Question

find the zeroes of 5x²+29x+20 and verify the relationship between the zeroes and the coefficients.​

Answer 1

5x² + 25x + 4x + 20

5x ( x + 5 ) + 4 (x + 5 )

x = -5 or - 4/5

a + b = -29/5

- 5 - 4/5

-29/5 = 20/5

-5 × -4/5 = 20/5

A × b = 20/5

Answer 2

Answer:

The zeros of $5x^2+29x+20$ are $-5$ and $-4/5$. Also, the relation is verified.

Step-by-step explanation:

Consider the polynomial as follows:

$5x^2+29x+20=0$

Here, $a=5$, $b=29$ and $c=20$

Step 1 of 3

Finding the zeros of the polynomial as follows:

$5x^2+25x+4x+20=0$

⇒ $5x(x+5)+4(x+5)=0$

⇒ $(x+5)(5x+4)=0$

⇒ $x=-5$ and $x=-4/5$

Therefore, the zeros are $-5$ and $-4/5$.

Step 2 of 3

Verifying the sum of zeros is equal to $-b/a$, i.e., $\alpha +\beta=-\frac{b}{a}$.

Let $\alpha = -5$ and $\beta=-\frac{4}{5}$. Then,

Sum of zeros = $-\frac{b}{a}$, where $a$ and $b$ are coefficients of polynomial.

⇒ $\alpha+\beta=-\frac{b}{a}$

⇒ $-5+(\frac{-4}{5} )$ = $-\frac{29}{5}$ (Since $a=5$, $b=29$ )

⇒ $\frac{-25-4}{5} = -\frac{29}{5}$

⇒ $-\frac{29}{5} = -\frac{29}{5}$

Hence verified.

Step 3 of 3

Verifying the product of zeros is equal to $c/a$, i.e., $\alpha \beta=\frac{c}{a}$.

Let $\alpha = -5$ and $\beta=-\frac{4}{5}$. Then,

Product of zeros = $\frac{c}{a}$,  where $a$ and $c$ are coefficients of polynomial.

⇒ $\alpha\beta=\frac{c}{a}$

⇒ $(-5)(-\frac{4}{5} )$ = $\frac{20}{5}$ (Since $a=5$, $c=20$ )

⇒ 4 = 4

Hence verified.

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