Step-2

Title: Find the roots of the quadratic equation

Grade: 9-a Lesson: S1-L7

Explanation: Hello Students, time to practice and review the steps for the problem.

Lesson Steps

Step Type Explanation Answer

1

Problem

Find the roots of the quadratic equation:
\$ (-2x)(2x+3) = 6 \$

2

Step

Now converted to

\$ax^2+bx+c=0\$

3

Step

after converting

\$ -4x^2 - 6x - 6 = 0 \$

4

Step

By taking "-2" has common multiple we get

\$ -2(2x^2 + 3x + 3) = 0 \$

5

Step

After eleminating "-2" we get

\$ 2x^2 + 3x + 3 = 0 \$

6

Step

Since the equation is in the form of

\$ax^2+bx+c=0\$

7

Formula:

formula

\$ x = (-b \pm \sqrt (b^2 - 4ac))/(2a) \$

8

Step

Substitute the values

\$ a = 2, b = 3 , c = 3\$

9

Step

converting the equation into formula ny substituting the values

\$ x = (-3 \pm \sqrt (3^2 - 4(2)(3)))/(2(2)) \$

10

Step

Simplifing the equation

\$ x = (-3 \pm \sqrt (-15))/4 \$

11

Step

We know that [i^2 = -1]

\$ x = (-3 \pm \sqrt (i^2 15))/4 \$

12

Step

After simplification we get

\$ x = (-3 \pm i\sqrt (15))/4\$

13

Answer

B

Tutor: Questions

Seq Type Question Audio

1

Problem

What did you learn from this problem?

2

Clue

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3

Hint

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4

Step

What did you learn from the Steps?

5

Step

How can we improve the Steps?


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