DiscussA quadratic equation in a variable x is an equation of the form ax2 + bx + c = 0, where a, b, and c are real numbers and a0.
In other words, the equation in which the highest power of a variable, i.e, degree of equation is 2, is called a quadratic equation.
In this equation,
‘a’ is known as the coefficient of x2
‘b’ is known as the coefficient of x
‘c’ is known as the constant term
Some examples of quadratic equations are:
x2+3x+2=0,
x2-4=0,
(x+3)(x-1)=0
Note: “ax2 + bx + c = 0“ is known as the general form of quadratic equation.
Q. Check whether the following equations are quadratic or not?
i) (x+1)2 = 2(x-3)
Sol. x2+2x+1 = 2x-6
x2+2x-2x = -6-1
x2= -7
x2+7 = 0
So, it is a quadratic equation.
ii) x2-2x = (-2)(3-x)
Sol. x2-2x = -6+2x
x2-2x-2x = -6
x2-4x+7 = 0
So, it is a quadratic equation.
iii) (x-2)(x+1) = (x-1)(x+3)
Sol. x2+x-2x-2 = x2=3x-x-3
x2-x-2 = x2+2x-3
x2-x2-x-2x-2+3 = 0
-3x+1 = 0
So , it is not a quadratic equation.
Q. Represent the following situations in the form of quadratic equations.
The area of the rectangular plot is 528m2. The length of the plot is one more than twice its breadth. We need to find the length and breadth of the plot.
Sol. Let the breadth of the rectangular plot be x.
According to the given conditions,
Breadth = x meter
Length = (2x+1) meter
∴ area of rectangular plot = length breadth
528 = (2x+1)x
528 = 2x2 + x
2x2+x-528 = 0, is the required quadratic equation.
The product of two consecutive integers is 306. We need to find the integers.
Sol. Let the first integer be x.
∴ consecutive integer will be x+1
Now, according to the question.
x(x+1) = 306
x2+x = 306
x2+x-306 = 0, will be the required solution.
Rohan’s mother is 26 years older than him. The product of their ages(in years) 8 years from now will be 360. We would like to find the present age of the rohan.
Sol. Let Rohan’s present age be x years
∴ his mother’s age = (x+26) years
3 years from now,
Rohan’s age = x+3
Age of Rohan’s mother will be = (x+26) +3
= x+29
According to the question,
The product of the ages 3 years from now will be 360
(x+3)(x+29) = 360
x2+29x+3x+87 = 360
x2+32x-273 = 0, is the required quadratic equation.
Till now, we have a basic understanding about what quadratic equation is and how ro form a quadratic equation for a problem statement.
Now, we will learn about some methods on how to solve a quadratic equation.
Methods to solve Quadratic Equations.
I. Factorisation:
Factorisation is the process of reducing the bracket of a quadratic equation, instead of expanding the bracket and converting the equation to a product of factors which cannot be reduced further. For example, factorizing (x²+5x+6) to (x+2) (x+3). Here, (x+2) (x+3) is the factorisation of a polynomial (x²+5x+6).
To solve a quadratic equation using factoring, following steps are taken.
Transform the equation using standard form in which one side is zero.
Factor the non-zero side.
Set each factor to zero
Solve each resulting equation.
Note: If the product of two numbers is ‘0’ then either the first number or the second number is ‘0’.
Q. Solve the following quadratic equations using the factorisation method.
2x2+x-6 = 0
2x2+4x-3x-6 = 0
2x(x+2) -3(x+2) = 0
(x+2)(2x-3)=0
Now, (x+2) = 0 or (2x-3) = 0
x = -2 or x = 3/2
This is the required solution for the given quadratic equation.
2x2+7x+52 = 0
2x2+5x+2x+52 = 0
x(2x+5)+2(2x+5) = 0
(2x+5)(x+2) = 0
Now, 2x+5 = 0 or x+2 = 0
x = -5/2 or x = -2
This is the required solution for the given quadratic equation.
Find two consecutive positive integers, sum of whose squares is 365.
Sol. let two consecutive positive integers be, x and (x+1)
According to the question,
x2 + (x+1)2 = 365
x2+ x2+2x+1 = 365
2x2+2x-364 = 0
x2+x-182 = 0
x2+14x-13x-182 = 0
x(x+14) -13(x+14) = 0
(x+14) (x-13) = 0
Now, (x+14) = 0
x -14 (not possible, because in the question it is given positive integers)
or x-13 = 0
x = 13
Hence, two consecutive integers are 13 and 14.
II. Completing the Square
Completing the square is a method in algebra that is used to write a quadratic expression in a way such that it contains the perfect square. In simple words, we can say that completing the square is a process where we consider a quadratic equation of the ax2 + bx + c = 0 and change it to write it in the form (x + p)2 = 0. This method is generally used to find the solution of the quadratic equation.
III. Sridharacharya equation:
Sridharacharya Formula is also known as the quadratic formula or Sridharacharya Method. Sridharacharya Method is used to find solutions to quadratic equations of the form ax2 + bx + c = 0, a ≠ 0 and is given by:
Important Notes on Sridharacharya Formula:
The Sridharacharya formula is commonly known as the Quadratic formula.
Sridharacharya Method is used to find solutions to quadratic equations of the form ax2 + bx + c = 0, a ≠ 0.
In case, a quadratic equation is not of the form ax2 + bx + c = 0, then we convert the equation to this standard form and then apply the Sridharacharya formula.
Now Let us solve some examples based on the above formula.
1. Solve the quadratic equation 2x2 - x - 3 = 0 using the Sridharacharya Method.
Sol. As we compare the given quadratic equation 2x2 - x - 3 = 0 with ax2 + bx + c = 0,
we have a = 2, b = -1, c = -3.
Substituting the values of a, b, c into the Sridharacharya Formula, we have
x = [-(-1) ± √{(-1)2 - 4×2×(-3)}] / 2 × 2
x = [1 ± √(1+24)]/4
x = [1 ± √25]/4
x = (1 ± 5)/4
x = 6/4, or -4/4
x = 3/2 or -1
Hence, the roots of quadratic equation 2x2 - x - 3 = 0 are 3/2 and -1 using the
Sridharacharya Formula.
2. Determine the roots of the quadratic equation 5x2 + x = 0 using the Sridharacharya
Formula.
Sol. As we compare the given quadratic equation 5x2 + x = 0 with ax2 + bx + c = 0, we
have a = 5, b = 1, c = 0.
Substituting the values of a, b, c into the Sridharacharya Formula, we have
x = (-1 ± √(12 - 4×2×0)) / 2 × 5
x = [-1 ± √(1 - 0)]/10
x = [-1 ± √1]/10
x = (-1 ± 1)/10
x = 0/10, or -2/10
x = 0 or -1/5
Hence the roots of quadratic equation 5x2 + x = 0 are 0 and -1/5 using the
Sridharacharya Method.
Nature of Roots
What is b2 - 4ac in Sridharacharya Formula? What is its importance?
The expression b2 - 4ac in Sridharacharya Formula is known as discriminant. Since it is under the square root, we can use it to determine the nature of roots of a quadratic equation.
If b2 - 4ac > 0 then the roots are real and distinct.
If b2 - 4ac = 0 then the roots are real and equal.
If b2 - 4ac < 0 then the roots are imaginary and distinct.
Note: Discriminant is generally represented by the symbol ‘’.
Q. Find the nature of roots of the following quadratic equation. If real roots exist, find
them.
i) 2x2-3x+5 = 0
We know that,
discriminant, = b2-4ac
= (-3)2 -4(2)(5)
= 9-40
= -31
Since the determinant is negative, it has no real roots.
ii) 3x2-43x+4 = 0
We know that,
discriminant, = b2-4ac
= (43)2 -4(3)(4)
= 48-48
= 0
The determinant is negative, hence the roots are real and equal.
iii) Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800m2. If so, find its length and breadth.
Sol. Let the breadth of the rectangular grove be x.
∴ according to question, length will be 2x
Now, Area = 800m2
(x)(2x) = 800
2x2 = 800
x2 = 400
x = 400
x = 20
Since, length cannot be negative.
So, x = 20
Yes, it is possible.
Length = 40 meter, breadth = 20 meter.
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