QUADRATIC EQUATIONS PART TWO

QUADRATIC EQUATIONS PART TWO

9th - 11th Grade

12 Qs

quiz-placeholder

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QUADRATIC EQUATIONS PART TWO

QUADRATIC EQUATIONS PART TWO

Assessment

Quiz

Mathematics, Specialty

9th - 11th Grade

Hard

Created by

Karoll Frenz Romeo Papna-Jardeleza

Used 12+ times

FREE Resource

12 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

45 sec • 1 pt

What polynomial are we aiming in the third step of completing the square method?

Perfect Square Monomial

Perfect Square Binomial

Perfect Square Trinomial

Perfect Square Multinomial

2.

MULTIPLE CHOICE QUESTION

45 sec • 1 pt

The first step in completing the square method is

Transposing the constant term to the right side of the equation.

Dividing each term of the equation by the numerical coefficient of the quadratic term.

Applying the Square Root Property for both sides of the equation.

Dividing the coefficient of the linear term by 2, squaring it, then adding the answer to both sides of the equation.

3.

MULTIPLE CHOICE QUESTION

45 sec • 1 pt

What formula is used to complete the trinomial into a perfect square trinomial?

b2a\frac{b}{2a}

ba\frac{b}{a}

(b2a)2 \left(\frac{b}{2a}\right)^{2\ }

(ba)2\left(\frac{b}{a}\right)^2

4.

MULTIPLE CHOICE QUESTION

45 sec • 1 pt

Excluding the checking, how many steps do completing the square method have?

3

4

5

6

5.

MULTIPLE CHOICE QUESTION

45 sec • 1 pt

Where is quadratic formula derived from?

Derivation

Completing the Square

Factoring

Change of Polynomials

6.

MULTIPLE CHOICE QUESTION

45 sec • 1 pt

The second step in completing the square method is

Transposing the constant term to the right side of the equation.

Dividing each term of the equation by the numerical coefficient of the quadratic term.

Applying the Square Root Property for both sides of the equation.

Dividing the coefficient of the linear term by 2, squaring it, then adding the answer to both sides of the equation.

7.

MULTIPLE CHOICE QUESTION

45 sec • 1 pt

What is the formula of the quadratic formula?

x=b±b4ac2ax=\frac{-b\pm\sqrt{b-4ac}}{2a}

x=b±b4ac2ax=\frac{b\pm\sqrt{b-4ac}}{2a}

x=b±b24ac2ax=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

x=b±b24ac2ax=\frac{b\pm\sqrt{b^2-4ac}}{2a}

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