Question

if alpha and beta are zeroes of 2x^2-4x+5, find:
a) 1/alpha+1/beta
b) $( { \alpha - \beta )}^{2}$
Please quickly help!​

Answer 1

EXPLANATION.

α, β are the zeroes of the polynomial.

⇒ 2x² - 4x + 5.

As we know that,

Sum of the zeroes of the quadratic equation.

⇒ α + β = -b/a.

⇒ α + β = -(-4)/2 = 2.

Products of the zeroes of the quadratic equation.

⇒ αβ = c/a.

⇒ αβ = 5/2.

To find :

(1) = 1/α + 1/β.

⇒ β + α/αβ.

⇒ 2/5/2.

⇒ 2/1 x 2/5 = 4/5.

(2) = (α - β)².

⇒ α² + β² - 2αβ.

As we know that,

Formula of :

⇒ (x - y)² = x² + y² - 2xy.

⇒ (x² + y²) = (x + y)² - 2xy.

Using this formula in the equation, we get.

⇒ (α + β)² - 2αβ - 2αβ.

⇒ (α + β)² - 4αβ.

⇒ (2)² - 4(5/2).

⇒ 4 - 10 = -6.

Value of :

(1) = 1/α + 1/β = 4/5.

(2) = (α - β)² = -6.

                                                                                                                           

MORE INFORMATION.

Nature of the roots of the quadratic expression.

(1) = Real and unequal, if b² - 4ac > 0.

(2) = Rational and different, if b² - 4ac is a perfect square.

(3) = Real and equal, if b² - 4ac = 0.

(4) = If D < 0 Roots are imaginary and unequal Or complex conjugate.

                                                     

Answer 2

Given :-

2x² - 4x + 5

To Find :-

1/α + 1/β

(α - β)²

Solution :-

We know that

Sum of zeroes = -b/a

Here

b = -4

a = 2

-(-4)/2

4/2

2/1

2

Product of zeroes = c/a

c = 5

a = 2

5/2

Now

1/α + 1/β = α + β/αβ

= 2/(5/2)

= 2/5 × 2/1

= 4/5

b)

We know that

(α - β)² = (α + β)² - 4αβ

(2)² - 4(5/2)

4 - 20/2

8 - 20/2

-12/2

-6

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