Quadratic equations are the equations where polynomial has the degree two. Quadratic equations are the equations of type ax^{2} + bx + c = 0 where x is unknown and a, b, c are known real numbers and a should not be zero. If a=0 then the equation will not remain quadratic, it will be then linear as a = 0 will eliminate x^{2 }term. As the quadratic equation has the highest degree two, so this equation has two roots, or we can say that we will find two values of x for a quadratic equation.
Method 1: The roots of the quadratic equations can be found by the Shridharacharaya formula.
x = [b±√(b^{2 }– 4ac)]/2a
Example: The length of sides of a rectangle is given by x – 3 and x – 5 and the area of the rectangle is 3 unit^{2}. Find the sides of the rectangle.
Solution:
Area of rectangle = length*breadth = (x – 3)(x – 5) = 3
Area = x^{2} – 8x + 15 = 3
= x^{2 }– 8x + 12 = 0
Discriminant = b^{2} – 4ac = 64 – (4(1)(12)) = 64 – 48 = 16
x = [b ± √(b^{2}4ac)]/2a = [(8) ± √16]/2 = (8±4)/2
x = 12/2 or 4/2
x = 2 or 6
When x is 2, sides are x – 3 = 2 – 3 = 1 and x – 5 = 2 – 5 = 3.
Since length of sides cannot be equal therefore x = 2 is not a valid ans.
When x is 6, sides are x – 3 = 6 – 3 = 3 and x – 5= 6 – 5 =1.
Therefore, x = 6 is the valid answer and the sides are 3 and 1.
Method 2. The other way is the factorizing method. A quadratic equation can be considered a factor of two terms. Like ax^{2 }+ bx + c = 0 can be written as (x – x_{1})(x – x_{2}) = 0 where x_{1} and x_{2} are roots of quadratic equation.
Steps:
Example: Let be the quadratic equation x^{2} + 3x = 18
x^{2} + 3x – 18 = 0
Step:
1. 6 and 3 are the numbers whose sum is equal to b and product is equal ac.
2. x^{2} + (63)x – 18 = x^{2} + 6x 3x – 18 =0
3. x(x + 6) – x(x + 6) = 0
4. taking (x + 6) as common.
(x + 6)(x – 3) = 0
x = 6 or x = 3
In a factorizing method it is not necessary that you will always find these two numbers easily(especially in the case when roots are imaginary or irrational) so it is better to use the quadratic formula.
The nature of roots depends on the discriminant of the quadratic equation. The discriminant of a quadratic equation is given by b^{2 }– 4ac. It is so because in quadratic formula square root of discriminant is there.
Root 1: If b^{2 }– 4ac > 0 roots are real and different. As the discriminant is >0 then the square root of it will not be imaginary. It has two cases.
Example: Let the quadratic equation be x^{2}5x+6=0.
Then the discriminant of the given equation is b^{2} – 4ac=(5)^{2} – 4*1*6 = 2524 = 1
According to Shridharacharaya formula
x = [b±√(b^{2}4ac)]/2a = x = [(5) ± √1]/2
x_{1} = [(5) + √1]/2 = 6/2 = 3
x_{2} = [(5) – √1]/2 = 4/2 = 2
Therefore, the roots are 3,2. Both are rational and different.
Example: Let the quadratic equation be x^{2}7x+8 = 0.
Then the discriminant of the given equation is
b^{2} – 4ac=(7)^{2 }– 4*1*8 = 4932 = 17
According to Shridharacharaya formula
x = [b±√(b^{2}4ac)]/2a = x = [(7) ± √17]/2
x_{1} = [(7) + √17]/2 = [7 + √17]/2
x_{2} = [(7) – √17]/2 = [7 – √17]/2
Therefore, the roots are [7 + √17]/2,[7 – √17]/2. Both are irrational and in pairs.
Root 2: If b^{2} – 4ac = 0 roots are real and equal.
Example: Let the quadratic equation be 3x^{2}6x+3=0.
Then the discriminant of the given equation is
b^{2} – 4ac=(6)^{2} – 4*3*3 = 36 – 36 = 0
According to Shridharacharaya formula
x = [b±√(b^{2}4ac)]/2a = x = [(6) ± √0]/[(2)(3)]
x_{1} = [(6) + √0]/2 = 6/6 = 1
x_{2} = [(6) – √0]/2 = 6/6 = 1
Therefore, the roots are 1,1. Both are real and equal.
Root 3: If b^{2} – 4ac < 0 roots are imaginary, or you can say complex roots. It is imaginary because the term under the square root is negative. These complex roots will always occur in pairs i.e, both the roots are conjugate of each other.
Example: Let the quadratic equation be x^{2}+6x+11=0.
Then the discriminant of the given equation is
b^{2} – 4ac=(6)^{2} – 4*1*11 = 3644 = 8
According to Shridharacharaya formula
x = [b±√(b^{2}4ac)]/2a = x = [(6) ± √(8)]/2
x_{1} = [(6) + √(8)]/2 = [6 + √8i]/2 = 2[3 + √2i]/2 = 3 + √2i
x_{2} = [(6) – √(8)]/2 = [6 – √8i]/2 = 2[3 – √2i]/2 = 3 – √2i
Therefore, the roots are 3,2. Both are imaginary and conjugate of each other(in pair).
The maximum/minimum value of quadratic function is found at x = b/2a
Proof:
We get maxima or minima when d(f(x))/dx = 0.
On differentiating quadratic function f(x) = ax^{2} + bx + c.
We get,
2ax + b = 0
x = b/2a
This x is either maxima(a<0) or minima(a>0).
1. When b^{2}– 4ac > 0
2. When b^{2}– 4ac = 0
3. When b^{2}– 4ac < 0
Question 1. The height of a triangle is less than 4 cm than the base. The area of triangle is 30 cm^{2}. Find the height and base of triangle.
Solution:
Let the base of triangle be x cm then height is x4 cm
Area of triangle = 1/2*height*base = 1/2*(x)(x – 4)=30
Area = x^{2} – 4x = 30*2
= x^{2 }– 4x = 60
= x^{2} – 4x – 60 = 0
Discriminant = (4)^{2} – 4(1)(60) = 16+240 = 256
x = [b±√(b^{2 }– 4ac)]/2a = [(4)±√256]/2 = (4±16)/2
x = 20/2 or 12/2
x = 10 or 6
As side cannot be negative, therefore 6 is not correct.
So, when x is 10, base =10 cm and height =x – 4 = 10 – 4 = 6 cm
Therefore, x = 10 is the valid answer and the base and height of the triangle are 10 and 6 respectively.
Question 2. The volume of a box is 600 inch^{2}. The length of box is 2 inches less than the width. The height of box is 5 inches. Find the dimensions of box.
Solution:
Let the width of box be x inches then length = x – 2 inches.
Volume of box =Length* Width *Height = (x2)(x)5= 600
x^{2} – 2x = 120 => x^{2} – 2x 120 = 0
x^{2} + 10x – 12x – 120 = 0
x(x + 10) – 12(x + 10)=0
(x – 12)(x + 10)=0
x = 12 or x = 10
As width can not be negative, therefore, 10 is not correct.
When x = 12, width = 12 inches, length = x – 2 = 12 – 2 = 10 inches, height = 5 inches
Question 3. A ball is thrown from the top of a building . its height in meters above the ground as a function of time is given by h(t) = 4t^{2} + 24t + 3. a) How much time it take to reach the maximum height and what is the maximum height. b) Find also the time at which ball hits the ground.
Solution:
a) Since a<0, therefore the time to reach maximum height is = b/2a. (refer graph part)
t = 24/(2(4)) = 24/8
t = 3 sec.
Height = h(t) = 4(3)^{2} +24(3)+3 = 39 meters.
b) When the ball hits the ground h(t)=0.
4t^{2 }+ 24t + 3 = 0
The discriminant of the given equation is
b^{2} – 4ac=(24)^{2} – 4*(4)*3 = 576 + 48 = 624
According to Shridharacharaya formula
x = [b±√(b^{2}4ac)]/2a = x = [(24) ± √624]/[(2)(4)]
x_{1} = [24 + √624]/8= 4 [6 + √39]/8 = [6 + √39]/2 = [6 – √39]/2 = 0.122499 = 0.1225(approx)
x_{2} = [24 – √624]/8= 4 [6 – √39]/8 = [6 – √39]/2 = [6 + √39]/2 = 6.122499 = 6.1225(approx)
As time can not be negative, therefore, [6 – √39]/2 sec is not correct.
Therefore, the ball hits the ground at [6 + √39]/2 = 6.122499 = 6.1225 sec
126 videos477 docs105 tests

1. What is the nature of roots of a quadratic equation? 
2. How can we determine the nature of roots using the discriminant of a quadratic equation? 
3. What does it mean if the discriminant of a quadratic equation is negative? 
4. Can a quadratic equation have only one real root? 
5. How can we determine the nature of roots of a quadratic equation without calculating the discriminant? 

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