Question

find the number of zeros √3x^2+20x+7√3​

Answer 1

Solve Quadratic Equation by Completing The Square

4.2 Solving 3x2-20x-7 = 0 by Completing The Square .

Divide both sides of the equation by 3 to have 1 as the coefficient of the first term :

x2-(20/3)x-(7/3) = 0

Add 7/3 to both side of the equation :

x2-(20/3)x = 7/3

Now the clever bit: Take the coefficient of x , which is 20/3 , divide by two, giving 10/3 , and finally square it giving 100/9

Add 100/9 to both sides of the equation :

On the right hand side we have :

7/3 + 100/9 The common denominator of the two fractions is 9 Adding (21/9)+(100/9) gives 121/9

So adding to both sides we finally get :

x2-(20/3)x+(100/9) = 121/9

Adding 100/9 has completed the left hand side into a perfect square :

x2-(20/3)x+(100/9) =

(x-(10/3)) • (x-(10/3)) =

(x-(10/3))2

Things which are equal to the same thing are also equal to one another. Since

x2-(20/3)x+(100/9) = 121/9 and

x2-(20/3)x+(100/9) = (x-(10/3))2

then, according to the law of transitivity,

(x-(10/3))2 = 121/9

We'll refer to this Equation as Eq. #4.2.1

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of

(x-(10/3))2 is

(x-(10/3))2/2 =

(x-(10/3))1 =

x-(10/3)

Now, applying the Square Root Principle to Eq. #4.2.1 we get:

x-(10/3) = √ 121/9

Add 10/3 to both sides to obtain:

x = 10/3 + √ 121/9

Since a square root has two values, one positive and the other negative

x2 - (20/3)x - (7/3) = 0

has two solutions:

x = 10/3 + √ 121/9

or

x = 10/3 - √ 121/9

Note that √ 121/9 can be written as

√ 121 / √ 9 which is 11 / 3

Solve Quadratic Equation using the Quadratic Formula

4.3 Solving 3x2-20x-7 = 0 by the Quadratic Formula .

According to the Quadratic Formula, x , the solution for Ax2+Bx+C = 0 , where A, B and C are numbers, often called coefficients, is given by :

- B ± √ B2-4AC

x = __________________

2A

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Answer 2

Answer:

x:7/9 because the zeroes are much than its whole wssigned