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Quadratic Equation comes under Algebra which is known to have a high weightage in the Mathematics section in JEE Main. This chapter is considered to be a complex chapter. However, with adequate practice, students will find that questions from Quadratic Equation can be quite scoring. Read JEE Main Mathematics Syllabus
Around 2-3 questions are asked from this chapter in JEE Main which bear a total of about 8 marks. Therefore, the total weightage of Quadratic Equation in JEE Main is 2-3%. Some of the topics tested under this chapter are - Identity, Completing the Square, Discriminant, Repeated Roots, Factorization, and Formation of New Equations.
Read the article to find notes for Quadratic Equation for JEE Main preparation.
The discriminant of a quadratic equation is described as the number D= b2-4ac and is found from the coefficients of the equation ax2+bx+c =0. The discriminant shows the nature of roots that an equation has.
Note: b2 – 4ac is found from the quadratic formula
The table given below lists the different kinds of roots that are associated with the values of the determinant.
|D < 0||Two roots that are complex conjugates|
|D = 0||One real root of multiplicity two|
|D > 0||Two distinct real roots|
|D = positive perfect square||Two distinct rational roots (with the assumption that a, b, and c are rational)|
For example, look at the quadratic equation y = 3x2+9x+5. Find its discriminant.
Solution: The quadratic equation given is y = 3x2+9x+5.
The formula of discriminant is D = b2-4ac.
This means, a= 3, b= 9 and c= 5.
Therefore, the discriminant is,
D = 92-4.3.5 = 31.
A polynomial equation is an equation that can be expressed in the following form -
axn + bxn-1 + . . . + rx + s = 0,
where a, b, . . . , r and s are constants. The largest exponent of x that appears in a non-zero term of a polynomial is known as the degree of that polynomial.
It is quite obviously a polynomial of degree 3 and will therefore have three roots. The equation could be factored as (x-1) (x-4) (x-4) =0. This indicates that the roots of the equation are x=1, x=4, x=4. Hence, the root x=4 is repeated.
Descartes Rule of Signs is a very popular rule that aids in getting an idea of the roots of a polynomial equation. This rule states that the number of positive real roots of Pn(x) = 0 cannot be higher than the number of sign changes. In a similar manner, the number of negative roots cannot be higher than the number of sign changes in Pn(-x).
For example, consider the equation P5(x) = x5+2x4-x3+x2-x+2 = 0.
As given above, this equation has four sign changes. This means that it can have a maximum of four positive real roots. Now, it is seen that P5(-x) has only one sign change as,
P5(-x) = -x5+2x4+x3+x2+x+2 = 0
Hence, it can hold only one negative real root.
Sturm theorem states that the number of real roots of the equation f(x) = 0 at [a, b] is equal to the difference between the number of sign changes in the Sturm sequence at x = a and x= b, if f (a) ≠ 0 and f (b) ≠ 0.
It is known that any nth degree polynomial has exactly n roots. This makes it obvious that the number of complex roots is equal to the total roots – number of real roots.
Hence, the number of complex roots = n - number of real roots, where a real root of multiplicity r has to be counted r times. In this case, the coefficients of the polynomial are real, the complex roots are α ± iβ, and therefore, the total number of complex roots are even.
Look at the question (x-3) (x-4)2 (x-5)3 = 0.
The first step must be to make each factor equal to zero. Then, we get,
(x-3) = 0, (x-4)2 = 0, (x-5)3 = 0
Explanation: When the value 6 is entered, a positive sign is obtained. Draw the curve on the top side of the axis.
Now, 5 is touched. It is obtained from the factor (x-5)3. The exponent of this factor is odd. Therefore, the side of the curve must be changed. Again, for the next factor (x-4)2, the power is even and therefore, the side of the wavy curve must be retained. It was below the axis and will remain there only. The curve now looks like this -
The next factor is (x-3). Its power is 1 which is again odd. The side of the curve must be changed. At first, it was below, now it should be above the line.
Therefore, this is how the curve will look like. It is evident that the function is positive in the intervals when the curve rests above the axis and negative in the intervals where the curve rests below the axis.
α + β = -b/a
α.β = c/a
α – β = √D/a
In the equation ax2 + bx + c = 0, where a, b and c ∈Q and a ≠ 0, then,
x2 – (Sum of roots)x + (Product of roots) = 0.
If α and β are the roots of equation, then the quadratic equation is,
x2 – (α + β)x + α β = 0
f(x) = a0xn +a1xn-1 + a2xn-2 + ……. + an-1x + an, then,
Σ α1α2α3 ……αn= (-1)n an/a0
Question 1: Suppose a, b, and c are the sides of a triangle ABC in a way that
x2 – 2(a + b + c)x + 3μ (ab + bc + ca) = 0 has real roots, then,
Solution: Given, the roots are real and hence, D ≥ 0
This means, 4(a+b+c)2 - 12μ(ab+bc+ca) ≥ 0
Thus, (a+b+c)2 ≥ 3μ (ab+bc+ca)
So, a2+b2+c2 ≥ (ab+bc+ca)( 3μ-2)
Hence, 3μ-2 ≤ (a2+b2+c2)/(ab+bc+ca)
Also, cos A = b2 + c2 - a2 /2bc
cos A < 1
This gives, b2 + c2 - a2 < 2bc
Likewise, c2 - a2 - b2 < 2ca
And a2 + b2 - c2 < 2ab
So, a2 + b2 + c2 < 2(ab + bc + ca)
Hence, (a2+b2+c2)/(ab+bc+ca) < 2
Therefore, using the equations obtained,
3μ -2 < 2
It gives μ < 4/3.
Question 2: Suppose x2 – 10ax -11b = 0 has ‘c’ and ‘d’ as its roots and the equation x2 – 10cx -11d = 0 has its roots a and b , find the value of a+b+c+d.
Solution: Given, x2 – 10ax -11b = 0
And x2 – 10ax -11b = 0
Using the concepts of sum of roots,
a+b = 10c and c+d = 10a
(a-c) + (b-d) = 10(c-a)
(b-d) = 11(c-a) ….. (1)
As ‘c’ is a root of x2 – 10ax -11b = 0
Hence, c2 – 10ac -11b = 0 …..(2)
Likewise, ‘a’ is a root of the equation x2 – 10cx -11d = 0
So, a2 – 10ca -11d = 0 …… (3)
Subtracting equation (3) from (2),
(c2 - a2) = 11(b-d) …….. (4)
(c+a) (c-a) = 11.11(c-a) …. (From eq(1))
This implies that c+a = 121
Hence, a+b+c+d = 10c + 10a
Therefore, the required value of (a+b+c+d) = 1210.
Question 3: For what value of k will both the quadratic equations 6x2 – 17x + 12 = 0 and 3x2 – 2x + k = 0 have a common root.
Solution: Given, one of the roots of quadratic equations a1x2 + b1x + c1 and a2x2 + b2x + c2 is common.
Then, (a1b2 – a2b1)(b1c2– b2c1) = (a2c1 – a1c2)2 . . . . . . . . . . . . . (1)
a1 = 6, b1 = -17, c1 = 12, a2 = 3, b2 = -2 and c2 = k
Substituting these values in equation (1),
[(6×-2) – (3×-17)] × [-17k – (-2×12)] = (3×12 – 6k)2
-663k + 936 = 1296 + 36k2 – 432k
36k2 + 231k + 360 = 0
12k2 + 125k + 120 = 0
(4k + 15) (3k + 8) = 0
Therefore, the values of k are -154, -83.
Question 4: Suppose the coefficient of x in the quadratic equation x2 + bx + c =0 was taken to be 17 instead of 13, its roots were found to be -2 and -15. Find the roots of the original quadratic equation.
Solution: As there is no change in the coefficient of x2 and c, the product of zeros will stay the same for both the equations.
The product of zeros (c) = -2 × -15 = 30,
As the original value of b is 13.
∴ Sum of zeros = -b/a = -13.
Hence, the original quadratic equation is:
x2 – (Sum of Zeros)x + (Product of Zeros) = 0
x2 + 13x + 30 = 0
∴ (x + 10) (x + 3) = 0
Therefore, the roots of original quadratic equations are -3 and -10.
Question 5: What is the set of all real numbers x for which x2 - |x + 2| + x > 0?
Solution: Given, x2 - |x + 2| + x > 0
The following two cases are possible -
Case 1: When (x+2) ≥ 0.
x2 - x - 2 + x > 0
So, x2 – 2 > 0
Either x < - √2 or x > √2.
Hence, x ∈ [-2, -√2) ∪ (√2, ∞) ……. (1)
Case 2: When (x+2) < 0
Then, x2 + x + 2 + x > 0
So, x2 + 2x + 2 > 0
This means (x+1)2 + 1 > 0 and this is true for every x
Hence, x ≤ -2 or x ∈ (-∞, -2) ...…… (2)
From equations (2) and (3), x ∈ (-∞, -√2) ∪ (√2, ∞).
Question 1: Determine the values of k for which the quadratic expression (x – a) (x – 10) + 1 = 0 has integral roots.
Trick to solve this question: Rewrite the given equation as x2 – (10 + k)x + 1 + 10k = 0. Determine the value of the discriminant and form an equation with k as (k – 10)2 – D = 4. Now, find the values of k.
Solution: k = 8 and 12.
Question 2: Determine the maximum or minimum value of the quadratic equation -4(x – 2)2 + 2.
Trick to solve this question: The value of a is negative. Therefore, the quadratic equation given is sure to have a maximum value. Now, find the maximum value.
Question 3: How many real roots are there in the equation 2x5 + 2x4 – 11x3 + 9x2 – 4x + 2 = 0?
Trick to solve this question: There are 4 sign changes in the given equation which means that it can have a maximum of 4 positive real roots. Note that for f(-x), there is only one sign change. Now, find the number of real roots in the equation.
Solution: The equation has only one negative real root.
For all candidates who will be appearing for JEE Main 2020 April session in the month of July, we hope that this article helps you in understanding the basic concepts of the topic and thus, pioneering the section.