Question

If alpha and beta are the zeroes of polynomial p(x) = x² - 7x + 10, find the quadratic polynomial with zeroes (-alpha) and (-beta).​

Answer 1

Answer:

$\rm = {x}^{2} + 6x + 10$

Explanation:

Given polynomial,

$\rm \implies \: p(x) = {x}^{2} - 7x + 10$

Whose zeros are α and β

We need to find the polynomial whose zeros are -α and -β

Let's find α and β first :-

On comparing p(x) with the general form of a quadratic equation i.e., ax² + bx + c = 0

We get,

• a = 1
• b = -7
• c = 10

By quadratic formula :-

$\rm \implies \: x = \dfrac{ - b \pm \sqrt{ {(b)}^{2} - 4ac } }{2a}$

$\rm \implies \: x = \dfrac{ - ( - 7) \pm \sqrt{ {( - 7)}^{2} - 4(1)(10) } }{2(1)}$

$\rm \implies \: x = \dfrac{7\pm \sqrt{ 49 - 40 } }{2}$

$\rm \implies \: x = \dfrac{7\pm \sqrt{ 9 } }{2}$

$\rm \implies \: x = \dfrac{7\pm 3 }{2}$

$\rm \implies \: x = \dfrac{7 + 3 }{2} \: and \: \dfrac{7 - 3}{2}$

$\rm \implies \: x = \dfrac{10}{2} \: and \: \dfrac{4}{2}$

$\rm \implies \: x = 5 \: and \: 2$

$\rm \therefore \: \alpha = 5 \: and \: \beta = 2$

Now,

New zeros :-

$= - \alpha$

$= - 5$

And also,

$= - \beta$

$= - 2$

The required polynomial is given by,

$\rm = {x}^{2} - (sum \: of \: zeros)x + (product \: of \: zeros)$

$\rm = {x}^{2} - [ ( - 5) + ( - 2)]x + ( - 5)( - 2)$

$\rm = {x}^{2} - [ - 5 - 2]x + 10$

$\rm = {x}^{2} - [ - 6]x + 10$

$\rm = {x}^{2} + 6x + 10$