QUADRATIC FORMULA SOLVER - SOLVED EXAMPLES AND PRACTICE PROBLEMS

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Please study
Derivation of Quadratic Formula before Quadratic Formula Solver
if you have not already done so.

There, We presented the Derivation
Of Quadratic Formula in a lucid way.
An Example in applying the formula
is given. Another problem for practice
is also given.

Here we present some more Solved Examples.

We also give problems for practice in Exercise.

Example 1 : Quadratic Formula Solver

Solve the following equation using Quadratic Formula

3x2 + 2x - 8 = 0

To solve 3x2 + 2x - 8 = 0 using Quadratic Formula
Comparing this equation with ax2 + bx + c = 0, we get
a = 3, b = 2 and c = -8
We know by Quadratic Formula, x = {(-b) ± √(b2 - 4ac)}⁄2a
Applying this Quadratic Formula here, we get
x = {(-b) ± √(b2 - 4ac)}⁄2a
= [ (-2) ± √{(2)2 - 4(3)(-8)}]⁄2(3)
= [ (-2) ± √{4 + 96}]⁄6 = [ (-2) ± √{100}]⁄6 = [(-2) ± 10]⁄6
= (-2 + 10)⁄6, (-2 - 10)⁄6 = 8⁄6, -12⁄6 = 4⁄3, -2 Ans. Great Deals on School & Homeschool Curriculum Books

Example 2 : Quadratic Formula Solver

Solve the following equation using Quadratic Formula

8 - 5x2 - 6x = 0

To solve 8 - 5x2 - 6x = 0 using Quadratic Formula
Multiplying the given equation by -1, we get
5x2 + 6x - 8 = 0 x -1 = 0
Comparing this equation with ax2 + bx + c = 0, we get
a = 5, b = 6 and c = -8
We know by Quadratic Formula, x = {(-b) ± √(b2 - 4ac)}⁄2a
Applying this Quadratic Formula here, we get
x = {(-b) ± √(b2 - 4ac)}⁄2a
= [ (-6) ± √{(6)2 - 4(5)(-8)}]⁄2(5)
= [ (-6) ± √{36 + 160}]⁄10 = [ (-6) ± √{196}]⁄10 = [(-6) ± 14]⁄10
= (-6 + 14)⁄10, (-6 - 14)⁄10 = 8⁄10, -20⁄10 = 4⁄5, -2 Ans.

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Example 3 : Quadratic Formula Solver

Solve the following equation using Quadratic Formula

6x2 - 13x - 63 = 0

To solve 6x2 - 13x - 63 = 0 using Quadratic Formula
Comparing this equation with ax2 + bx + c = 0, we get
a = 6, b = -13 and c = -63
We know by Quadratic Formula, x = {(-b) ± √(b2 - 4ac)}⁄2a
Applying this Quadratic Formula here, we get
x = {(-b) ± √(b2 - 4ac)}⁄2a
= [ -(-13) ± √{(-13)2 - 4(6)(-63)}]⁄2(6)
= [ (+13) ± √{169 + 1512}]⁄12 = [ (13) ± √{1681}]⁄12 = [(13) ± 41]⁄12
= (13 + 41)⁄12, (13 - 41)⁄12 = 54⁄12, -28⁄12 = 9⁄2, -7⁄3 Ans.

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Example 4 : Quadratic Formula Solver

Solved Example 4 on Quadratic Formula Solver :

Solve the following equation using Quadratic Formula
√(x + 1) + √(x - 2) = √(x + 3)

Solution of Example 4 of Quadratic Formula Solver :

The given equation is √(x + 1) + √(x - 2) = √(x + 3)
Squaring both sides, we get
{√(x + 1) + √(x - 2)}2 = {√(x + 3)}2
⇒ {√(x + 1)}2 + {√(x - 2)}2 + 2{√(x + 1)}{√(x - 2)} = (x + 3)

⇒ (x + 1) + (x - 2) + 2[√{(x + 1)(x - 2)] = (x + 3)

⇒ (2x - 1) - (x + 3) = -2[√{(x + 1)(x - 2)] ⇒ 2x - 1 - x - 3 = -2[√{(x + 1)(x - 2)] ⇒ x - 4 = -2[√{(x + 1)(x - 2)]
Squaring both sides, we get
(x - 4)2 = 4(x + 1)(x - 2) ⇒ x2 - 8x + 16 = 4(x2 - x - 2) = 4x2 - 4x - 8
x2 - 4x2 - 8x + 4x + 16 + 8 = 0 ⇒ -3x2 - 4x + 24 = 0
⇒ 3x2 + 4x - 24 = 0
Comparing this equation with ax2 + bx + c = 0, we get
a = 3, b = 4 and c = -24
We know by Quadratic Formula, x = {(-b) ± √(b2 - 4ac)}⁄2a
Applying this Quadratic Formula here, we get
x = {(-b) ± √(b2 - 4ac)}⁄2a
= [ -(4) ± √{(4)2 - 4(3)(-24)}]⁄2(3)
= [ (-4) ± √{16 + 288}]⁄6 = [-4 ± √(304)]⁄6 = [-4 ± √{4(76)}]⁄6
= [-4 ± 2{√(76)}]⁄6 = [-2 ± {√(76)}]⁄3. Ans.

Exercise : Quadratic Formula Solver

Problems on Quadratic Formula Solver :

  1. Solve the following equations using Quadratic Formula
    1. x2 + 16x + 48 = 0
    2. 10x2 - 7x - 12 = 0
    3. 16x - 15 - 4x2 = 0
  2. Solve the following equations using Quadratic Formula
    √(x + 8) - √(x + 3) = √x
For Answers See at the bottom of the Page.

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Answers to Exercise : Quadratic Formula Solver

Answers to Problems on Quadratic Formula Solver :

    1. -4, -12
    2. -4⁄5, 3⁄2
    3. 5⁄2, 3⁄2
  1. 1, -25⁄3