Question

If alpha beta are the zeros of polynomial 2 y square + 7y + 5 is equals to zero write the value of Alpha square plus beta square

Answer 1

Given :-

alpha and beta are the zeroes of the polynomial 2y² + 7y + 5.

here, we've to find the value of alpha² + beta²

so first of all, we needa find the zeroes of the given polynomial.

by splitting the middle term method, we get

➡ 2y² + 7y + 5 = 0

➡ 2y² + 5y + 2y + 5 = 0

➡ y(2y + 5) + 1(2y + 5) = 0

➡ (2y + 5) (y + 1) = 0

» y = -5/2 , y = -1

therefore,

• alpha = -5/2

• beta = -1

hence, the value of alpha² + beta²

= (-5/2)² + (-1)²

= (-5 × -5)/(2 × 2) + (-1 × -1)

= 25/4 + 1

taking LCM of 4 and 1, we get

= 25/4 + 4/4

= 29/4 or 7.25 FINAL ANSWER

Answer 2

$\huge\bf{Answer:-}$

The answer answer is 29/4 (or) 7.25

• 2y² + 7y + 5 is the given alpha and beta zeroes of the polynomial.Let me add up alpha² + beta² to get the value .Then, now I will in the zeros of the polynomial

Steps:-

$= > \: (2y {}^{2} + 7y + 5 = 0)$

$= > \: (2y {}^{2} + 5y + 2y + 5 = 0)$

$= > \: y(2y + 5) + 1(2y + 5) = 0 \\ \\ </p><p> = \: > (2y + 5) (y + 1) = 0 \\ \\ </p><p> = > \: (y = \frac{ - 5}{2} ) \: and \: ( y = -1)$

Hence, finding the alpha and beta is -5/2 is the alpha I got and = -1 is the beta is hot.

So, the Value I got form (alpha² + beta²)

$= > ( \frac{ - 5}{2}) {}^{2} + (-1) {}^{2}$

$= > \: \frac{(-5 × -5)}{(2 × 2)} + (-1 × -1) \\ </p><p>= > \: \frac{25}{4} + 1$

Now I take the least common multiple for 4 and 1, we get

$= > ( \frac{ 25}{4})+\frac{4}{4})$

Therefore,The answer answer is

$\frac{ 29}{4}$(or) 7.25