Question

If a and Beta are the zero of the the polynomial x² + 6x + 5 then a² + B² + 2aB =

Answer 1

Answer:

f(x)=x2-6x+k;

by comparing this eq. with ax2+bx+c = 0;

we get a = 1, b = -6  and c = k;

Suppose the roots of the equation is alpha and beta:

alpha + beta = -b/a

alpha+beta = 6/1;

alpha*beta = c/a;

alpha*beta = k

we know that:

(a+b)2 = a2+b2+2ab ;

Not putting the values in above eq.;

(6)2 = 40 + 2k;

36 - 40 = 2k;

-4 = 2k;

k= -2

Step-by-step explanation:

Answer 2

Answer:

The answer is 36.

Step-by-step explanation:

We are given the equation $x^2+6x+5$ whose roots are $\alpha$ and $\beta$

We have to find $\alpha^{2}+2\alpha\beta+\beta^{2}$

Recall the algebraic identity $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$

Hence, $\alpha^{2}+2\alpha\beta+\beta^{2}$ can be written as $(\alpha+\beta)^2$

Here, $\alpha+\beta$ is the sum of zeroes, and the sum of zeroes of a quadratic equation is $\frac{-b}{a}$.

Hence, our required answer must be $(\frac{-b}{a})^2$

That is, $(-6)^2$

Which happens to be 36.

Hence, your required answer is 36.

Hope it helps you.