Question

If α and β are the zeros of polynomial x² - 7x + 10, then α² + β² =​

Answer 1

Answer:

The value of α² + β² is 29

Step-by-step explanation:

Given polynomial = x² – 7x + 10

Zeros of the polynomial are α and β

According to the relationship between zeros and coefficients,

• a = 1
• b = (–7)
• c = 10

$\sf{\longrightarrow} \: \alpha + \beta = \dfrac{ - b}{a}$

$\sf{\longrightarrow} \: \alpha + \beta = \dfrac{ - ( - 7)}{1}$

$\sf{\longrightarrow} \: \alpha + \beta = 7$

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

$\sf{\longrightarrow} \: \alpha \beta = \dfrac{10}{1}$

$\sf{\longrightarrow} \: \alpha \beta = 10$

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Value of α² + β²,

$\sf{\longrightarrow} \: { \alpha }^{2} + { \beta }^{2}$

$\sf{\longrightarrow} \: ( { \alpha + \beta )}^{2} - 2 \alpha \beta$

$\sf{\longrightarrow} \: (7)^{2} - 2(10)$

$\sf{\longrightarrow} \:49 - 20$

$\sf{\longrightarrow} \:29$

α² + β² = 29

Therefore, the value of α² + β² is 29

Answer 2

Step-by-step explanation:

Given :-

$\\ \tt{:}\Rrightarrow \alpha \:and\:\beta\:are\:zeros\:of\:polynomial \:x^2-7x+10=0$

To find:-

$\\ \tt{:}\Rrightarrow \alpha^2+\beta^2$

Solution:-

here

• a=1
• b=-7
• c=10

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

$\\ \tt{:}\Rrightarrow \dfrac{-(-7)}{1}$

$\\ \tt{:}\Rrightarrow \dfrac{7}{1}$

$\\ \tt{:}\Rrightarrow 7$

and,

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

$\\ \tt{:}\Rrightarrow \dfrac{10}{1}$

$\\ \tt{:}\Rrightarrow 10$

Now,

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

$\\ \tt{:}\Rrightarrow (7)^2=\alpha^2+\beta^2+2(10)$

$\\ \tt{:}\Rrightarrow -(\alpha^2+\beta^2)=20-49$

$\\ \tt{:}\Rrightarrow -(\alpha^2+\beta^2)=-29$

$\\ \tt{:}\Rrightarrow \underline{\boxed{\boldsymbol{\alpha}^2+\boldsymbol{\beta}^2=29}}$