Question

If and are the zeroes of the polynomical p(x) = 2 x square - 3 x + 1 then find

the value of ( Alpha upon beta + beta upon Alpha
can you explain me please​

Answer 1

Answer:

5/2

Step-by-step explanation:

p(x) = 2x^2 - 3x + 1

(it is of the form ax^2+bx+c)

alpha and beta are the roots of p(x).

Therefore, alpha + beta = -b/a = - (-3)/2 = 3/2

alpha × beta = c/a = 1/2

Now, alpha/beta + beta/alpha

= ( alpha^2 + beta^2) / (alpha × beta)

= ( (alpha + beta)^2 - 2alpha.beta ) / (alpha × beta)

= ( 9/4 -1 ) / (1/2)

= 5/4 × 2

= 5/2

Hope it helps you.

Answer 2

Step-by-step explanation:

Let alpha=a and Beta=b.

Given:. a and b are zeroes of the polynomial-

2x²-3x+1.

To prove:

$\frac{a}{b} + \frac{b}{a} = ?$

By factorisation method:

2x²-3x+1=0

2x²-2x-x+1=0

2x(x-1)-(x-1)=0

(x-1)(2x-1)=0

x=1,

x=½.

Now,

(a+b)=-b/a=-(-3/1)=3.

ab=c/a=1/1=1.

$\frac{ a}{b} + \frac{b}{a} = > \frac{ {a}^{2} + {b}^{2} }{ab} = > {a}^{2} + {b }^{2} /ab= > \\ {(a + b)}^{2} - 2ab$ /ab

$= > {(1 + \frac{1}{2}) }^{2} - 2 \times 1 \times \frac{1}{2}$/1/2

$= > {( \frac{2 + 1}{2}) }^{2} - 1$/1/2

$= > {( \frac{3}{2}) }^{2} - 1 /1/2= > \frac{9}{4} - 1 = > \frac{9 - 4}{4}\times 2$

$= > \frac{5}{4}\times2$

=>5/2.