if alfa and beta are zeros of polynomial p(x)=3x²-10x+7find value of alfa²+beta²
Answers
Step-by-step explanation:
let a and b are two zeros of polynomial
a+b=10/3
ab=7/3
squaring (a+b) on both sides
a^2+b^2+2ab=100/9
a^2+b^2=100/9-14/3=100-42/9=58/9
Answer:
plzzzz mark this as the brainliest answer
Step-by-step explanation:
Product of zeros = αβ= constant / coefficient of x^2 = 4/k
Sum of zeros =α+β = - coefficient of x / coefficient of x^2= -4/k
Given
(\alpha)^2 + (\beta)^2 = 24(α)
2
+(β)
2
=24
(\alpha)^2 + (\beta)^2(α)
2
+(β)
2
can be written as (\alpha)^2 + 2(\alpha)(\beta) + (\beta)^2(α)
2
+2(α)(β)+(β)
2
if we add \pm 2 (\alpha)(\beta)±2(α)(β) in the above equation.
(\alpha)^2 + 2(\alpha)(\beta) + (\beta)^2 -2(\alpha)(\beta)(α)
2
+2(α)(β)+(β)
2
−2(α)(β)
(\alpha + \beta)^2 -2(\alpha)(\beta)(α+β)
2
−2(α)(β)
Putting values of αβ and α+β
\begin{gathered}(\frac {-4}{k})^2 -2( \frac {4}{k}) = 24\\\frac {16}{k^2} - \frac {8}{k} = 24\\Multiplying\,\, the \,\, equation\,\, with\,\, 8K^2\\ 2 - k= 3K^2\\3k^2-2+k=0\\or\\3k^2+k-2=0\\3k^2+3k-2k-2=0\\3k(k+1)-2(k+1)=0\\(3k-2)(k+1)=0\\3k-2=0 \,\,and\,\, k+1 =0\\k= 2/3 \,\,and\,\, k=-1\end{gathered}
(
k
−4
)
2
−2(
k
4
)=24
k
2
16
−
k
8
=24
Multiplyingtheequationwith8K
2
2−k=3K
2
3k
2
−2+k=0
or
3k
2
+k−2=0
3k
2
+3k−2k−2=0
3k(k+1)−2(k+1)=0
(3k−2)(k+1)=0
3k−2=0andk+1=0
k=2/3andk=−1
The values of k are 2/3 and -1