Question

form the quadratic polynomial whose one of zero is 8 and the product of the zeros is -56

Answer 1

Concept

The polynomial equations with degree 2 in one variable are said to be quadratic equations .

Given

1) One zero is 8

2) Product of zeroes = -56

Find

Form the quadratic equation

Solution

Let the other zero be x

One zero is 8

Second zero is x

Product of zeroes = -56

8x = -56

x = -56/8

x = -7

Second zero is -7

Quadratic equation is

(x-a)(x-b) = 0 where a and b are the zeroes of the equation

(x-8)(x+7) = 0

x² - 8x + 7x -56 = 0

x² - x + 56 = 0

The quadratic equation whose one zero is 8 and product of zeroes is -56 is x² - x + 56 .

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Answer 2

Answer:

The required quadratic polynomial is $x^2-x-56=0$.

Step-by-step explanation:

Let the zeros of the quadratic polynomial $ax^2+bc+c=0$ be $\alpha$ and $\beta$.

Quadratic polynomial is also written in the form,

$x^2+(\alpha+\beta)x+\alpha\beta=0$ . . . . . (1)

Step 1 of 2

It is given that the one of the zero of quadratic polynomial is 8.

So, let $\alpha=8$.

And the product the zeros is -56.

So, $\alpha\beta=-56$

⇒ $8\times \beta=-56$

⇒ $\beta = -56/8$

⇒ $\beta = -7$

Thus, $\alpha=8$ and $\beta=-7$.

Step 2 of 2

Sum of zeros,

= $\alpha+\beta$

= 8 + (-7)

= 1

Product of zeros, $\alpha\beta$ = -56 (Given)

Substitute the values 1 for $\alpha+\beta$ and -56 for $\alpha\beta$ in the equation (1) as follows:

⇒ $x^2+(1)x+(-56)=0$

⇒ $x^2-x-56=0$

Therefore, the required quadratic polynomial is $x^2-x-56=0$.

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