10. Quadratic Equations

. Quadratic Equations

(1) A polynomial of degree 2 is called a quadratic polynomial. The general form of a quadratic polynomial is ax² + bx + c, where a, b, c are real number such that a ≠0 and x is a real variable. For Example: x² + 5x + 3, where a=1, b=5, c=3 are real number. So given equation is quadratic polynomial.

(2) If p(x) = ax² + bx + c , a ≠ 0 is a quadratic polynomial and α is a real number, then p(α) = aα² + bα + c is known as the value of the quadratic polynomial p(α)

For Example: p(α) = α² + 5α + 3 in that equation if α=3 then p(α)=27. So 27 is a value of quadratic polynomial

(3) A real number α is said to be a zero of quadratic polynomial p(x) = ax² + bx + c, if p(α)=0. For Example: p(x) = x² + 6x + 5

If x=(−5) then p(x)=0, So −5 is the zero of polynomial.

(4) If p(x) = ax² + bx + c is a quadratic polynomial, then p(x)=0 i.e., ax² + bx + c = 0, a ≠ 0 is called a quadratic equation.

For Example: p(x) = x² − 8x + 16, a = 1, is called a quadratic equation.

(5) A real number α is said to be a root of the quadratic equation ax² + bx + c = 0. In other words, α is a root of ax² + bx + c = 0 if and only if α is a zero of the polynomial p(x) = ax² + bx + c.

For Example: Suppose quadratic equation is 2x² − x − 6 = 0.

If we put x=2 then p(x)=0, So 2 is a root of that given equation so here α=2.

(6) If ax² + bx + c = 0, a ≠ 0 is factorizable into a product of two linear factors, then the roots of the quadratic equation ax² + bx + c = 0 can be found by equating each factor to zero.

For Example: The Given equation is 9x² − 3x − 2 = 0

Now, Ansving the above equation using factorization method.

⇒ 9x² − 6x + 3x − 2

⇒ 3x(3x − 2) + 1(3x − 2)

⇒(3x+1)=0 or (3x-2)=0

⇒ 3x = −1 or (3x − 2) = 0

⇒ x = −

The given equation is of the form of ax2 + bx + c = 0, where a = 2, b = - 2√6, c = 3 Therefore, the discriminant- D = b2 – 4ac

= (− 2√6)2 – 4 x 2 x 3

= 24 − 24 = 0

Since, D = 0

Therefore, the roots of the given equation are real.The real and equal roots are given

by − b 2a

The given equation is of the form of ax2 + bx + c = 0, where a = 1, b = 1, c = 2 Therefore, the discriminant

D = b2 – 4ac

D= (1)2 – 4 x 1 x 2

D= 1 - 8

D= -7

Since, D < 0

Therefore, the given equation has not real roots.

Exercise 5.1

Exercise 5.2

Complete the square method
Dothis (pg.no 177)

Exercise 5.3

Question 1:
Find the roots of the following quadratic equations, if they exist.

Question 1(i):
2x2 + x – 4 = 0
Solution :

Question 1(ii):
$4{ x }^{ 2 }+4\sqrt { 3 } x+3=0$
Solution :

Question 1(iii):
5x2 – 7x – 6 = 0
Solution :

Question 1(iv):
x2 + 5 = -6x
Solution :
x2 + 5 = -6x
x2 + 6x = -5
x2 + 6x + 32 = -5 + 32 (Adding in 32 on both sides)
(x + 3)2 = -5 + 9
(x + 3)2 = 4
(x + 3) = ±2
x = -3 + 2 or x = -3 – 2
x = -1 or x = -5.

Question 2:
Find the roots of the quadratic equations given in Q. 1 above by applying the quadratic formula.
Solution :

Question 3:
Find the roots of the following equations:

Solution :

Multiplying throughout by x, we get x2 – 1 = 3x
i.e., x2 – 3x – 1 = 0, which is a quadratic equation.
Here, a = 1, b = -3, c = -1
So, b2 – 4ac = 9 + 4 = 13 > 0

As x ≠ -4, 7,

Multiplying the equation by (x + 4)(x – 7), we get

30((x – 7) – (x + 4)) = 11(x + 4)(x – 7)
-30 = (x2 – 7x + 4x – 28)
-30 = (x2 – 3x – 28)
-30 = x2 – 3x – 28
x2 – 3x – 28 + 30 = 0
x2 – 3x + 2 = 0
(x – 1)(x – 2) = 0
x = 1 and 2 are roots of the given equation

Question 4:
The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now 1/3. Find his present age.
Solution :

x=-3 and x=7
Since age cannot be negative,
Therefore discarding x = -3,we get,
x = 7
Hence, Rehman’s present age is 7 years.

Question 5:
In a class test, the sum of Moulika’s marks in Mathematics and English is 30. If she got 2 marks more in Mathematics and 3 marks less in English, the product of her marks would have been 210. Find her marks in the two subjects

Solution :

Let the marks in Mathematics be x and in English be y
As given in question,
x + y = 30, ….(1)
If she got 2 marks more in mathematics and 3 marks less in English
Then, x + 2 and y – 3 and, product of her marks,
x y = 210.
(x + 2)(y – 3) = 210 ….(2)
From equation (1)=

i.e., y = 18 or 17
Hence, x = 12 or 13
Thus, Moulika got 12 in Mathematics and 18 in English or She got 13 in Mathematics and 17 in English.

Question 6:
The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
Solution :
Let the shorter side of the rectangle be ‘x’, longer side be ‘y’ and diagonal be ‘z’.
According to given condition,
z = x + 60——–(1)
y = x + 30——–(2)
Since the angles of a rectangle are right angles,
Therefore, applying Pythagoras theorem
z2 = x2 + y2 (Since one part of a rectangle is the triangle with diagonal as the hypotenuse)
Substituting (1) and (2) in the above equations (x + 60)2 = x2 + (x + 30)2
x2 + 2(x)(60) + (60)2 = x2 + x2 + 2(x)(30) + (30)2
x2 + 120x + 3600 = 2x2 + 60x + 900
x2 – 60x – 2700 = 0
(x – 90)(x + 30) = 0
x = 90 or -30
x cannot be negative, hence, x = 90 m and y = 120 m

Question 7:
The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
Solution :

Question 8:
A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Solution :

i.e., x = 45 or -40
But speed can’t be negative so the speed is 45. Hence the speed of the train = 45 – 5 = 40 km/hr.

Question 9:
Two water taps together can fill a tank in $9\frac { 3 }{ 8 }$ The tap of larger diameter takes 10
hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Solution :

Question 10:
An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangaluru (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.
Solution :
Let speed = x and time = y
For the express train,
Speed= x + 11 and time = y – 1
For passenger train, xy = 132

i.e., x = 44 or x = -33
But speed cannot be negative, so the speed is 44.
The speed of the express train is 44 km/hr
The speed of the passenger train is 33 km/hr.

Question 11:
Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.
Solution :
Let the side of the 1st square (sq1) be ‘x’ and side of the 2nd square (sq2) be ‘y’.
Since, Perimeter(square) = 4 × (side)
Therefore,
P(sq1) = 4 × x and P(sq2) = 4 × y
From the above condition,
P(sq1) – P(sq2) = 24 (Assuming ‘x’ is greater than ‘y’)
(4x) – (4y) = 24
4(x – y) = 24
(x – y) = 24/4
x – y = 6 ————-(1)
Also, Area(square) = (side)2
Therefore,
A(sq1) = x2and A(sq2) = y2
From the above condition,
x2 + y2 = 468 ——-(2)
But, x2 + y2 = (x – y)2 + 2xy——(3)
468 = (6)2 + 2(xy)
468 = 36 + 2(xy)
432 = 2(xy)
2(xy) = 432
xy = 432/2
xy = 216—————(4)
x = 216/y————–(5)
Substituting (5) in (1)
(216/y) – y = 6
(216 – y2)/y = 6
216 – y2 = 6y
y2 + 6y – 216 = 0
Therefore,
y = 12 or y = -18
But, since ‘y’ is the length of the square it cannot be
negative.
Hence, discarding y = -18
Therefore, y = 12——–(6)
Substituting (6) in (5),we get,
x = 216/12
x = 18
Side of square1 = 18 m
Side of square2 = 12 m

Question 12:
A ball is thrown vertically upward from the top of a building 96 m tall with an initial velocity 80 m/second. The distance ‘s’ of the ball from the ground after t seconds is S = 96 + 80t – 4.9t2. After how may seconds does the ball strike the ground.
Solution :
Given,
16t2 – 80t – 96 = 0
t2 – 5t – 6 = 0
(t – 6)(t + 1) = 0
Hence, t = 6 or t = -1
But, time cannot be negative so, t= 6seconds.

Question 13:
If a polygon of ‘n’ sides has 1/2 n (n-3) diagonals. How many sides will a polygon having 65 diagonals? Is there a polygon with 50 diagonals?
Solution :

But the side should be an integer, so a polygon cannot have 50 diagonals.

Exercise 5.4

Note