Solving Quadratic Equations by Formula Method :
In this section, how we solve quadratic equation using formula.
How to use quadratic formula to solve 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
Question 1 :
Solve by using quadratic formula
x2 – 7 x + 12 = 0
Solution :
By comparing the given quadratic equation with general form of a quadratic equation,
ax2 + bx + c = 0
a = 1, b = -7 and c = 12
b2 – 4ac = (-7)2 - 4(1) (12)
= 49 - 48
= 1
x = -b ± √(b² – 4ac)/2a
x = [ 7 ± √1 ] / 2(1)
x = [ 7 ± √1 ] / 2
x = (7 + 1)/2 x = 8/2 x = 4 |
x = (7 - 1)/2 x = 6/2 x = 3 |
Hence the solution is { 3, 4 }.
Question 2 :
Solve by using quadratic formula
15x2 – 11 x + 2 = 0
Solution :
By comparing the given quadratic equation with general form of a quadratic equation,
ax2 + bx + c = 0
a = 15, b = -11 and c = 2
b2 – 4ac = (-11)2 - 4(15) (2)
= 121 - 120
= 1
x = -b ± √(b² – 4ac)/2a
x = [ 11 ± √1 ] / 2(15)
x = [ 11 ± √1 ] / 30
x = (11 + 1)/30 x = 12/30 x = 2/5 |
x = (11 - 1)/30 x = 10/30 x = 1/3 |
Hence the solution is { 1/3, 2/5 }.
Question 3 :
Solve by using quadratic formula
x + (1/x) = 2 1/2
Solution :
(x2 + 1)/x = 5/2
2(x2 + 1) = 5x
2x2 + 2 = 5x
2x2 – 5x + 2 = 0
By comparing the given quadratic equation with general form of a quadratic equation,
ax2 + bx + c = 0
a = 2, b = -5 and c = 2
b2 – 4ac = (-5)2 - 4(2) (2)
= 25 - 16
= 9
x = -b ± √(b² – 4ac)/2a
x = [ 5 ± √9 ] / 2(4)
x = [ 5 ± 3 ] / 8
x = (5 + 3)/8 x = 8/8 x = 1 |
x = (5 - 3)/8 x = 2/8 x = 1/4 |
Hence the solution is { 1, 1/4 }.
Question 4 :
Solve by using quadratic formula
3a²x² – abx - 2b² = 0
Solution :
3 a²x² – a b x - 2b2 = 0
By comparing the given quadratic equation with general form of a quadratic equation,
ax2 + bx + c = 0
a = 3a², b = -ab and c = -2b2
b2 – 4ac = (-ab)2 - 4(3a²) (- 2b2)
= a2b2 + 24a2b2
= 25a2b2
x = -b ± √(b² – 4ac)/2a
x = [ ab ± √(25a2b2) ] / 2(3a²)
x = [ ab ± 5ab ] / 6a²
x = (ab + 5ab)/6a2 x = 6ab/6a² x = b/a |
x = (ab - 5ab)/6a2 x = -4ab/6a² x = -2b/3a |
After having gone through the stuff given above, we hope that the students would have understood how to solve quadratic equations using quadratic formula.
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