JEE Quadratic Equations

This page will teach you how to master JEE quadratic equations. We highlight the main concepts, provide a list of examples with solutions, and include problems for you to try. Once you are confident, you can take the quiz to establish your mastery.

An equation of the form $ax^2 + bx + c = 0$ with $a\neq 0$ and $a,b,c \in \mathbb C$ is called a quadratic equation, where $a,b,$ and $c$ are the coefficients of $x^2$ and $x,$ and the constant term, respectively. The roots of the quadratic equation $ax^2+bx+c=0$ are given by $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$.

As per JEE syllabus, the main concepts under quadratic equations are identity, discriminant, completing the square, factorization, repeated roots, and formation of new equations.

Identity

• $f(x)=g(x)$ is an identity if $f(x)$ and $g(x)$ have the same value for every real value. An equation with arbitrary coefficients may be an identity under certain conditions.
• $ax^2+bx+c=0 \implies a=b=c=0$

Discriminant

• $b^2-4ac>0:$ real and distinct roots
• $b^2-4ac=0:$ real and equal roots
• $b^2-4ac<0:$ non-real roots

Completing the Square

• Completing the square: $ax^2+bx+c=a\big(x-\frac{b}{2a}\big)^2+\frac{4ac-b^2}{4a}$

Factorization and Repeated Roots

• $x-\alpha$ is a factor of the polynomial $f(x).$ $\Leftrightarrow$ $\alpha$ is a root of the equation $f(x)=0$, i.e. $f(\alpha)=0.$
• $ax^2+bx+c=a(x-\alpha)(x-\beta),$ $\alpha$ and $\beta$ being roots of $ax^2+bx+c=0$
• $ax^2+bx+c=a(x-\alpha)^2,$ $\alpha$ being repeated root of $ax^2+bx+c=0$

Formation of New Equations

• $x^2-\text{(sum of roots)}x + \text{(product of roots)}=0$
• Transformation of equation

For what real values of $p$, will the equation $$\big(p^2-16\big)x^2-\big(4+5p+p^2\big)x-p^3-4p^2-4p-16=0$$ have more than two solutions?

$$\begin{array} { l l } A) \, \text{No such value of } \ p \ \text{exists} & \quad \quad \quad \quad \quad & B) \, 4,-4 \\ C) \, -4 & & D) \, 4,-4,-1 \\ \end{array}$$

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Concepts tested: Identity

Answer: C) $-4$

Solution:
For the quadratic equation $\big(p^2-16\big)x^2-\big(4+5p+p^2\big)x-p^3-4p^2-4p-16=0$ to have more than two solutions, it must be an identity. Therefore, $$p^2-16=4+5p+p^2=-p^3-4p^2-4p-16=0.$$ Solving these, we get
$$\begin{aligned} p^2-16&=0 \implies p=\pm 4&\qquad (1) \\ 4+5p+p^2&=0 \implies p=-1,-4&\qquad (2) \\ -p^3-4p^2-4p-16&=0 \implies (p+4)\big(p^2+4\big)=0 \implies p=-4.&\qquad (3) \end{aligned}$$ Taking the intersection of (1), (2), and (3), we get $p=-4.$

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Common mistakes:

• If you take the union of the values of $p$ which you got after equating the coefficients to zero, you will end up with option D), which is wrong.

Let $a,b$ be the roots of the equation $x^2+13x-2=0$, then find the quadratic equation whose roots are $\frac{a}{a+10}$ and $\frac{b}{b+10}$.

$$\begin{array} { l l } A) \, -32x^2+134x-2=0 & \quad \quad \quad \quad \quad \quad & B) \, 32x^2+134x-2=0 \\ C) \, 111x^2-9x-2=0 & & D) \, 111x^2+9x+2=0 \\ \end{array}$$

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Concepts tested: Transformation of equation

Answer: A) $-32x^2+134x-2=0$

Solution:
Let $y=\frac{x}{x+10}$, where $x$ can take values $a,b$. Then $xy+10y=x \implies x=\frac{10y}{1-y}$. On substituting the value of $x$ in $x^2+13x-2=0$, we get
$$\begin{aligned} \left( \frac{10y}{1-y}\right)^2+13\left( \frac{10y}{1-y}\right)-2&=0\\ 100y^2+13(10y)(1-y)-2(1-y)^2&=0\\ -32y^2+134y-2&=0. \end{aligned}$$ Writing this in terms of the variable $x$, we get $-32x^2+134x-2=0$.

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Common mistakes:

• If you tried to find the new equation by using the exact roots of the given equation, it will lead you to a lot of additional, complicated calculations.

If $P(x)=ax^2+bx+c$ and $Q(x)=ax^2+ax-c$, provided that $ac\neq 0$, then which of the following is true about the roots of the equation $P(x).Q(x)=0?$

$$\begin{array} { l l } A) \, \text{ four real roots} & \quad \quad \quad \quad \quad & B) \, \text{ exactly two real roots } \\ C) \, \text{ at most two real roots }& & D) \, \text{ either two or four real roots } \\ \end{array}$$

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Concepts tested: Discriminant

Answer: D) either two or four real roots

Solution:
$P(x).Q(x)=0$ implies either $P(x)=0$ or $Q(x)=0.$ Let $D_1=b^2-4ac$ be the discriminant of the equation $P(x)=0$ and $D_2=a^2+4ac$ be the discriminant of the equation $Q(x)=0$. If $4ac>0,$ then $D_2>0,$ which implies $Q(x)=0$ has two real and distinct roots; if $4ac <0,$ then $D_1>0,$ which implies $P(x)=0$ has two real and distinct roots. Hence, in any case, the equation $P(x).Q(x)=0$ has either two or four real roots.

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Common mistakes:

• If you expanded the product of $P(x)$ and $Q(x)$, you would be left with a $4^\text{th}$-degree equation and we don't know how to easily solve a $4^\text{th}$-degree equation.

Find the sum of all possible values of $c$ for which the expression $x^3+3x^2-9x+c$ can be expressed as the product of three factors, two of which are identical and monic.

$$\begin{array} { l l } A) \, 2 & \quad \quad \quad \quad \quad \quad \quad \quad \quad & B) \, -22 \\ C) \, -2 & & D) \, -27 \\ \end{array}$$

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Concepts tested: Factorization

Answer: B) $-22$

Solution:
Step 1: Figure out the appropriate factors
Let the identical and monic factor be $x+a$ and then the remaining factor will be $x+\frac{c}{a^2}$. So, we can say that $x^3+3x^2-9x+c=(x+a)^2(x+\frac{c}{a^2})$.

Step 2: Comparing the coefficients and finding the values of $c$
On comparing the coefficients, we get two equations : $\frac{c}{a^2}+2a=3$ and $\frac{2c}{a} +a^2=-9$. On finding the value of $\frac c a$ from first equation i.e. $\frac c a=3a-2a^2$ and substituting in the second one, we get $2(3a-2a^2)+a^2=-9 \Rightarrow 3a^2-6a-9=0 \Rightarrow a=3,-1$ and thus the corresponding values of $c$ are $-27,5$.

Step 3: Summing all the values
Hence the sum of the possible values of $c$ is $-27+5=-22$.

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Common mistakes:

• If you didn't compare the coefficients carefully, you will obtain incorrect equations and hence incorrect values of $c$.

Suppose the inequality$x^2+y^2+xy+1 \geq h(x+y)$ holds for all real $x$ and $y$, then find the sum of all possible integral values of $h$.

$$\begin{array} { l l } A) \, 1 & \quad \quad \quad \quad \quad \quad \quad \quad \quad & B) \, -1 \\ C) \, \infty & & D) \, 0 \\ \end{array}$$

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Concepts tested: Discriminant

Answer: D) $0$

Solution:
Step 1: Finding sign of the quadratic equation in terms of $x$
$x^2+x(y-h)+y^2-hy+1 \geq 0 \ \forall x \ \in \mathbb R \\ \Rightarrow (y-h)^2-4(y^2-hy+1) \leq 0 \ \ (\text{ Discriminant } \leq 0) \\ \Rightarrow -3y^2+2hy+h^2-4 \leq 0$
$\therefore 3y^2-2hy+4-h^2 \geq 0 \ \ \forall y \ \in \mathbb R$

Step 2: Finding the sign of quadratic inequality in terms of $y$
$3y^2-2hy+4-h^2 \geq 0 \\ \Rightarrow 4h^2-4 \cdot 3(4-h^2) \ \leq 0 \ \ (\text{Discriminant } \leq 0) \\ \Rightarrow h^2-3(4-h^2) \leq 0 \\ \Rightarrow 4a^2-12 \leq 0 \Rightarrow h \in [-\sqrt 3, \sqrt 3]$

Step 3: Finding and summing the values of $h$
The integral values of $h$ lying in the interval $[-\sqrt 3, \sqrt 3]$ are $-1,0,1$. Hence the sum of all suitable values of $h$ is $0$.

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Common mistakes:

• If you thought that the coefficient of $xy$ is 1, while when considered as a function in $x$, the coefficient is $y$.

Once you are confident of Quadratic Expressions, move on to JEE Quadratic Roots or JEE Complex Numbers.