Question

if alpha and beta are the zeros of polynomial 2 x square - 7 x + 3 find the value of Alpha - beta and alpha square - beta square​

Answer 1

Given :

α & β are the zeros of polynomial 2x² - 7x + 3.

To find :

• Value of α - β and α² - β²

Solution :

• Given polynomial

→ p(x) = 2x² - 7x + 3

• Split middle term

→ 2x² - 6x - x + 3 = 0

→ 2x(x - 3) - 1(x - 3) = 0

→ (x - 3)(2x - 1) = 0

Either

→ x - 3 = 0

→ x = 3

Or

→ 2x - 1 = 0

→ x = ½

•°• 3 & ½ are the zeros of the given polynomial.

Now,

• α = 3
• β = ½

Value of α - β

→ α - β

→ 3 - ½

→ 6 - 1/2

→ 5/2

• Value of α² - β²

→ α² - β²

→ (3)² - (½)²

→ 9 - ¼

→ 36 - 1/4

→ 35/4

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Answer 2

Given :-

If alpha and beta are the zeros of polynomial

$\bf 2x^{2} - 7x +3$

Need to find :-

Value of

$\sf \alpha -\beta$

$\sf \alpha ^2 - \beta^2$

Solution :-

Sum of zeroes

$\bf \alpha +\beta =\dfrac{-b}{a}$

$\sf \alpha +\beta =\dfrac{-(-7)}{2}$

$\sf \alpha +\beta = \dfrac{7}{2}$

Product of zeroes

$\sf \alpha \beta =\dfrac{c}{a}$

$\sf \alpha \beta =\dfrac{3}2$

By using identity

$\sf a^2+b^2=a^2+2ab+b^2$

$\sf \alpha^2 + \beta^2 = \alpha ^2+2\alpha \beta +\beta ^2$

$\sf {\bigg(\dfrac{7}{2}\bigg)}^2 + 2\bigg(\dfrac{3}{2}\bigg)$

$\sf \dfrac{49}{4} - \dfrac{9}4$

$\sf\dfrac{40}4$

$\sf 10$

Now

Finding alpha - beta

By factorization

$\sf 2x^2 - 6x - x +3=0$

$\sf 2x(x-3)-1(x-3)$

Taking x as common

$\sf (x - 3)(2x - 1)$

So,

x = 3

and,

x = 1/2

$\sf 3 -\dfrac{1}{2}$

$\sf \dfrac{6 - 1}{2}$

$\sf\dfrac{5}{2}$