Question

if alpha and beta are the zeros of polynomial p(X)=3x^2-14x+15 find the alpha^2 +beta^2.​

Answer 1

alpha²+beta²=(alpha+beta) ²-2alphabeta

sum of roots=-b/a

product of roots=c/a

therefore,

alpha²+beta²=(14/3)²-2*15/3

alpha²+beta²=196/9 -10

alpha²+beta²=106/9

hope it helps

:-)

Answer 2

The value of $\alpha^2+\beta^2=\frac{106}{9}$

Step-by-step explanation:

Polynomial :  $p(x) = 3x^2 - 14x +15$  whose $\alpha$ and $\beta$ are zeroes .

General equation : $ax^2+bx+c=0$

Where,

Sum of zeroes =  $-\frac{b}{a}$

Product of zeroes =  $\frac{c}{a}$

Sum of zeroes of given equation : $\alpha+\beta=\frac{14}{3}$

Product of zeroes of given equation : $\alpha\beta =\frac{15}{3}=5$

Formula used,

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

The identity form is

$(\alpha+\beta)^2-2 \alpha \beta=\alpha^2+\beta^2$

Substitute the values,

$(\frac{14}{3})^2-2(5)=\alpha^2+\beta^2\\\\\frac{106}{9}=\alpha^2+\beta^2$

Therefore, the value of $\alpha^2+\beta^2=\frac{106}{9}$

#Learn more:

If alpha and beta are the zeroes of the polynomial p(x) = 3x2 - 14x +15 ,find the value of alpha square + beta square

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