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If a nand b are the zeros of the polynomial p(x) = 3x2 - 14x + 15,find the value of A2 + b2 |
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Answer» Answer: The value of \alpha^2+\beta^2=\frac{106}{9}α 2 +β 2 = 9 106
Step-by-step explanation: p(x) = 3x^2 - 14x +15p(x)=3x 2 −14x+15 \alphaα and \betaβ are zeroes General equation : ax^2+bx+c=0ax 2 +bx+c=0 a −b
Product of zeroes = \frac{c}{a} a c
Sum of zeroes of given equation : \alpha+\beta=\frac{14}{3}α+β= 3 14
Product of zeroes of given equation : \alpha\beta =\frac{15}{3}=5αβ= 3 15
=5 we are supposed to find \alpha^2+\beta^2α 2 +β 2
Formula : \begin{gathered}(a+b)^2=a^2+b^2+2ab\\(a+b)^2-2ab=a^2+b^2\end{gathered} (a+b) 2 =a 2 +b 2 +2ab (a+b) 2 −2ab=a 2 +b 2
(\alpha+\beta)^2-2 \alpha \beta=\alpha^2+\beta^2(α+β) 2 −2αβ=α 2 +β 2
Substitute the values: \begin{gathered}(\frac{14}{3})^2-2(5)=\alpha^2+\beta^2\\\\\frac{106}{9}=\alpha^2+\beta^2\end{gathered} ( 3 14
) 2 −2(5)=α 2 +β 2
9 106
=α 2 +β 2
Hence the value of \alpha^2+\beta^2=\frac{106}{9}α 2 +β 2 = 9 106
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