SOLVING QUADRATIC EQUATIONS BY QUADRATIC FORMULA EXAMPLES

Solving Quadratic Equations by Quadratic Formula Examples :

Here we are going to see how we use the formula to solve a quadratic equation.

The formula to find a quadratic equation ax2 + bx + c = 0 is x = 

Solving Quadratic Equations by Quadratic Formula Examples 

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.

Hence the solutions are 2 and 1/2.

Let us look into the next problem on "Solving Quadratic Equations by Quadratic Formula Examples".

(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]/22

x  =  2(3 + √3)/2√2

x  =  (3 + √3)/√2

x  =  [6 - 2√3]/22

x  =  2(3 - √3)/2√2

x  =  (3 - √3)/√2

Hence the solutions are (3 + √3)/√2 and (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]/6

x = 46/6

x = 23/3

x  =  [20 - 26]/6

x  =  -6/6

x = -1

Hence the solutions are 23/3 and -1.

(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

Hence the solutions are (a + b)/6 and (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 travelled by the ball is 11.25 feet.

Solution :

Distance travelled by the ball = 11.25 feet.

d = 11.25

11.25  =  t2 −0.75t

t2 −0.75t - 11.25  =  0

t2 −0.75t  =  11.25

Multiply the entire equation by 100

100t2 −75t  =  1125

100t2 −75t - 1125  =  0

Divide by 25, we get 

4t2 - 3t - 45  =  0

4t2 + 12t - 15t - 45  =  0

4t(t - 3) - 15(t + 3)  =  0

(4t - 15) (t + 3)  =  0

4t  =  15 and t = -3

t = 15/4  =  3.75 seconds

After having gone through the stuff given above, we hope that the students would have understood, "Solving Quadratic Equations by Quadratic Formula Examples". 

Apart from the stuff given in this section "Solving Quadratic Equations by Quadratic Formula Examples"if you need any other stuff in math, please use our google custom search here.

Widget is loading comments...