Question

If alpha and beta are the zeroes of the polynomial x2-6x+k find the value k such that alpha2 + beta2 =40

Answer 1

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Answer 2

Answer:

k=-2

Step-by-step explanation:

The given equation is:

$x^{2}-6x+k$, comparing this equation with $ax^{2}+bx+c=0$, we have a=1, b=-6, c=k.

Now, if α and β arethe two zeroes of the given polynomial, then α+β=$\frac{-b}{a}=6$ and αβ=$\frac{c}{a}$=$k$

Also, it is given that ${\alpha}^{2}+{\beta}^{2}=40$

⇒${\alpha}^{2}+{\beta}^{2}=({\alpha}+{\beta})^{2}-2{\alpha}{\beta}$

⇒$40=(6)^{2}-2k$

⇒$40-36=-2k$

⇒$k=-2$