IIT JEE preparation tips for Location Of Roots
IIT JEE preparation tips for this post will be the roots of quadratic equations:
Let’s start by the beginning, what are the roots of a quadratic equation?
The root for a quadratic equation are the values of x for which f(x) equals zero.
In general, a polynomial equation can have at the most as many roots as the highest exponent of the independent variable in the equation.
Quadratic equations are a particular case of polynomial equations for which the order of the highest exponent of the independent variable is two, coincidentally they are also called “second order” equations.
Therefore a quadratic equation has, at most, two roots.
That is the same as saying there are at most two values of x that could satisfy f(x) = 0.
There are some second order equations for which there are not two real numbers that satisfy f(x) = 0.
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We will soon explain why is that, but first lets see the general expresion of a quadratic function.
The general form of a second order equation is as follows:
f(x) = ax2 + bx + c
Since we said root are values of x for which f(x) = 0, then a root can also be expressed for the equation as ax2 + bx + c = 0 and the solution, must be a number (real or complex), which satisfies the equation.
The process of solving a quadratic equation is also called factorizing or factoring and
the roots of a quadratic equation are given by “resolvente” method, as long as “a” is not zero.
We will now describe the “resolvente” method or general resolution formula.
This method can be applied to calculate the solutions for any second order equation with one unknown variable. These equation is commonly refereed to as “general solving method”.
The “general solving method” or “completing the square” :
Lets consider a generalization of a second order equation with one unknown,
ax2 + bx + c = 0 for which we will find its roots.
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We will start, as a first step, by substracting the independent term (c) on both sides of the equation obtaining:
1) ax2 + bx = -c
Then, as a second step, we will divide both sides by “a” obtaining:
2) x2 + b/a x = -c/a
Now,as a third step, we will change “b/a x” in the equation for “2b/2a x”, we obtain:
and at the same time we add to both sides of the equation “ (b/2a)2”, we will observe that on the left side of the equation a perfect binomial quadratic appears.
X2 + 2(x) (b/2a) + (b/2a)2 = -c/a + (b/2a)2
Which is equivalent to :
x2 + 2(x) (b/2a) + (b/2a)2 = -c/a + b2/4a2
And then, we factorize the first term of the equation and re write the algebraic fraction of the second term to obtain:
(x + b/2a)2 = (b2– 4ac) / 2a2
We apply a square root to both sides to obtain:
x + b/2a = (b2 – 4ac)1/2 / 2a
Note that this equation is just one of the roots of (x + b/2a)2 another equally valid equation would be :
– x – b/2a = (b2 – 4ac)1/2 / 2a
Finally we substract b/2a to both sides and rearrange to obtain:
x = (- b ± (b2 – 4ac)1/2 ) / 2a
The ± symbol is meant to represents both of the possible roots of (b2 – 4ac)1/2 we mentioned.
By solving the equation for both +(b2 – 4ac)1/2 and -(b2 – 4ac)1/2 we will obtain the two possible roots or resolutions for the general quadratic equation.
We must remember, that while there will be two roots, either none or both of them could actually be imaginary. We will call the two roots X1 and X2.
Taking in consideration the above explained method for solving second order equations, it is possible now to understand that there will be three quite distinct possible situations.
Nature of roots:
When b2 – 4ac > 0 , there will be 2 distinct and real roots.
When b2 – 4ac = 0 , there will be 1 root (or 2 equal) and real roots.
When b2 – 4ac < 0 , there will be 2 distinct and non-real roots.
Once the roots of a second order equation are known, it is possible to represent the equation factorized as :
f(x) = a (x – X1 ) (x – X2)
Perhaps this factorized representation makes it more easy to see that,
when x = X1 or x = X2, f(x) = 0.
In some problems we might want to produce a second order equation such that the roots of the equation ax2 + bx + c = 0 lie in a given interval. For this we will impose conditions on a, b, and c. Let f(x) = ax2+ bx + c. There are five posible restrictions on the location of roots:
– Both roots greater than a specific number k.
– Both roots smaller than a specific number k.
– A specific number lies between both the roots.
– Both the roots of f(x) =0 are confined between two specific numbers k1 and k2
- Exactly one root of f(x)=0 lies between two specific numbers k1 and k2 Lets discuss each situation: (i) If both the roots are positive (X1 > 0 and X2 > 0 ), then the sum of the roots as well as the product of the roots must be positive. For this to happen, it is required that: (- b ± (b2 - 4ac)1/2 ) / 2a > 0 Similarly, if both the roots are negative (X1 < 0 and X2 > < 0 ), then the sum of the roots will be negative and the product of the roots must be positive. (ii) Both the roots are greater than a given number k if the following three conditions are satisfied: D>0, (- b / 2a) > k and a f(k) > 0. (iii) Both the roots will be less than a given number k if the following conditions are satisfied: D>0, (- b / 2a) < k and a f(k) > 0. (iv)Both the roots will lie in the given interval (k1, k2) if he following conditions are satisfied: D>0, k1< (- b / 2a) < k2 and a f(k) > 0. (v) Exactly one of the roots lies in the given interval (k1, k2) if f(k1). f(k2) < 0. (vi) A given number k will lie between the roots if a.f.(k) < 0. In particular, the roots of the equation will be of opposite signs if 0 lies between the roots and a.f(0) < 0.
For more about Quadratic equation you can buy Algebra For JEE Mains & Advanced by G. Tewani.
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