Question

if alpha and beta are zeroes of quadratic polynomial ax²+bx+c, evaluate
(i)
$\alpha^{2} + \beta^{2}$
(ii)
$\alpha \div \beta \times \beta \div \alpha$
(iii)
$\alpha ^{3} + \beta {}^{3}$
​

Answer 1

Answer:

i) α² + ß² = b²/a² - 2c/a

ii) α/ß + ß/α = b²/ac - 2

iii) α³ + ß³ = 3bc/a² - b³/a³

Note:

★ The possible values of variable for which the polynomial becomes zero are called its zeros.

★ A quadratic polynomial can have atmost two zeros.

★ If α and ß are the zeros of the quadratic polynomial ax² + bx + c , then ;

Sum of zeros , (α + ß) = -b/a

Product of zeros , αß = c/a

Solution:

Here,

The given quadratic polynomial is :

ax² + bx + c .

Also,

α and ß are the zeros of the given quadratic polynomial , thus ;

α + ß = -b/a

αß = c/a

Now,

=> (α + ß)² = α² + ß² + 2αß

=> α² + ß² = (α + ß)² - 2αß

= (-b/a)² - 2(c/a)

= b²/a² - 2c/a

Hence,

α² + ß² = b²/a² - 2c/a

Now,

α/ß + ß/α = (α² + ß²)/αß

= (b²/a² - 2c/a)/(c/a)

= (b²/a² - 2c/a)×(a/c)

= b²/ac - 2

Hence,

α/ß + ß/α = b²/ac - 2

Now ,

α³ + ß³ = (α + ß)(α² - αß + ß²)

= (α + ß)(α² + 2αß + ß² - 3αß)

= (α + ß)[ (α + ß)² - 3αß ]

= (-b/a)[ (-b/a)² - 3(c/a) ]

= (-b/a)(b²/a² - 3c/a)

= -b³/a³ + 3bc/a²

= 3bc/a² - b³/a³

Hence,

α³ + ß³ = 3bc/a² - b³/a³