Math, asked by bhakti007, 11 months ago


If alpha and beta are the zeroes of the quadratic polynomial x^2 - 7x - 18, then find the value of alpha^2+ ß^2?

Answers

Answered by Anonymous
1

Step-by-step explanation:

alpha + beta = 7

alpha beta = -18

alpha² + beta²

= (alpha + beta)² - 2alpha*beta

= (7)² - 2(-18)

= 49 + 36

= 85

Answered by Delta13
3

Given:

  • Polynomial :- x² -7x -18
  • α and β are zeroes of the polynomial.

To find:

The value of α² + β²

Solution:

Comparing with standard form

ax² + bx +c =0

So,

a = 1

b = -7

c = -18

Now,

We know that

 \texttt{Sum of zeroes} =  \frac{ - coefficient \: of \: x}{coefficient \: of \:  {x}^{2} }  \\  \\  \:  \:  \:  \:  \:  \:  \:   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  =  \frac{ - b}{ \: a}

Therefore,

 \alpha  +  \beta  =  \frac{ - ( - 7)}{1}  \\  \\    \implies \: \blue{  \alpha  +  \beta  = 7}

Also,

 \texttt{Product of zeroes} \:  =  \frac{constant \: term}{coefficient \: of \:  {x}^{2} }  \\  \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  =  \frac{c}{a}

So,

 \alpha  \beta  =  \frac{ - 18}{1}  \\  \\  \implies \:  \blue{ \alpha  \beta  =  - 18}

Now using identity we will find the value of a²+b²

(a + b)² = a² + b² +2ab

=> a² + b² = (a + b)² -2ab

Similarly,

α² + β² = ( α+β )² -2αβ

Substituting values

α²+β² = (7)² -2(-18)

= 49 + 36

  \:  \:  \:  \:  \: =  \boxed{ \red{ 85}} \\

Hence, the value of α² + β² = 85

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