Question

if α and β are two zeros of the quadratic polynomial
$2 {x }^{2} - 3x + 7$
,find
a)
$\frac{1}{ \alpha } + \frac{1}{ \beta }$
b)
$\alpha {}^{2} + \beta {}^{2}$
​

Answer 1

$\huge\bold\red{ANSWER}$

2x²-3x+7

HERE a=2,b=-3,c=7

D=b²-4ac

=>(-3)²-4×2×7

=>9-56

=>47

by quadratic formula

x= -b+-√D/2a

6+-√47/4

alpha=6+√47/4

beta=6-√47/4

$\frac{1}{ \alpha } + \frac{1}{ \beta }$

????????????????

yaar aage ka kud solve karlo please type nhi kar paunga

$\huge\fbox\bold\pink{DIVYANSH}$

Answer 2

Step-by-step explanation:

Given -

• α and β are zeroes of polynomial p(x) = 2x² - 3x + 7

To Find -

• Value of 1/α + 1/β and α² + β²

Now,

As we know that :-

• αβ = c/a

→ αβ = 7/2

And

• α + β = -b/a

→ α + β = -(-3)/2

→ α + β = 3/2

Squaring both sides :-

(α + β)² = (3/2)²

→ α² + 2αβ + β² = 9/4

→ α² + β² = 9/4 - 2×7/2

→ α² + β² = 9/4 - 7

→ α² + β² = 9-28/4

→ α² + β² = -19/4

And

1/α + 1/β

→ β + α/αβ

→ 3/2 ÷ 7/2

→ 3/2 × 2/7

→ 3/7

Hence,

The value of α² + β² is -19/4

and

value of 1/α + 1/β is 3/7