Question

find the zeros of the quadratic polynomial 3x^2-75and verify the relation between the zeros and coefficients​

Answer 1

Step by step solution :

STEP

1

:

Equation at the end of step 1

3x2 - 75 = 0

STEP

2

:

STEP

3

:

Pulling out like terms

3.1 Pull out like factors :

3x2 - 75 = 3 • (x2 - 25)

Trying to factor as a Difference of Squares:

3.2 Factoring: x2 - 25

Theory : A difference of two perfect squares, A2 - B2 can be factored into (A+B) • (A-B)

Proof : (A+B) • (A-B) =

A2 - AB + BA - B2 =

A2 - AB + AB - B2 =

A2 - B2

Note : AB = BA is the commutative property of multiplication.

Note : - AB + AB equals zero and is therefore eliminated from the expression.

Check : 25 is the square of 5

Check : x2 is the square of x1

Factorization is : (x + 5) • (x - 5)

Equation at the end of step

3

:

3 • (x + 5) • (x - 5) = 0

STEP

4

:

Theory - Roots of a product

4.1 A product of several terms equals zero.

When a product of two or more terms equals zero, then at least one of the terms must be zero.

We shall now solve each term = 0 separately

In other words, we are going to solve as many equations as there are terms in the product

Any solution of term = 0 solves product = 0 as well.

Equations which are never true:

4.2 Solve : 3 = 0

This equation has no solution.

A a non-zero constant never equals zero.

Solving a Single Variable Equation:

4.3 Solve : x+5 = 0

Subtract 5 from both sides of the equation :

x = -5

Solving a Single Variable Equation:

4.4 Solve : x-5 = 0

Add 5 to both sides of the equation :

x = 5

Two solutions were found :

x = 5

x = -5

hope this helps you...plzz Mark as BRAINLIEST answer

Answer 2

Answer:

Step-by-step explanation:

$3x^{2} - 75=0$

$3 [ x^{2} -25] = 0$

$3[ x^{2} - 5^{2} ]$$3[x+5][x-5]$

$[x+5][x-5]$ = 0

either,

x+5 = 0

x = -5

OR

x-5 = 0

x = 5

Verification,

$sum of zeros = \frac{-b}{a}$

$5+[-5] = \frac{0}{3}$

0 = 0

$product of zeros = \frac{c}{a}$

$5 * [-5] = \frac{-75}{3}$

$-25 = -25$

hope it helps .

plz mark as brainliest