- Trending Categories
- Data Structure
- Networking
- RDBMS
- Operating System
- Java
- MS Excel
- iOS
- HTML
- CSS
- Android
- Python
- C Programming
- C++
- C#
- MongoDB
- MySQL
- Javascript
- PHP
- Physics
- Chemistry
- Biology
- Mathematics
- English
- Economics
- Psychology
- Social Studies
- Fashion Studies
- Legal Studies

- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who

# Are the following statements 'True' or 'False'? Justify your answers.

If the zeroes of a quadratic polynomial $ a x^{2}+b x+c $ are both positive, then $ a, b $ and $ c $ all have the same sign.

To do:

We have to find whether the given statements are true or false.

Solution:

(i) Let $\alpha$ and $\beta$ be the zeroes of the quadratic polynomial \( a x^{2}+b x+c \).

If the zeroes of a quadratic polynomial \( a x^{2}+b x+c \) are both positive, then

$\alpha+\beta=-\frac{b}{a}$

$\alpha \beta=\frac{c}{a}$

This implies,

$c>0, a>0$ and $b<0$ OR $c<0, a<0$ and $b>0$

All of them do not have the same sign.

Hence, the given statement is false.

(ii) We know that,

A quadratic polynomial may touch the X-axis at exactly one point or intersects X-axis at exactly two points or does not touch the X-axis.

Therefore,

If the graph of a polynomial intersects the \( X \)-axis at only one point, then it cannot be a quadratic polynomial.

Hence, the given statement is true.

(iii) We know that,

If the graph of a polynomial intersects the X-axis at exactly two points, then it may or may not be a quadratic polynomial.

A polynomial of degree more than 2 is also possible which intersects the X-axis at exactly two points when it has two real roots and other imaginary roots.

Hence, the given statement is true.

(iv) Let $\alpha, \beta$ and $\gamma$ be the zeroes of a cubic polynomial p(x).

Given that two of the zeroes are zero.

Let $\alpha=\beta=0$ and $\gamma=a$

Therefore,

$p(x)=(x-\alpha)(x-\beta)(x-\gamma)$

$=(x-0)(x-0)(x-a)$

$=x^{3}-a x^{2}$ which does not have linear and constant terms.

Hence, the given statement is true.

(v) Let $p(x)=x^{3}+a x^{2}+b x+c$ is a cubic polynomial and $\alpha, \beta, \gamma$ be the roots of $p(x)$.

This implies,

Sum of the roots $=\alpha+\beta+\gamma=-a$

Sum of negative numbers is negative.

This implies,

$a$ is positive.

Product of the roots taken two at a time $=\alpha \cdot \beta+\alpha \cdot \gamma+\gamma \cdot \beta=b$

Product of two negative numbers is positive and sum of positive numbers is positive.

This implies,

$b$ is positive

Product of the roots $=\alpha \beta \gamma=-c$

Product of three negative numbers is negative

This implies,

$c$ is positive.

Therefore, the sign of all three coefficients will be positive.

Hence, the given statement is true.

(vi) Let $\alpha, \beta$ and $ \gamma$ be the zeroes of cubic polynomial $x^{3}+a x^{2}-b x+c$

This implies,

Product of zeroes $=\alpha \beta \gamma=-\frac{\text { constant term }}{\text { Coefficient of } \mathrm{x}^{3}}$

$=\frac{-c}{1}$

$\alpha \beta \gamma=-c$

Given that, all three zeroes are positive.

This implies,

The product of all three zeroes is also positive.

$\alpha \beta \gamma>0$

$-c>0$

$c<0$

Sum of the zeroes $=\alpha+\beta+\gamma=-\frac{\text { coefficient of } \mathrm{x}^{2}}{\text { Coefficient of } \mathrm{x}^{3}}$

$=\frac{-a}{1}$

$=-a$

But $\alpha, \beta$ and $\gamma$ are all positive.

This implies, its sum is also positive.

$\alpha+\beta+\gamma>0$

$-a>0$

$a<0$

Sum of the product of two zeroes at a time $=\frac{\text { coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{3}}$

$=\frac{-b}{1}$

$= -b$

Therefore, the cubic polynomial $x^{3}+a x^{2}-b x+c$ has all three zeroes which are positive only when all constants $a, b$ and $c$ are negative.

Hence, the given statement is false.

(vii) Let $f(x) = kx^2 + x + k$

For equal roots, discriminant of $f(x)$ should be zero.

$D = b^2 - 4ac = 0$

Therefore,

$D=1^2-4(k)(k) = 0$

$1=4k^2$

$k^2=\frac{1}{4}$

$k =\sqrt{\frac{1}{4}}$

$k=\pm \frac{1}{2}$

So, for two values of $k$, the given quadratic polynomial has equal zeroes.

Hence, the given statement is false.