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
\[\begin{align} p^2-16&=0 \implies p=\pm 4&\qquad (1) \\ 4+5p+p^2&=0 \implies p=-1,-4&\qquad (2) \\ -p^3-4p^2-4p-16&=0 \implies (p+4)\big(p^2+4\big)=0 \implies p=-4.&\qquad (3) \end{align}\] 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
\[\begin{align} \left( \frac{10y}{1-y}\right)^2+13\left( \frac{10y}{1-y}\right)-2&=0\\ 100y^2+13(10y)(1-y)-2(1-y)^2&=0\\ -32y^2+134y-2&=0. \end{align}\] 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.