**Solving quadratic equation by using quadratic formula :**

We use the formula given below to find the exact roots of the given quadratic equation.

(i) Compare the given quadratic equation with the general form of quadratic equation ax² + bx + c = 0

(ii) Here the coefficient of x² is a, coefficient of x is b and the constant term is c.

(iii) Then apply those values in the formula

-b ± √ (b²-4ac)/2a

Let us see some example problems on solving quadratic equation by using quadratic formula.

**Example 1 :**

Solve the following quadratic equations using quadratic formula

x^{2} - 2x - 24 = 0

**Solution :**

By comparing the above equation with the general form of a quadratic equation ax^{2} + bx + c = 0, we get

a = 1 b = -2 c = -24

= -b ± √ (b = -(-2) ± √(4-(-96))/2(1) = 2 ± √(4+96)/2 = (2 ± √100)/2 = (2 ± √(10 x 10))/2 = (2 ± 10)/2 |
b 4ac = 4(1)(-24) = -96 |

= (2 + 10)/2 = 12/2 = 6 |
= (2 - 10)/2 = -8/2 = -4 |

**Hence the solutions are 6 and -4.**

**Example 2 :**

Solve the following quadratic equations using quadratic formula

x^{2} - 2x - 15 = 0

**Solution :**

By comparing the above equation with the general form of a quadratic equation ax^{2} + bx + c = 0, we get

a = 1 b = -2 c = -15 = -b ± √ (b b = 4 + 60 = 64 = -(-2) ± √64/2(1) = (2 ± 8)/2 |
b 4ac = 4(1)(-15) = -60 |

= (2 + 8)/2 = 10/2 = 5 |
= (2 - 8)/2 = -4/2 = -2 |

**Hence the solutions are 5 and -2.**

**Example 3 :**

Solve the following quadratic equations using quadratic formula

x + (1/x) = 2 ½

**Solution :**

(x² + 1)/x = 5/2

2 (x² + 1) = 5 x

2 x² + 2 = 5 x

2 x² - 5 x + 2 = 0

By comparing the above equation with the general form of a quadratic equation ax^{2} + bx + c = 0, we get

x = 8/4 , 2/4

x = 2 , 1/2

Hence the solutions are 2 and 1/2

**Example 4 :**

Solve the following quadratic equations using quadratic formula

3 a²x² - ab x - 2b² = 0

**Solution :**

**By comparing the above equation with the general form of a quadratic equation ax ^{2}**

**a = 3 a² b = - ab c = - 2b²**

x = 6ab/6a² , -4ab/6a²

x = b/a , x = -2b/a

Hence the solutions are b/a and -2b/a

**Example 5 :**

Solve the following quadratic equations using quadratic formula

a (x² + 1) = x(a² + 1)

**Solution :**

a x² + a = x(a² + 1)

a x² - x(a² + 1) + a = 0

**By comparing the above equation with the general form of a quadratic equation ax ^{2}**

**a = a b = -(a² + 1) c = a **

x = 2a²/2a , 2/2a

= a , 1/a

Hence the solutions are a and 1/a.

**Example 6 :**

Solve the following quadratic equations using quadratic formula

[(x - 1)/(x + 1)] + [(x - 3)/(x - 4)] = 10/3

**Solution :**

[(x - 1)(x - 4) + (x - 3)(x + 1)]/(x + 1)(x- 4) = 10/3

(x - 1)(x - 4) = x² - 5 x + 4 ---(1)

(x - 3)(x + 1) = x² - 2 x - 3 -----(2)

(x + 1)(x- 4) = (x² - 3 x - 4) -----(3)

(1) + (2) ==>x²-5x+4+x²-2x-3 = 2x²-7x+1 -----(4)

(4)/(3) ==> (2x²-7x+1)/(x² - 3 x - 4) = 10/3

3 (2x² - 7 x + 1) = 10(x² - 3 x - 4)

10 x² - 6 x² - 30 x + 21 x - 40 - 3 = 0

4 x² - 9 x - 43 = 0

**By comparing the above equation with the general form of a quadratic equation ax ^{2}**

a = 4 b = -9 c = -43

= -b ± √ (b b = 81 + 688 b = -(-9) ± √769/2(1) = -(-9) ± √769/2(1) = (9 ± √769)/2 |
b 4ac= 4(4)(-43) = -688 |

After having gone through the stuff given above, we hope that the students would have understood, "Solving quadratic equation by using quadratic formula".

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