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

ax2 + 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) 2x2 −5x + 2 = 0

Solution :

By comparing the given quadratic equation with the general form of quadratic equation  ax2 + bx + c = 0, we get

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

x  =  [-b ± √b2 - 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) 2f2 −6f + 32 = 0

Solution :

By comparing the given quadratic equation with the general form of quadratic equation  ax2 + bx + c = 0, we get

a = 2, b = -6 and c = 32

x  =  [-b ± √b2 - 4ac] / 2a

x  =  [-(-6) ± √(-6)2 - 4(2)(32)] / 2(2)

x  =  [6 ± √(36 - 24)] / 22

x  =  [6 ± √12] / 22

x  =  [6 ± 2√3] / 22

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

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

(iii) 3y2 −20y - 23 = 0

Solution :

By comparing the given quadratic equation with the general form of quadratic equation  ax2 + bx + c = 0, we get

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

x  =  [-b ± √b2 - 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] / 6x = 46 / 6x = 23 / 3 x  =  [20 - 26] / 6x  =  -6 / 6x = -1

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

(iv) 36y2 −12ay + (a2 −b2) = 0

Solution :

By comparing the given quadratic equation with the general form of quadratic equation  ax2 + bx + c = 0, we get

a = 36, b = -12a and c = (a2 −b2)

x  =  [-b ± √b2 - 4ac] / 2a

x  =  [-(-12a) ± √(-12a)2 - 4(36)(a2 −b2)] / 2(36)

x  =  [12a ± √(144a2 - 144a2 + 144b2)] / 72

x  =  [12a ± √144b2] / 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 = t2 −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  =  t2 −0.75t

t2 −0.75t - 11.25  =  0

t2 −0.75t  =  11.25

Multiply each side by 100.

100t2 −75t  =  1125

100t2 −75t - 1125  =  0

Divide each side by 25.

4t2 - 3t - 45  =  0

Comparing

4t2 - 3t - 45  =  0

and

ax2 + 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 ± √b2 - 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] / 8t  =  30 / 8t  =  15 / 4t  =  3.75 t  =  [3 - 27] / 8t  =  -24 / 8t  =  -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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