# Quadratic Equation – Quiz 6

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# Quadratic Equation – Quiz 6

### Introduction

Quadratic Equation is one of important topic in Quantitative Aptitude Section. In Quadratic Equation – Quiz 6 article candidates can find questions with answer. By solving this questions candidates can improve and maintain, speed, and accuracy in the exams. Quadratic Equation - Quiz 6 questions are very useful for different exams such as IBPS PO, Clerk, SSC CGL, SBI PO, NIACL Assistant, NICL AO, IBPS SO, RRB, Railways, Civil Services etc.

### Q1

If ${x}^{2} -3x + 1 = 0$, then the value of x + $\frac {1}{x}$ is :
A. 0 B. 2 C. 1 D. 3

D
${x}^{2} -3x + 1 = 0 \Rightarrow {x}^{2} + 1 = 3x$
$\Rightarrow \frac {{x}^{2} + 1}{x} = 3$
$\Rightarrow x + \frac {1}{x}$ = 3

### Q2

The roots of $2 {x}^{2} - 6x + 3 = 0$ are :
A. real B. real, unequal & irrational C. unequal & irrational D. irrational

C
D = $[{(6)}^{2} - 4 \times 2 \times 3] = (36 - 24) = 12$
Thus, D>0 and not a perfect square
i.e, Roots are real, unequal and irrational

### Q3

Find out the relationship between x and y. I. ${x}^{2} - 7x + 10 = 0$ II. ${y}^{2} - 14y + 45 = 0$
A. if x > y B. if x = y C. if y > x D. x $\leq$ y

D
I. ${x}^{2} - 7x + 10 = 0$
$\Rightarrow {x}^{2} - 5x -2x + 10 = 0$
$\Rightarrow x(x - 5) - 2(x - 5) = 0$
$\Rightarrow x = 2, 5$
II. ${y}^{2} - 14y + 45 = 0$
$\Rightarrow {y}^{2} - 9y - 5y + 45 = 0$
$\Rightarrow y(y -9) -5(y -9) = 0$
$\Rightarrow y = 9, 5$
i.e, x $\leq$ y

### Q4

If a, b are the two roots of a quadratic equation such that a + b = 24 and a - b = 8, then the quadratic equation is :
A. $2{x}^{2} + 8x + 9 = 0$ B. ${x}^{2} - 4x + 8 = 0$ C. ${x}^{2} - 24x + 128 = 0$ D. ${x}^{2} + 2x + 8 = 0$

C
On solving a + b = 24 and a- b = 8, we get a = 16, b = 8
i.e, ab = 128
i.e, Required equation is ${x}^{2} - (a + b)x + ab = 0$
i.e, ${x}^{2} - 24x + 128 = 0$

### Q5

Find out the relationship between x and y. I. $3{x}^{2} + 5x + 4 = 0$ II. ${y}^{2} + 9y + 20 = 0$
A. if x > y B. if x = y C. if y > x D. if y = x

A
I. $3{x}^{2} + 5x + 4 = 0$
$3{x}^{2} + 3x + 2x + 4 = 0$
$\Rightarrow 3x(x + 1) + (x + 1) = 0$
$\Rightarrow = -1. \frac{-2}{3}$
II. ${y}^{2} + 9y + 20 = 0$
$\Rightarrow y(y + 9) + 3(y + 9) = 0$
$\Rightarrow = -3, -9$
i.e, x>y