Quadratic Equations is an important topic in the Mathematics section in VITEEE exam. Practising this topic will increase your score overall and make your conceptual grip on VITEEE exam stronger.
This article gives you a full set of VITEEE PYQs for Quadratic Equations with explanations for effective preparation. Practice of VITEEE Mathematics PYQs including Quadratic Equations questions regularly will improve accuracy, speed, and confidence in the VITEEE 2026 exam.
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VITEEE PYQs for Quadratic Equations with Solutions
1.
If $\alpha$ and $\beta$ are the roots of $x^{2}-ax+b=0$ and if $\alpha^{n}+\beta^{n}=V_{_n},$ then- $V_{_{n+1}}= aV_{_n}+ bV_{_{n-1}}$
- $V_{_{n+1}}= aV_{_n}+ aV_{_{n-1}}$
- $V_{_{n+1}}= aV_{_n}- bV_{_{n-1}}$
- $V_{_{n+1}}= aV_{_n-1}- bV_{_{n}}$
2.
If $\frac{x^{2}+2x+7}{2x+3} < 6, x\,\in\,R, $ then- $x >11$ or $x < -\frac{3}{2}$
- $x > 11$ or $x < -1$
- $ -\frac{3}{2} < x < -1$
- $ -1 < x < 11$ or $x < -\frac{3}{2}$
3.
The ratio of the sum of two numbers to their difference is 5:1. If the sum of the numbers is 18, find the numbers.10.8, 7.2
- 10, 8
- 9, 9
- 14, 4
4.
If $z=\frac{7-i}{3-4i} \, then\, z^{14}=$- $2^{7}$
- $2^{7} i$
- $2^{14} i$
- $-2^{7}i$
5.
If one root is square of the other root of the equation $x^2 + px + q = 0$, then the relations between p and q is- $p^3 - (3p - 1) q + q^2 = 0$
- $p^3 - q (3p + 1) + q^2 = 0$
- $p^3 + q (3p - 1) + q^2 = 0$
- $p^3 + q (3p + 1) + q^2 = 0$
6.
The value of x obtained from the equation $\begin{vmatrix}x+\alpha&\beta&\gamma\\ \gamma&x+\beta&\alpha\\ \alpha&\beta&x+\gamma\end{vmatrix}=0$ will be- 0 and $-\left(\alpha+\beta+\gamma\right)$
- 0 and $\alpha+\beta+\gamma$
- 1 and $\left(\alpha-\beta-\gamma\right)$
- $0 and \alpha^{2}+\beta+\gamma^{2}$
7.
Find the sum of the roots of the quadratic equation \( 2x^2 - 3x - 5 = 0 \).- \( \frac{3}{2} \)
- \( \frac{5}{2} \)
- \( \frac{-5}{2} \)
- \( \frac{-3}{2} \)
8.
Solve for \( x \) in the equation \( \frac{1}{x} + \frac{1}{x+2} = \frac{5}{6} \).- \( x = 1 \)
- \( x = 2 \)
- \( x = 3 \)
- \( x = 4 \)
9.
Find the solution of the quadratic equation \( 2x^2 - 3x - 5 = 0 \).- \( x = 1, x = -2 \)
- \( x = -1, x = 2 \)
- \( x = \frac{5}{2}, x = -1 \)
- \( x = \frac{-5}{2}, x = 1 \)
10.
If the roots of the quadratic equation $$ (a^2 + b^2) \, x^2 - 2 \, (bc + ad) \, x + (c^2 + d^2) = 0 $$ are equal, then:- \( \frac{a}{b} = \frac{c}{d} \)
- \( \frac{a}{c} + \frac{b}{d} = 0 \)
- \( \frac{a}{d} = \frac{b}{c} \)
- \( a + b = c + d \)
11.
If $\alpha \, and \, \beta$ are the roots of the equation ax+bx+c=0, then the value of $\alpha^{3} \, + \, \beta^{3}$ is- $\frac{3abc \, + \, b^{3}}{a^3}$
- $\frac{a^3 \, + \, b^3}{3abc}$
- $\frac{3abc \, - \, b^3}{a^3}$
- $\frac{-(3abc \, + \, b^3)}{a^3}$
12.
If b$^2 \ge 4 ac$ for the equation $ax^4 + bx^2 + c = 0$, then all the roots of the equation will be real if- b>0, a<0, c>0
- b<0, a>0, c>0
- b>0, a>0, c>0
- b>0, a>0, c<0
13.
If the roots of the equation $x^2 + ax + b = 0$ are $c$ and $d$, then one of the roots of the equation $x^{2}+\left(2c+a\right)x+c^{2}+ac+b=0$ is- $c$
- $d-c$
- $2\,d$
- $2\,c$
14.
For what values of m can the expression $2x^2 + mxy + 3y^2 - 5y - 2$ be expressed as the product of two linear factors- 0
- $\pm\, 1$
- $\pm\, 7$
- 49
15.
The product of all values of $(\cos \alpha + i \sin \alpha)^{3/5}$ is equal to- 1
- $\cos \alpha + i \sin \alpha $
- $\cos 3 \alpha + i \sin 3 \alpha $
- $\cos 5 \alpha + i \sin 5 \alpha $
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