Question

If α and β are the zeroes of the polynomial x2−8x+k such that α2+β2=40, find ′k′.

Answer 1

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

• p(x) = x² - 8x + k
• α² + β² = 40

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

• Value of k.

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

it is given that α and β are the zeroes of the given polynomial.

• »»p(x) = x² - 8x + k
• a = 1
• b = -8
• c = k

≫ Sum of zeroes = - b/a

⠀⠀⠀⠀⠀»» α + β = -(-8)/1

⠀⠀⠀⠀⠀»» α + β = 8

≫ product of zeroes = c/a

⠀⠀⠀⠀⠀»» αβ = k

we know that,

★ (a + b) ² = a² + b² + 2ab

≫ (a + b)² - 2ab = a² + b²

So,

⠀⠀⠀⠀⠀»» (α + β)² - 2ab = a² + b²

⠀⠀⠀⠀⠀»» 8² -2× k = 40

⠀⠀⠀⠀⠀»» 64 - 2k = 40

⠀⠀⠀⠀⠀»» - 2k = 40 - 64

⠀⠀⠀⠀⠀»» - 2k = -24

⠀⠀⠀⠀⠀»» k = -24/-2

⠀⠀⠀⠀⠀»» k = 12

✦ Value of k is 12.

Answer 2

$\huge\underline\mathtt\color{pink}Given:$

• p(x)=x² - 8x + k
• ${ \alpha }^{2} + { \beta }^{2} = 40$

$\huge\underline\mathtt\color{pink}ToFind:$

• The value of K

$\rule{300}{2}$

$\huge\underline\mathtt\color{pink}Solution:$

$\alpha \: and \: \beta$are the zeros of the polynomial x² - 8x + k.

Here, a = 1 ,b = - 8 and c = k

Sum of the zeros

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

$\sf{\implies \alpha + \beta = \frac{ - 8}{1}}$

$\sf{\implies \alpha + \beta = - 8}$

Product of the zeros

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

$\sf{\implies \alpha \beta = \frac{k}{1} }$

$\sf{\implies \alpha \beta = k}$

Now,

$\sf{ ({a + b})^{2} = {a}^{2} + 2ab + {b}^{2} }$

$\sf{\implies {a + b}^{2} - 2ab = {a}^{2} + {b}^{2} }$

$\rule{300}{2}$

The required value of K is

$\sf{ ({a + b})^{2} - 2ab = {a}^{2} + {b}^{2} }$

$\sf{ {\implies - 8}^{2} - 2.k = 40}$

$\sf{\implies 64 - 2k = 40}$

$\sf{\implies - 2k = 40 - 64}$

$\sf{\implies - 2k = - 24}$

$\sf{\implies k = \cancel\frac{ - 24}{ - 2} }$

$\sf{\implies k = 12}$

Hence,the value of k is 12.

$\huge\underline\mathtt\color{pink}ThankYou$