JEE Quadratic Roots

This page will teach you how to master JEE Quadratic Roots. 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.

A root of the equation $ax^2+bx+c=0$ is a number (real or complex), say $\alpha$, which satisfies the equation i.e. $a\alpha^2+b\alpha+c=0$. The roots of the quadratic equation $ax^2+bx+c=0$ with $a\neq 0$ are given by $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$.

As per JEE syllabus, the main concepts under Quadratic Roots are nature of roots, common roots, Vieta's theorem and symmetric function of roots, Newton's theorem, and location of roots.

Nature of roots

• $b^2-4ac>0$: real and distinct roots
• $b^2-4ac=0$: real and equal roots
• $b^2-4ac<0$: non-real roots
• Conditions for rational, integral, and irrational, roots

Common roots

• Finding the conditions that the equations $a_1x^2+b_1x+c_1=0$ and $a_2x^2+b_2x+c_2=0$ have at least one common root or both the roots common.
• At least one common root: $\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}=\frac{b_1c_2-b_2c_1}{c_1a_2-c_2a_1}$
• Both common roots: $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

Vieta's theorem and symmetric function of roots

• Vieta's theorem: If $\alpha,\beta$ are the roots of the quadratic equation $ax^2=bx+c=0$, then $\alpha+\beta=-\frac ba$ and $\alpha \beta=\frac ca.$
• Relation between roots and coefficients
• Symmetric function of roots: $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha \beta$

Newton's theorem

• If $\alpha$ and $\beta$ are the roots of $ax^2+bx+c=0$ and $S_n=\alpha^n+\beta^n$, then for any natural number $n>2,$ we have $aS_n+bS_{n-1}+cS_{n-2}=0.$

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 $x_1$ and $x_2.$
• Exactly one root of $f(x)=0$ lies between two specific numbers $x_1$ and $x_2.$

Find the value of $k$ such that the equations $x^2+kx+k+2=0$ and $x^2+(1-k)x+3-k=0$ have exactly one common root.

$\begin{array} { l l } A) \, \text{No such value of } \ k \ \text{exists} & \quad \quad \quad \quad \quad & B) \, \frac12 \\ C) \, -\frac12 & & D) \, 1 \\ \end{array}$

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Concepts tested: Common roots

Answer: A) No such value of $k$ exists

Solution:
Using the condition of common root $(c_1a_2-c_2a_1)^2=-(a_1b_2-a_2b_1)(b_2c_1-b_1c_2),$ we have

$\begin{aligned} \left[ (k+2) \cdot 1 - 1 \cdot (3-k) \right]^2&=-\left[ k \cdot (3-k)-(k+2)(1-k) \right] \left[ (1-k) \cdot 1 - k \cdot1\right]\\ (2k-1)^2&=-2(1-2k)(2k-1)\\ (2k-1)^2&=0\\ \Rightarrow k&=\frac12. \end{aligned}$

But on putting $k=\frac 12$ in the given equations $x^2+kx+k+2=0$ and $x^2+(1-k)x+3-k=0,$ we get the two equations as $2x^2+x+5=0$ and $2x^2+x+5=0,$ respectively. Here we see that these two equations have become identical, i.e. both these equations will have both the roots in common.
Hence $k=\frac 12$ is neglected. And hence there is no such value of $k$ for which the equations $x^2+kx+k+2=0$ and $x^2+(1-k)x+3-k=0$ have exactly one common root.

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Common mistakes:

• If you didn't check whether $k=\frac12$ is associated with exactly one common root or both common roots, then you will end up saying that the given equations have exactly one common root for $k=\frac12$.

Find all the values of $a$ for which both the roots of the equation $(a-2)x^2+2ax+(a+3)=0$ lie in the interval $(-2,1)$.

$\begin{array} { l l } A) \, (-\infty,2) \cup (4,\infty) & \quad \quad \quad \quad \quad & B) \, \left( -\infty,-\frac14 \right) \cup (5,6] \\ C) \, (5,6] & & D) \, \left( -\frac14,\infty \right) \\ \end{array}$

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Concepts tested: Location of roots

Answer: B) $\left( -\infty,-\frac14 \right) \cup (5,6]$

Solution:

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Common mistakes:

If $\alpha$ and $\beta$ are the roots of the equation $x^2+3x+1=0$, then find the value of $\left( \frac{\alpha}{1+\beta}\right)^2+\left( \frac{\beta}{1+\alpha} \right)^2$.

$\begin{array} { l l } A) \, \ 18 & \quad \quad \quad \quad \quad & B) \, 19 \\ C) \, 20 & & D) \, 21 \\ \end{array}$

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Concepts tested: Vieta's theorem

Answer: A) $18$

Solution:

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Common mistakes:

The following is a problem from JEE-Mains 2015:

The following is a problem from JEE-Mains 2014:

Find all possible parameters $a$ for which $f(x)=(a^2+a-2)x^2-(a+5)x-2$ is non-positive for every $x \in [0,1]$.

$\begin{array} { l l } A) \, [-3,3] & \quad \quad \quad \quad \quad & B) \, [-4,4] \\ C) \, [-3,-2) \cup (1,3] & & D) \, (-2,-1) \\ \end{array}$

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Concepts tested: Location of roots

Answer: A) $[-3,3]$

Solution:

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Common mistakes:

The following is a problem from JEE-Advanced 2015:

The following is a problem from JEE-Advanced 2014:

The following is a problem from JEE-Advanced 2014:

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