**Solving Quadratic Equations by Quadratic Formula Examples :**

In this section, you will learn how to solve quadratic equation using quadratic formula.

To use quadratic formula, the quadratic equation must be in the form of

ax^{2} + bx + c = 0

We can substitute the values of a, b and c into the formula shown below and solve the quadratic equation given.

**Question 1 :**

Solve the following quadratic equations by formula method

(i) 2x^{2} −5x + 2 = 0

**Solution :**

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

a = 2, b = -5 and c = 2

x = [-b ± √b^{2} - 4ac] / 2a

x = [-(-5) ± √(-5)^{2} - 4(2)(2)] / 2(2)

x = [5 ± √(25 - 16)] / 4

x = [5 ± √9] / 4

x = [5 ± 3] / 4

x = (5 + 3) /4 and x = (5 - 3)/4

x = 8/4 and x = 2/4

x = 2 and x = 1/2

Therefore, the solution is {1/2, 2}.

(ii) √2f^{2} −6f + 3√2 = 0

**Solution :**

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

a = √2, b = -6 and c = 3√2

x = [-b ± √b^{2} - 4ac] / 2a

x = [-(-6) ± √(-6)^{2} - 4(√2)(3√2)] / 2(√2)

x = [6 ± √(36 - 24)] / 2√2

x = [6 ± √12] / 2√2

x = [6 ± 2√3] / 2√2

x = [6 + 2√3] / 2√2 x = 2(3 + √3) / 2√2 x = (3 + √3) / √2 |
x = [6 - 2√3] / 2√2 x = 2(3 - √3) / 2√2 x = (3 - √3) / √2 |

Therefore, the solution is {(3 + √3)/√2, (3 - √3)/√2}.

(iii) 3y^{2} −20y - 23 = 0

**Solution :**

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

a = 3, b = -20 and c = -23

x = [-b ± √b^{2} - 4ac] / 2a

x = [-(-20) ± √(-20)^{2} - 4(3)(-23)] / 2(3)

x = [20 ± √(400 + 276)] / 6

x = [20 ± √676] / 6

x = [20 ± 26] / 6

x = [20 + 26] / 6 x = 46 / 6 x = 23 / 3 |
x = [20 - 26] / 6 x = -6 / 6 x = -1 |

Therefore the solution is {-1, 23/3}.

(iv) 36y^{2} −12ay + (a^{2} −b^{2}) = 0

**Solution :**

^{2} + bx + c = 0, we get

a = 36, b = -12a and c = (a^{2} −b^{2})

x = [-b ± √b^{2} - 4ac] / 2a

x = [-(-12a) ± √(-12a)^{2} - 4(36)(a^{2} −b^{2})] / 2(36)

x = [12a ± √(144a^{2} - 144a^{2} + 144b^{2})] / 72

x = [12a ± √144b^{2}] / 72

x = [12a ± 12b] / 72

x = 12(a ± b) / 72

x = (a ± b) / 6

x = (a + b) / 6 |
x = (a - b) / 6 |

Therefore, the solution is {(a + b)/6, (a - b)/6}.

**Question 2 :**

A ball rolls down a slope and travels a distance d = t^{2} −0.75t feet in t seconds. Find the time when the distance traveled by the ball is 11.25 feet.

**Solution :**

Let t be the time in seconds when the distance traveled by the ball is 11.25 feet.

Distance travelled by the ball = 11.25 feet.

d = 11.25

Then,

11.25 = t^{2} −0.75t

t^{2} −0.75t - 11.25 = 0

t^{2} −0.75t = 11.25

Multiply each side by 100.

100t^{2} −75t = 1125

100t^{2} −75t - 1125 = 0

Divide each side by 25.

4t^{2} - 3t - 45 = 0

Comparing

4t^{2} - 3t - 45 = 0

and

ax^{2} + bx + c = 0

we get

a = 4, b = -3 and c = -45

Substitute the above values of a, b and c into the quadratic formula.

t = [-b ± √b^{2} - 4ac] / 2a

t = [-(-3) ± √(-3)^{2} - 4(4)(-45)] / 2(4)

t = [3 ± √(9 + 720)] / 8

t = [3 ± √729] / 8

t = [3 ± 27] / 8

t = [3 + 27] / 8 t = 30 / 8 t = 15 / 4 t = 3.75 |
t = [3 - 27] / 8 t = -24 / 8 t = -3 |

Amount of time taken can never be a negative value.

So, we can ignore t = -3.

Therefore,

t = 3.75 seconds

After having gone through the stuff given above, we hope that the students would have understood how to solve quadratic equations by quadratic formula.

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