Question

alpha and beta are zeros of x square + 3 x minus 10 then Alpha square plus beta​

Answer 1

$\large\bf\underline {To \: find:-}$

• We need to find the value of α² + β²

$\large\bf\underline{Given:-}$

• p(x) = x² + 3x - 10

$\huge\bf\underline{Solution:-}$

p(x) = x² + 3x - 10

a = 1

b = 3

c = -10

≫Sum of zeroes = -b/a

➛ α + β = -3/1

➛ α + β = -3 ........1)

≫ Product of zeroes = c/a

➛ αβ = -10/1

➛ αβ = -10 ......2)

We know that,

➻ (a + b)² = a² + b² + 2ab

➻ a² + b² = (a + b)² - 2ab

Now, finding value of α² + β²

➺ α² + β² = (α + β)² - 2αβ

putting value of α + β and αβ from 1) and 2)

➺ α² + β² = (-3)² - 2 × -10

➺ α² + β² = 9 + 20

➺ α² + β² = 29

hence,

✍ Value of α² + β² is 29

$\rule{200}3$

Answer 2

$\blue{\bold{\underline{QueStion}}}$

$\alpha \: and \ \: \beta \: are \: zeroes \: of \: {x}^{2} + 3x - 10 \: then \: { \alpha }^{2} + { \beta }^{2}$

$\blue{\bold{\underline{AnsWer}}}$

29

$\blue{\bold{\underline{To FiNd}}}$

${ \alpha }^{2} + { \beta }^{2}$

$\blue{\bold{\underline{GiVen}}}$

quadratic equations is x²+3x-10

general formula for quadratic equations is ax²+by+c=0

• a=1
• b=3
• c=-10

$\blue{\bold{\underline{SolUtion}}}$

sum of polynomial=-b/a

$\alpha + \beta = \frac{ - b}{a}$

$\alpha + \beta = \frac{ - 3}{1}$

$\alpha + \beta = - 3$

product of polynomial=c/a

$\alpha \beta = \frac{c}{a}$

$\alpha \beta = \frac{ - 10}{1}$

$\alpha \beta = - 10$

Now,

by using:

${ \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta )}^{2} - 2 \alpha \beta$

${ \alpha }^{2} + { \beta }^{2} = {( - 3)}^{2} - 2( - 10)$

${ \alpha }^{2} + { \beta }^{2} = 9 + 20$

${ \alpha }^{2} + { \beta }^{2} = 29$

$\red{\bold{\underline{\underline{Important}}}}$

${ \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta )}^{2} - 2 \alpha \beta$

${( \alpha - \beta )}^{2} = {( \alpha + \beta )}^{2} - 4 \alpha \beta$