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Are the following statements ‘True’ or ‘False’? Justify your answers.

(i)If the zeroes of a quadratic polynomial ax^2 + bx + c are both positive, then a, b and c all have the same sign

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

(iii) If the graph of a polynomial intersects the x-axis at exactly two points, it need not be a quadratic polynomial.

(iv) If two of the zeroes of a cubic polynomial are zero, then it does not have linear and constant terms.

(v) If all the zeroes of a cubic polynomial are negative, then all the coefficients and the constant term of the polynomial have the same sign.

(vi) If all three zeroes of a cubic polynomial x^3 + ax^2 - bx + c are positive, then at least one of a, b and c is non-negative.

(vii) The only value of k for which the quadratic polynomial kx^2 + x + k has equal zeros is\frac{1}{2}.

 

 

 

 

 

 

Answers (1)

(i) Answer. [false]

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

Quadratic polynomial : when the degree of polynomial is two then the polynomial is called quadratic polynomial.

If a quadratic polynomial is p(x)=ax^2 + bx + c then

Sum of zeroes =-\frac{b}{a}

Product of zeroes =\frac{c}{a}

Here two possibilities can occurs:

b > 0 and a < 0, c < 0

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

Here we conclude that a, b and c all have nor same sign is given statement is false.

(ii) Answer. [False]

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

Quadratic polynomial : when the degree of polynomial is two then the polynomial is called quadratic polynomial.

We know that the roots of a quadratic polynomial is almost 2, Hence the graph of a quadratic polynomial intersects the x-axis at 2 point, 1 point or 0 point.

For example :

x^2 + 4x + 4 = 0                       (a quadratic polynomial)

(x + 2)^2 = 0

x + 2 = 0

x =-2

Here only one value of x exist which is –2.

Hence the graph of the quadratic polynomial x^2 + 4x + 4 = 0  intersect the x-axis at x = –2.

Hence, we can say that if the graph of a polynomial intersect the x-axis at only one point can be a quadratic polynomial.

Hence the given statement is false.

(iii)

Answer.  [True]

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

Quadratic polynomial : when the degree of polynomial is two then the polynomial is called quadratic polynomial.

Let us take an example :-

(x -1)^2 (x -2) = 0

It is a cubic polynomial

If we find its roots then x = 1, 2

Hence, there are only 2 roots of cubic polynomial (x -1)^2 (x -2) = 0 exist.

In other words we can say that the graph of this cubic polynomial intersect x-axis at two points x = 1, 2.

Hence we can say that if the graph of a polynomial intersect the x-axis at exactly two points, it need not to be a quadratic polynomial it may be a polynomial of higher degree.

Hence the given statement is true.

(iv)

Answer.   [True]

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

Cubic polynomial: when the degree of polynomial is three then the polynomial is called cubic polynomial.

Let \alpha _1 , \alpha _2 ,\alpha _3 be the zeroes of a cubic polynomial.

It is given that two of the given zeroes have value zero.

i.e. \alpha _1 = \alpha _2 = 0

Let p(y) = (y - \alpha _1)(y - \alpha _2)(y -\alpha _3)

p(y) = (y - 0) (y - 0) (y - \alpha _3)

p(y) = y^3 -y^2\alpha _3

Here, we conclude that if two zeroes of a cubic polynomial are zero than the polynomial does not have linear and constant terms.

Hence, given statement is true.

(v)

Answer.  [True]

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

Cubic polynomial : when the degree of polynomial is three then the polynomial is called cubic polynomial.

Let the standard equation of cubic polynomial is:

p(x) = ax^3 + bx^2 + cx + b

Let \alpha ,\betaand\gamma be the roots of p(x)

It is given that all the zeroes of a cubic polynomial are negative

i.e \alpha = -\alpha , \beta = -\beta and \gamma = -\gamma

Sum of zeroes =\frac{coefficent of x^{2}}{coefficent of x^{3}}

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

It is given that zeroes are negative then \alpha = -\alpha , \beta = -\beta and \gamma = -\gamma

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

-\alpha -\beta -\gamma =-\frac{b}{a}

-\left (\alpha +\beta +\gamma \right ) =-\frac{b}{a} 

\left (\alpha +\beta +\gamma \right ) =\frac{b}{a}   …….(1)

That is \frac{b}{a}>0

Sum of the products of two zeroes at a time =\frac{coefficent of x}{coefficent of x^{3}}

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

Replace \alpha = -\alpha , \beta = -\beta and \gamma = -\gamma

(-\alpha ) (-\beta) +(-\beta) (-\gamma) +(-\gamma) (-\alpha )=\frac{c}{a}

\alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}                …..(2)

That is \frac{c}{a}>0

Product of all zeroes =\frac{-constant terms}{coefficent of x^{3}}

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

Replace \alpha = -\alpha , \beta = -\beta and \gamma = -\gamma

(-\alpha) (-\beta) (-\gamma) =-\frac{d}{a}

-(\alpha \beta \gamma)=-\frac{d}{a}

\alpha \beta \gamma =\frac{d}{a}                              ……(3)

That is \frac{d}{a}>0

From equation (1), (2) and (3) we conclude that all the coefficient and the constant term of the polynomial have the same sign.

Hence, given statement is true.

(vi)

Answer.  [False]

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

Cubic polynomial : when the degree of polynomial is three then the polynomial is called cubic polynomial.

The given cubic polynomial is

p(x)=x^3 + ax^2 - bx + c

Let a, b and g are the roots of the given polynomial

Sum of zeroes =\frac{(coefficent of x^{2} )}{(coefficent of x^{3})}

\alpha +\beta +\gamma =-\frac{a}{1}=-a

We know that when all zeroes of a given polynomial are positive then their sum is also positive

But here a is negative

Sum of the product of two zeroes at a time =\frac{(coefficent of x)}{(coefficent of x^{3})}

\alpha \beta +\beta \gamma +\gamma \alpha =-\frac{b}{1}=-b

Also here b is negative

Product of all zeroes =\frac{-(constant terms)}{(coefficent of x^{3})}

\alpha \beta \gamma =\frac{-c}{1}=-c

Also c is negative

Hence if all three zeroes of a cubic polynomial x^3 + ax^2 - bx + care positive then a, b and c must be negative.

Hence given statement is false.

(vii)

Answer. [False]

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

Quadratic polynomial : when the degree of polynomial is two then the polynomial is called cubic polynomial.

Let p(x)=kx^2 + x + k

Here it is gives that zeroes of p(x) are equal and we know that when any polynomial having equal zeroes then their discriminate will be equal to zero.

i.e. d = 0

b^2 - 4ac = 0            (Q d = b2 – 4ac)

p(x)=kx^2 + x + k

Here, a = k, b = 1, c = k

b^2 - 4ac = 0

(1)2 -4 \times k \times k = 0

1 - 4k^2 = 0

1 = 4k^2

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

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

\therefore When +\frac{1}{2}  and -\frac{1}{2}  then the given quadratic polynomial has equal zeroes.

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