Question

if alpha and beta are the roots of x square + 5 x minus 1 is equals to zero then find Alpha Cube + beta cube and Alpha square plus beta square ​

Answer 1

Answer:

(@+B) = -b/a = (-5)

(@*B) = C/a = (-1)

(@²+B²) = (@+B)² - 2@B = (-5)² -2(-1) = 25+2 = 27 (Ans.)

(@³+B³) = (@+B)³ -3@B(@+B) = (-5)³ -3(-1)(-5) = (-125) -15 = (-140) (Ans.)

Answer 2

Correct Question: If $\alpha$ and $\beta$ are the roots of $x^2+5x-1=0$, then find $\alpha^3+\beta^3$ and $\alpha^2+\beta^2$.

Answer:

$\alpha^3+\beta^3=-140$ and $\alpha^2+\beta^2=27$

Step-by-step explanation:

Consider the quadratic polynomial as follows:

$x^2+5x-1=0$

Here, $a=1$, $b=5$ and $c=-1$.

Since $\alpha$ and $\beta$ are the roots of $x^2+5x-1=0$.

As we know,

Sum of roots = $-\frac{b}{a}$

⇒ $\alpha +\beta = -\frac{b}{a}$

⇒ $\alpha +\beta = -\frac{5}{1}$

Simplify as follows:

⇒ $\alpha +\beta = -5$ . . . . . (1)

Now, the product of roots = $\frac{c}{a}$

⇒ $\alpha \beta = \frac{c}{a}$

⇒ $\alpha \beta = \frac{-1}{1}$

⇒ $\alpha \beta = -1$ . . . . . (2)

a) Finding the value of $\alpha^3+\beta^3$.

From (1), we have

$\alpha +\beta = -5$

Cubing both the sides as follows:

⇒ $(\alpha +\beta)^3 = (-5)^3$

Using $(a+b)^3=a^3+b^3+3ab(a+b)$

⇒ $\alpha^3+\beta^3+3\alpha\beta(\alpha+\beta)=-125$

⇒ $\alpha^3+\beta^3+3(-1)(-5)=-125$ (Since $\alpha +\beta = -5$ and $\alpha \beta = -1$)

Further, simplify as follows:

⇒ $\alpha^3+\beta^3+15=-125$

⇒ $\alpha^3+\beta^3=-125-15$

⇒ $\alpha^3+\beta^3=-140$

Therefore, the value of $\alpha^3+\beta^3$ is $-140$.

b) Finding the value of $\alpha^2+\beta^2$.

From (1), we have

$\alpha +\beta = -5$

Squaring both the sides as follows:

⇒ $(\alpha +\beta)^2 = (-5)^2$

Using $(a+b)^2=a^2+b^2+2ab$

⇒ $\alpha^2+\beta^2+2\alpha\beta=25$

⇒ $\alpha^2+\beta^2+2(-1)=25$ (Since $\alpha \beta = -1$)

Further, simplify as follows:

⇒ $\alpha^2+\beta^2-2=25$

⇒ $\alpha^2+\beta^2=25+2$

⇒ $\alpha^2+\beta^2=27$

Therefore, the value of $\alpha^2+\beta^2$ is $27$.

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