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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**.

**Must Read:**

- The word ‘quadratic equation’ is derived from the Latin word ‘quadratus’ which means ‘a square’.
- A quadratic equation is any equation with the form,
, where x represents an unknown, and a, b, and c are constants, and a is not equal to 0.**ax2+bx+c =0** - If a = 0, then the equation becomes linear, not quadratic.
- The constants a, b, and c are known as the coefficients.
- ‘c’ expresses the constant term, b is the linear coefficient, and ‘a’ the quadratic coefficient.
- The quadratic equations include only one unknown and hence, are considered to be univariate.
- The quadratic equations are typically polynomial equations because they have non-integral powers of x.
- As the greatest power is two, they are known as second degree polynomial equations.

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.

Discriminant | Roots |
---|---|

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.

For example,

- Look at the equation 3x+1 = 0. The equation has degree 1 as the largest power of x which appears in the equation is 1. These equations are known as linear equations.
- x2 +x-3 = 0 has degree 2 because this is the largest power of x. Such degree 2 equations are known as quadratic equations or simply quadratics.
- Degree 3 equations like x3+2x2-4=0 are known as cubic.
- A polynomial equation of degree n has n roots, but a few of them could be multiple roots. For example, consider x3- 9x2+24x-16 =0.

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

- By equating all the terms to zero,various values of x which are known as the critical points are obtained. Here, 3, 4, and 5 are the critical points.
- Next, plot these points on a number line in an increasing order. After plotting the points, enter a value higher than the rightmost point in the whole function.
- Here, the rightmost point is 5. Enter a value higher than 5, say, 6 in the whole function. When you put x=6 in the whole function, you get a positive sign so you can start drawing the curve above the number line.
- Now is the crucial step. Look for the exponents of every term. When the power of the factor is odd, you should change the side of the wavy curve (if it is above, then make it below and vice versa). Otherwise, you should continue with the same side of the number line.

**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.

**Must Read:**

- To be able to solve a quadratic equation of the form ax2 + bx + c, the discriminant can be calculated using the formula D = b2 – 4ac.
- The solution of the quadratic equation ax2 + bx + c= 0 is expressed as x = [-b ± √ b2 – 4ac] / 2a
- Suppose α and β are the roots of the quadratic equation ax2 + bx + c = 0, then we will have the results given below for the sum and product of roots:

α + β = -b/a

α.β = c/a

α – β = √D/a

- It is impossible for a quadratic equation to hold three different roots and if, in any case it occurs, then the equation turns into an identity.
**Nature of Roots:**Look at the equation ax2 + bx + c = 0, where a, b, and c ∈ R and a ≠ 0, then the following cases are obtained:

- D > 0 when the roots are real and distinct, that is, the roots are unequal.
- D = 0 when the roots are real and coincident, that is, equal.
- D < 0 when the roots are imaginary.
- The imaginary roots typically happen in pairs, that is, if a+ib is one root of a quadratic equation, then the other root should be the conjugate, that is, a-ib, where a, b ∈ R and i = √-1.

In the equation ax2 + bx + c = 0, where a, b and c ∈Q and a ≠ 0, then,

- When D > 0 and is also a perfect square, then the roots are unequal and rational.
- When α = p + √q is a root of the equation, where ‘p’ is rational and √q is a surd, then the other root should be the conjugate of it i.e. β = p - √q and vice versa.

- When the roots of the quadratic equation are given, the quadratic equation could be created using the formula -

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

- In the quadratic expression y = ax2 + bx + c, where a, b, c ∈ R and a ≠ 0, the graph between x and y is usually a parabola.

- When a > 0, the shape of the parabola is concave upwards.
- When a < 0, the shape of the parabola is concave upwards.

- The inequalities of the form P(x)/ Q(x) > 0 could be easily solved using the method of intervals of number line rule.
- The maximum and minimum values of the form y = ax2 + bx + c happen at the point x = -b/2a according to whether a > 0 or a< 0.

- y ∈[(4ac-b2) / 4a, ∞] if a > 0
- When a < 0, y ∈ [-∞, (4ac-b2) / 4a]

- The quadratic function of the form f(x, y) = ax2+by2 + 2hxy + 2gx + 2fy + c = 0 could be divided into two linear factors given that it satisfies the following condition: abc + 2fgh –af2 – bg2 – ch2 = 0
- Generally, if α1,α2, α3, …… ,αn are the roots of the equation

f(x) = a0xn +a1xn-1 + a2xn-2 + ……. + an-1x + an, then,

- 1.Σα1 = - a1/a0
- 2.Σ α1α2 = a2/a0
- 3.Σ α1α2α3 = - a3/a0

……… ……….

Σ α1α2α3 ……αn= (-1)n an/a0

- Each equation of the nth degree consists of exactly n roots (n ≥1) and if it has more than n roots, then the equation turns into an identity.
- When there are two real numbers ‘a’ and ‘b’ such that f(a) and f(b) are of the opposing signs, then f(x) = 0 should have a minimum of one real root between ‘a’ and ‘b’.
- Each equation f(x) = 0 of the odd degree consists of a minimum of one real root of a sign which is opposite to that of its last term.

**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,**

**μ < 4/3****μ > 5/3****μ ∈ (4/3, 5/3)****μ ∈ (1/3, 5/3)**

**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

= 10(c+a)

= 1210.

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?**

**(-∞, -2) ∪ (2, ∞)****(-∞, -√2) ∪ (√2, ∞)****(-∞, -1) ∪ (1, ∞)****(√2, ∞)**

**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.

Solution: 2

**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.

- Dedicate more effort and time to Algebra and Calculus because they have the highest weightage in the Mathematics section.
- Thoroughly learn and revise all the formulae from calculus, probability, trigonometry, and geometry.
- Make use of NCERT, Arihant, and R.D Sharma books for JEE Main Mathematics preparation.
- Solve as many previous years’ question papers and mock tests as you can.

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.

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