Q&A - Ask Doubts and Get Answers

#2.1

Which of the following is not the graph of a quadratic polynomial?

Answer. [D]

Solution. Polynomial : It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s) and a quadratic polynomial is polynomial of degree 2.

If p(x) is quadratic polynomial then there are almost 2 values of x exists called roots.

Hence, the graph of p(x) cuts the x-axis almost 2 times.

But here we see that in option (D) the curve cuts the x-axis at 3 times

Hence, graph (D) is not a graph of a quadratic polynomial.

View Full Answer(1)

Posted by

infoexpert21

#2.1

If one of the zeroes of a quadratic polynomial of the form $x^{}2+ax + b$ is the negative of the other, then it

(A) has no linear term and the constant term is negative.

(B) has no linear term and the constant term is positive.

(C) can have a linear term but the constant term is negative.

(D) can have a linear term but the constant term is positive.

Answer. [A]

Solution. Polynomial : It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s) and a quadratic polynomial is polynomial of degree 2.

Here the given quadratic polynomial is $x^{}2+ax + b$ …..(1)

a = 1, b = a, c = b

Let x₁, x₂ are the zeroes of the equation (1)

According to question:

$x_2 = - x_1$

sum of zeroes = $x_2 + x_1=\frac{-b}{a}$

$x_1 - x_1=\frac{-b}{a}$ ( because x₂ = - x₁ )

$0=\frac{-a}{1}\Rightarrow a=0$ ( because b = a , a = 1)

Product of zeroes $= (x_1)(x_2) = \frac{c}{a}$

$= (x_1)(-x_1) = \frac{c}{a}$

$= -x_{1}^{2} =b$ ( because c= b, a = 1)

Put value of a and b in (1)

$x^{2}+ \left (-x_{1}^{2} \right )$

Hence, it has no linear term and the constant term is negative.

View Full Answer(1)

Posted by

infoexpert21

#2.1

If the zeroes of the quadratic polynomial $ax^2 + bx + c, c \neq 0$ are equal, then

(A) c and a have opposite signs

(B) c and b have opposite signs

(C) c and a have the same sign

(D) c and b have the same sign

Answer. [C]

Solution. Polynomial : It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s) and a quadratic polynomial is polynomial of degree 2.

Here the given polynomial is $ax^2 + bx + c, c \neq 0$

$a = a, b = b, c = c$

We know that if both the zeroes are equal there

$b^2 - 4ac = 0$

$b^2 =4ac$ ….(1)

(A) c and a have opposite sign

If c and a have opposite sign then R.H.S. of equation (1) is negative but L.H.S. is always positive. So (A) is not a correct one.

(B) c and b have opposite sign

If c is negative and b is positive L.H.S. is positive but R.H.S. of eqn. (1) is negative. Hence (B) is not correct one.

(C) c and a have same sign

If c and a have same sign R.H.S. of eqn. (1) is positive and L.H.S. is always positive hence it is a correct one.

(D) c and b have same sign

If c and b both have negative sign then R.H.S. of eqn. (1) is negative and L.H.S. is positive. So this is not correct.

Only one option i.e. (c) is correct one.

View Full Answer(1)

Posted by

infoexpert21

#2.1

The zeroes of the quadratic polynomial $x^2 + kx + k, k \neq 0$ ,

(A) cannot both be positive

(B) cannot both be negative

(C) are always unequal

(D) are always equal

Answer. [A]

Solution. Polynomial : It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s) and a quadratic polynomial is polynomial of degree 2.

Here the given quadratic polynomial is $x^2 + kx + k$

$a = 1, b = k, c = k$

$x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

$x=\frac{-k\pm \sqrt{k^{2}-4k}}{2}$

$x=\frac{-k\pm \sqrt{k(k-4)}}{2}$

$k(k - 4)$ must be grater then 0

Hence the value of k is either less than 0 or greater than 4.

If value of k is less than 0 only one zero is positive.

If value of k is greater than 4 only one zero is positive.

Hence both the zeroes can not be positive.

View Full Answer(1)

Posted by

infoexpert21

#2.1

The zeroes of the quadratic polynomial $x^2 + 99x + 127$ are

(A) both positive

(B) both negative

(C) one positive and one negative

(D) both equal

Answer. [B]

Solution. Polynomial : It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s) and a quadratic polynomial is polynomial of degree 2.

Here the given quadratic polynomial is $x^2 + 99x + 127$

$x^2 + 99x + 127$

$a = 1, b = 99, c = 127$

$x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

$x=\frac{-99\pm \sqrt{(99)^{2}-4(1)(127)}}{2(1)}$

$x=\frac{-99\pm \sqrt{9801-508}}{2(1)}$

$x=\frac{-99\pm \sqrt{9293}}{2}$

$x=\frac{-99\pm 96.4}{2}$

$x=\frac{-99+ 96.4}{2}$ $x=\frac{-99-96.4}{2}$

$x=\frac{-2.6}{2}=-1.3$ $x = -97.7$

The value of both the zeroes are negative.

View Full Answer(1)

Posted by

infoexpert21

#2.1

If one of the zeroes of the cubic polynomial $x^3 + ax^2 + bx + c$ is –1, then the product of the other two zeroes is

(A) b – a + 1

(B) b – a – 1

(C) a – b + 1

(D) a – b –1

Answer. [A]

Solution. Polynomial : It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s) and a quadratic polynomial is polynomial of degree 3.

Here the given cubic polynomial is $x^3 + ax^2 + bx + c$

Let $\alpha ,\beta ,\gamma$ are the zeroes of polynomial

$\alpha =-1$ (given) …..(1)

put x = –1 in $p(x)=x^3 + ax^2 + bx + c$

$p(-1) = (-1)^3 + a(-1)^2 + b(-1) + c$

$0 = -1 + a - b + c ( Q=-1 is zero)$

$c = 1 - a + b$

We know that

$\alpha \beta \gamma =\frac{-d}{a}$

Here, a = 1, b = a, c = b, d = c

So, $\alpha \beta \gamma =\frac{-c}{1}$

$\beta \gamma =\frac{-(1-a+b)}{\alpha }$

$\beta \gamma =\frac{-(1-a+b)}{-1 } = 1 -a + b$ ( using equation (1))

Hence the product of other two is b – a + 1.

View Full Answer(1)

Posted by

infoexpert21

#2.1

Given that one of the zeroes of the cubic polynomial $ax^3 + bx^2 + cx + d$ is zero, the product of the other two zeroes is

(A) $-\frac{c}{a}$

(B) $\frac{c}{a}$

(C) 0

(D) $-\frac{b}{a}$

Answer. [B]

Solution. Polynomial : It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s) and a quadratic polynomial is polynomial of degree 3.

Here the given cubic polynomial is $ax^3 + bx^2 + cx + d$

Let three zeroes are $\alpha ,\beta ,\gamma$

$\alpha =0$ (given) …..(1)

we know that

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

Put $\alpha =0$

$\beta \gamma =\frac{c}{a}$ ( using equation (1))

Hence the product of other two zeroes is $\frac{c}{a}$ .

View Full Answer(1)

Posted by

infoexpert21

#2.1

The number of polynomials having zeroes as –2 and 5 is

(A) 1

(B) 2

(C) 3

(D) more than 3

Answer. [D]

Solution. Polynomial : It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s).

Let the polynomial is $ax^2 + bx + c = 0$ ….(*)

We know that sum of zeroes $-\frac{b}{a}$

$- 2 + 5= -\frac{b}{a}$

$\frac{3}{1}=-\frac{b}{a}$

…..(1)

Multiplication of zeroes $=\frac{c}{a}$

$-2 \times 5=\frac{c}{a}$

$-\frac{10}{1}=\frac{c}{a}$ …..(2)

Form equation (1) and (2) it is clear that

a = 1, b = –3, c = –10

put value of a, b and c in equation (*)

$x^3 - 3x - 10 = 0$ ….(3)

But we can multiply of divide eqn. (3) by any real number except 0 and the zeroes remain same.

Hence, there are infinite number of polynomial exist with zeroes –2 and 5.

Hence the answer is more than 3.

View Full Answer(1)

Posted by

infoexpert21

#2.1

If the zeroes of the quadratic polynomial $x^2 + (a + 1) x + b$ are 2 and –3, then

(A) a = –7, b = –1

(B) a = 5, b = –1

(C) a = 2, b = – 6

(D) a = 0, b = – 6

Answer. [D]

Solution. Polynomial : It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s) and a quadratic polynomial is polynomial of degree 2.

If 2 and –3 are the zero of $p(x) = x^2 + (a + 1)x + b$ then p(2) = p(–3) = 0.

$p(2) = (2)^2 + (a + 1) (2) + b$

$0 = 4 + 2a + 2 + b$

$0 = 6 + 2a + b$

$2a + b = -6$ ….(1)

$p(-3) = (-3)^2 + (a + 1) (-3) + b$

$0 = 9 - 3a - 3 + b$

$3a - b = 6$ ….(2)

Add equation (1) and (2)

$2a + b + 3a - b = -6 + 6$

2a + 3a = 0

5a = 0

a = 0

put a = 0 in (1)

2(0) + b = –6

b = –6

Hence a = 0, b = –6.

View Full Answer(1)

Posted by

infoexpert21

#2.1

A quadratic polynomial, whose zeroes are –3 and 4, is

(A) $x^2 - x + 12$

(B) $x^2 + x + 12$

(C) $\frac{x^2}{2} - \frac{x}{2} -6$

(D) $2x^2 + 2x -24$

Answer. [C]

Solution. Polynomial : It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s) and a quadratic polynomial is polynomial of degree 2.

(A) $p(x) = x^2 - x + 12$