Question

If alpha , beta are the zeroes of the quadratic polynomial x²+5x-10,then find the value of alpha²+beta²..plzz answer with proper explanation!!

Answer 1

Answer:

sum of zeroe = -b/a = -5

product of zero = c/a = -10

(a+b)² = (a)²+(b)²+2ab

(-5)² = a²+b²+2(-10)

25= a²+b²-20

25+20 =a²+b²

55= a²+b²

hence value of a²+b² is 55

hope this is helpful

Answer 2

Answer:

The value of $\alpha^2+\beta^2$ is 45.

Step-by-step explanation:

Step 1 of 2

It is given that $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $x^2+5x-10$.

Consider the quadratic polynomial as follows:

$x^2+5x-10=0$

Here, $a=1$, $b=5$ and $c=-10$

Recall the relation between the roots of the quadratic polynomial and their coefficients,

Sum of zeros = -b/a

⇒ $\alpha+\beta=-\frac{b}{a}$ . . . . . (1)

Product of zeros = c/a

⇒ $\alpha\beta=\frac{c}{a}$ . . . . . (2)

Step 2 of 2

To find: The value of $\alpha^2+\beta^2$.

From (2), we have

⇒ $\alpha\beta=\frac{-10}{1}$ (Since $c=-10$ and $a=1$)

⇒ $\alpha\beta=-10$ . . . . . (3)

From (1), we have

⇒ $\alpha+\beta=-\frac{5}{1}$ (Since $b=5$ and $a=1$)

⇒ $\alpha+\beta=-5$

Squaring both the sides, we get

⇒ $(\alpha+\beta)^2=(-5)^2$

Simplify using the identity, $(a+b)^2=a^2+b^2+2ab$ as follows:

⇒ $\alpha^2+\beta^2+2\alpha\beta=25$

⇒ $\alpha^2+\beta^2+2(-10)=25$ (From (3))

⇒ $\alpha^2+\beta^2-20=25$

⇒ $\alpha^2+\beta^2=25+20$

⇒ $\alpha^2+\beta^2=45$

Therefore, the value of $\alpha^2+\beta^2$ is 45.

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