# Quadratic Equation – Quiz 5

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# Quadratic Equation – Quiz 5

### Introduction

Quadratic Equation is one of important topic in Quantitative Aptitude Section. In Quadratic Equation – Quiz 5 article candidates can find questions with an answer. By solving this questions candidates can improve and maintain, speed, and accuracy in the exams. Quadratic Equation – Quiz 5 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 one of the roots of the quadratic equation ${x}^{2}$ + mx + 24 = 0 is 1.5, then what is the value of m?
A. -22.5 B. 16 C. -105 D. -17.5

D
We know that the product of the roots of a quadratic equation a${x}^{2}$ + bx + c = 0 is $\frac{c}{a}$
In the given equation, ${x}^{2}$ + mx + 24 = 0, the product of the roots = $\frac{24}{1}$ = 24.
The question states that one of the roots of this equation = 1.5.
If ${x}_{1}$ and ${x}_{2}$ are the roots of the given quadratic equation and let ${x}_{1}$ = 1.5.
Therefore, ${x}_{2}$ = $\frac{24}{15}$= 16.
In the given equation, m is the co-efficient of the x term.
We know that the sum of the roots of the quadratic equation a${x}^{2}$ + bx + c = 0 is
$\frac{-b}{a}$ = $\frac{-m}{1}$ = -m
Sum of the roots = 16 + 1. 5 = 17 = -17.5.
Therefore, the value of m = -17.5

### Q2

Find the remainder when the polynomial ${x}^{4}$ - 3${x}^{2}$ + 7x - 10 is divided by (x - 2).
A. 8 B. -20 C. 18 D. 0

A
By remainder theorem: If a polynomial in one variable 'x' is divided by (x - a), where 'a' is any real number, then the remainder is the value of the polynomial at x = a.
Therefore, remainder = ${(2)}^{4}$ - 3${(2)}^{2}$ + 7(2) - 10 = 8

### Q3

If one of the roots of the quadratic equation 2${x}^{2}$ - 7x + q = 0 is 3, find the other root.
A. -3 B. $\frac {-1}{2}$ C. $\frac {1}{2}$ D. $\frac {-1}{4}$

C
Substituting x = 3 in the given equation we get
2(${(3)}^{2}$) - 7 (3) + q = 0. Therefore, q = 3
Now, the given equation becomes 2${x}^{2}$ - 7x + 3 = 0
By solving this we get (x - 3) (2x - 1) = 0
i.e., x = 3 & $\frac{1}{2}$. The second root is $\frac{1}{2}$.

### Q4

If ${(x + 2)}^{2}$ = 9 and ${(y + 3)}^{2}$ = 25, then the maximum value of x/y is ____.
A. $\frac{1}{2}$ B. $\frac{5}{2}$ C. $\frac{5}{8}$ D. $\frac{1}{8}$

C
${(x + 2)}^{2}$ = 9
$\rightarrow$ x + 2 = ± 3
$\rightarrow$ x = 1 or -5
and ${(y + 3)}^{2}$ = 25
$\rightarrow$ y + 3 = ± 5
$\rightarrow$ y = 2 or -8
Therefore, maximum value of $\frac{x}{y}$ = $\frac{-5}{-8}$ = $\frac{5}{8}$

### Q5

For what value of 'm' will the quadratic equation ${x}^{2}$ + mx + 4 = 0 have real and equal roots?
A. 4 B. -4 C. 4 or -4 D. 16

C
The question states that the roots of the equation are real and equal.
Note:
The roots of a quadratic equation of the form a${x}^{2}$ + bx + c = 0 will be real and equal if its discriminant D = ${b}^{2}$ - 4ac = 0
In this case, b = m, a = 4 and c = 4.
Therefore, ${m}^{2}$ - 4 * 4 * 1 = 0
Or ${m}^{2}$ = 16
Or m = +4 or m = -4.