x) is required Polynomial

$\dashrightarrow\sf f(x) = k\big(x^2- [7+(-2)] x+(7\times (-2)]\big)\\ \\ \dashrightarrow\sf k(x^2-(7-2)x-14)\\ \\ \dashrightarrow\sf k(x^2-5x-14)$

$\boxed{\sf{\purple{ f(x)= k(x^2-5x-14)}}}$