MADE BY:
MR AMIT KUMAR
T.G.T.(MATHEMATICS)
J.N.V.SURAJPUR (C.G)
JAWAHAR NAVODAYA VIDYALAYA
BASDEI, SURAJPUR(C.G.)
QUADRATIC EQUATIONS
CLASS: 10TH
Solution of a Quadratic Equation by Quadratic Formula
Hidden Quadratic Equations
TOPICS TO BE COVERED
Roots of Quadratic Equations
Solution of a Quadratic Equation by Factorization Method
Introduction to Quadratic Equations
INTRODUCTION
* A quadratic equation in the variable x is an equation of the form ax²+bx+c=0 where a, b, c are real numbers, a ≠ 0. For example 2x²+x-300=0 is a quadratic equation.
* Any equation of the form p(x)=0, where p(x) is a polynomial of degree 2, is a quadratic equation. But, when we write the terms of p(x) in descending order of their degrees, then we get the standard form of the equation, that is ax²+bx+c=0.
* The constants a, b, and c are called the quadratic coefficients, the linear coefficients and the constant terms respectively.
ROOTS OF QUADRATIC EQUATIONS
*A real number α is called a root of the quadratic equation ax²+bx+c=0, then we say that: x =α satisfies the equation, thus it is a solution of the equation ax²+bx+c=0.
*The root of a quadratic equation ax²+bx+c=0 are called zeroes of the polynomial ax²+bx+c.
NOTE:- The zeroes of the quadratic polynomial ax²+bx+c and the roots of the quadratic equation ax²+bx+c=0 are the same.
REMEMBER:- Any quadratic equation can have at most two roots.
MORE EXAMPLES OF QUADRATIC EQUATIONS
* 2x²+5x+3=0 In this one a=2, b=5 and c=3.
* x²-3x=0 This one is a little more tricky: where is a? In fact a=1, as we don’t usually write “1x²”, b=-3 and where is c? Well c=0, so it is not shown.
HIDDEN QUADRATIC EQUATIONS
So far we have seen the “standard form” of a quadratic equation: ax²+bx+c=0
But, sometimes a quadratic equation doesn’t looks like that.
Here are some examples of different forms:
IN DISGUISE | | IN STANDARD FORM | a, b and c |
x²=3x-1 | Move all terms to left hand side | x²-3x+1=0 | a=1, b=(-3) and c=1 |
2(w²-2w)=5 | Expand (undo the brackets), and move 5 to left | 2w²-4w-5=0 | a=2, b=(-4) and c=(-5) |
z(z-1)=3 | Expand, and move 3 to left | z²-z-3=0 | a=1, b=(-1) and c=(-3) |
5+1/x-1/x²=0 | Multiply by x² | 5x²+x-1=0 | a=5, b=1 and c=(-1) |
FACTORIZATION METHOD
Step 1: Find the product of first and last term (a*c).
Step 2: Find the factors of ac in such way that addition or subtraction of the factors of ac is the middle term b.
Step 3: (Splitting of middle term)
write the middle term using the sum of the two new factors, including the proper signs.
Step 4: Group the terms to form pairs and factor out the common binomial.
Step 5: Set the binomial found equal to 0 and solve the linear equation to find the zeroes of the quadratic equation.
USING THE FACTORIZATION METHOD
Q, Solve x²+3x-4=0 Q. Solve 100x²-20x+1=0
Sol. x²+3x-4=0 Sol. 100x²-20x+1=0
x²-x+4x-4=0 100x²-10x-10x+1=0
x(x-1)+4(x-1)=0 10x(10x-1)-1(10x-1)=0
(x+4) (x-1)=0 (10x-1) (10x-1)=0
Either (x+4)=0 or (x-1)=0 Either (10x-1)=0 or (10x-1)=0
x=(-4) and x=1 x=1/10 and x=1/10
QUADRATIC FORMULA METHOD
Formula:
Thus, ax²+bx+c=0 has two roots α and ß, given by
DISCRIMINANT
The expression under the square root in the quadratic formula is called the Discriminant and is denoted by “D” and is given by D= b²-4ac
Relationship Between Discriminant and Nature of Roots
VALUE OF D | NATURE OF ROOTS | ROOTS |
D>0 | Real and unequal | [(-b±√D)/2a] |
D=0 | Real and equal | Each root=(-b/2a) |
D<0 | No real roots | None |
USING THE QUADRATIC FORMULA
Example: Solve 5x²+6x+1=0
Coefficients are: a=5, b=6, c=1
Quadratic Formula:
Put the values of a, b and c:
Solve:
x=(-0.2) or (-1)
QUESTIONS RELATED TO QUADRATIC EQUATIONS
Q. If x=3, is one root of the quadratic equation x²-2kx-6=0, then find the value of k.
Sol. x²-2kx-6=0
since x=3 is a root,
3²-2(k)(3)-6=0
9-6k-6=0
3-6k=0
6k=3
K=1/2
Q. If -5 is a root of the quadratic equation 2x²+px-15=0 and the quadratic equation p(x²+x)+k=0 has equal roots, find the value of k.
Sol. Given, -5 is a root of the quadratic equation
2x²+px-15=0
2(-5)²+p(-5)-15=0
50-5p-15=0
35-5p=0
5p=35
p=7
Substituting the value of p in p(x²+x)+k=0
7(x²+x)+k=0
7x²+7x+k=0
The roots of the equation are equal
D=b²-4ac=0
Here, a=7, b=7, c=k
b²-4ac=0
7²-4(7)(k)=0
49-28k=0
k=49/28 or k=7/4
Q. Solve for x:
x²-(√3+1)x+ √3=0
Sol. x²-(√3+1)x+ √3=0
x²-√3x-x+√3=0
x(x-√3)-1(x-√3)=0
(x-√3)(x-1)=0
Either, x-√3=0 or x-1=0
So, x= √3 or x=1
-Albert Einstein
Politics is for the present, but an equation is for eternity.
THANK YOU��Hope that this presentation makes you understand the topic better.�