CBSE Class 10 Mathematics Polynomials VBQs Set 04

Question. If \( p(x) \) is a polynomial of at least degree one and \( p(k) = 0 \), then \( k \) is known as
(a) value of \( p(x) \)
(b) zero of \( p(x) \)
(c) constant term of \( p(x) \)
(d) none of the options
Answer: (b) zero of \( p(x) \)
Sol. Let \( p(x) = ax + b \)
Put \( x = k \)
\( p(k) = ak + b = 0 \)

\( \therefore \) \( k \) is zero of \( p(x) \).

Question. Ravi claims that the polynomial \( p(x) = mxa + x^{2b} \) has \( 4b \) zeroes. For Ravi’s claim to be correct, which of these must be true?
(a) \( a = 2b \) or \( a = 4b \)
(b) \( a = 2 \) or \( a = 4b \)
(c) \( m = 2b \)
(d) \( m = 4b \)
Answer: (a) \( a = 2b \) or \( a = 4b \)
Sol. only (a) satisfy the condition

Question. The polynomial having \( x = 3 \) as one of the zeroes is
(a) \( 2x^3 - 5x^2 - 4x + 3 \)
(b) \( x^2 + 5 \)
(c) \( x^3 + 9 \)
(d) \( x^2 - 12 \)
Answer: (a) \( 2x^3 - 5x^2 - 4x + 3 \)
Sol. Here \( p(x) = 2x^3 - 5x^2 - 4x + 3 \)

\( \therefore \) \( p(3) = 2(3)^3 - 5 \times (3)^2 - 4 \times 3 + 3 \)
\( = 54 - 45 - 12 + 3 = 0 \)
\( \because p(3) = 0 \)

\( \therefore \) \( x = 3 \) is a zero of \( p(x) \)

Question. The graph of \( y = f(x) \) is given, the number of zeroes of \( f(x) \) is/are
(a) 0
(b) 1
(c) 2
(d) 3
Answer: (a) 0
Sol. \( \because \) Graph \( y = f(x) \) does not intersect \( x \)-axis.

\( \therefore \) \( f(x) \) has no zeroes.

Question. If one zero of \( p(x) = ax^2 + bx + c \) is zero, then the value of \( c \) is
(a) 1
(b) 2
(c) 3
(d) 0
Answer: (d) 0
Sol. \( x = 0 \) is a zero of \( p(x) \)

\( \therefore \) \( p(0) = 0 \)
\( \implies \) \( a \times (0)^2 + b(0) + c = 0 \)
\( \implies \) \( c = 0 \)

Question. For what real value of \( k \), is 3 a zero of the polynomial \( 2x^2 + x + k^2 \)?
Answer: Sol. Since 3 is a zero of the polynomial \( p(x) = 2x^2 + x + k^2 \)

\( \therefore \) \( p(3) = 0 \)

\( \implies \) \( p(3) = 2(3)^2 + 3 + k^2 \)

\( \implies \) \( 0 = 18 + 3 + k^2 \)
\( \implies \) \( k^2 = - 21 \)
The square of any real number can’t be negative.

\( \therefore \) there is no real value of \( k \), for which ‘3’ is a zero of \( p(x) \).

Question. Zeroes of a polynomial can be determined graphically. Number of zeroes of a polynomial is equal to number of points where the graph of polynomial
(a) intersects y-axis
(b) intersects x-axis
(c) intersects y-axis or intersects x-axis
(d) none of the options
Answer: (b) intersects x-axis

Question. If graph of a polynomial does not intersect the x-axis but intersects y-axis in one point, then number of zeroes of the polynomial is equal to
(a) 0
(b) 1
(c) 0 or 1
(d) none of the options
Answer: (a) 0

Question. A polynomial of degree n has
(a) only 1 zero
(b) at least n zeroes
(c) atmost n zeroes
(d) more than n zeroes
Answer: (c) atmost n zeroes

Question. The graph of the polynomial \( p(x) \) cuts the x-axis at 2 places and touches it at 4 places. The number of zeroes of \( p(x) \) is
(a) 2
(b) 6
(c) 4
(d) 8
Answer: (b) 6

Question. The graph of \( y = x^3 - 4x \) cuts x-axis at (-2, 0), (0, 0) and (2, 0). The zeroes of \( x^3 - 4x \) are
(a) 0, 0, 0
(b) -2, 2, 2
(c) -2, 0, 2
(d) -2, -2, 2
Answer: (c) -2, 0, 2

Question. Is \( x = - 3 \) zero of the polynomial \( p(x) = 2x^2 + 5x + 3 \)?
Answer: No

Question. Show that \( x = - 2 \) is a zero of the polynomial \( p(x) = 3x^2 + 13x + 14 \)
Answer: \( p(-2) = 3(-2)^2 + 13(-2) + 14 = 12 - 26 + 14 = 0 \)

Question. If 1 is one of the zeroes of polynomial \( x^2 - x + k \), then find value of \( k \).
Answer: \( k = 0 \)

Question. If \( p \) is a zero of \( 2x^2 - 5x + 3 \), then find the value of \( p \).
Answer: \( p = 3/2 \) or \( p = 1 \)

Question. \( k \) is a zero of the polynomial \( p(x) = x^2 - 11x + 24 \). If \( k \) is a prime number, then find the value of \( k \).
Answer: \( k = 3 \)

Question. If \( p(x) = ax^2 + bx + c \), then \( -b/a \) is equal to
(a) 0
(b) 1
(c) product of zeroes
(d) sum of zeroes
Answer: (d) sum of zeroes
Sol. Sum of zeroes \( = -b/a \)

Question. If \( p(x) = ax^2 + bx + c \) and \( a + b + c = 0 \), then one zero is
(a) \( -b/a \)
(b) \( c/a \)
(c) \( c/b \)
(d) none of the options
Answer: (b) \( c/a \)
Sol. \( p(x) = ax^2 + bx + c \)

\( \implies \) \( p(1) = a(1)^2 + b(1) + c = a + b + c \)
So, \( p(1) = a + b + c = 0 \) \( \{ \because a + b + c = 0 \} \)

\( \therefore \) One of the zeroes of \( p(x) \) is \( 1 \), \( \alpha = 1 \)
\( \alpha\beta = \text{product of zeroes} = c/a \)

\( \implies \) \( 1 \cdot \beta = c/a \)

\( \implies \) \( \beta = c/a \)

\( \therefore \) zeroes are 1 and \( c/a \)

Question. If \( p(x) = ax^2 + bx + c \) and \( a + c = b \), then one of the zeroes is
(a) \( b/a \)
(b) \( c/a \)
(c) \( -c/a \)
(d) \( -b/a \)
Answer: (c) \( -c/a \)
Sol. \( p(-1) = a(-1)^2 + b(-1) + c = a - b + c = 0 \) (given),

\( \therefore \) One zero \( (\alpha) = -1 \)
\( \alpha\beta = \text{product of zeroes} = c/a \)

\( \implies \) \( (-1) \cdot \beta = c/a \)

\( \implies \) \( \beta = -c/a \)

Question. The quadratic polynomial, the sum of whose zeroes is –5 and their product is 6, is
(a) \( x^2 + 5x + 6 \)
(b) \( x^2 - 5x + 6 \)
(c) \( x^2 - 5x - 6 \)
(d) \( -x^2 + 5x + 6 \)
Answer: (a) \( x^2 + 5x + 6 \)
Sol. sum of zeroes = –5, product of zeroes = 6
Polynomial is,
\( x^2 - (\text{sum of zeroes})x + \text{product of zeroes} \)

\( \implies \) \( x^2 - (-5)x + 6 = x^2 + 5x + 6 \).

Question. If the product of the zeroes of \( x^2 - 3kx + 2k^2 - 1 \) is 7, then values of \( k \) are
(a) \( \pm 1 \)
(b) \( \pm 2\sqrt{2} \)
(c) \( \pm 2 \)
(d) \( \pm 4 \)
Answer: (c) \( \pm 2 \)
Sol. Product of zeroes = 7

\( \implies \) \( 2k^2 - 1 = 7 \)

\( \implies \) \( 2k^2 = 8 \) \( \implies \) \( k^2 = 4 \) \( \implies \) \( k = \pm 2 \)

Question. The product of the zeroes of \( - 2x^2 + kx + 6 \) is
(a) 1
(b) 2
(c) -3
(d) -4
Answer: (c) -3
Sol. Here \( a = - 2 \), \( b = k \), \( c = 6 \)
Product of zeroes \( = c/a \)
i.e., \( \alpha \times \beta = \frac{6}{-2} = - 3 \)

Question. The sum of the zeroes of the given quadratic polynomial \( -3x^2 + k \) is
(a) 1
(b) \( -1/3 \)
(c) 3
(d) 0
Answer: (d) 0
Sol. Since polynomial is \( -3x^2 + 0x + k \)

\( \therefore \) \( a = -3 \), \( b = 0 \), \( c = k \)
and sum of zeroes \( = -b/a \)
i.e., \( \alpha + \beta = \frac{-b}{a} \) \( \implies \) \( \alpha + \beta = \frac{0}{-3} = 0 \)

Question. If one zero of the polynomial \( x^2 - 4x + 1 \) is \( 2 + \sqrt{3} \), then the other zero is
(a) \( -\sqrt{3} \)
(b) \( 2 - \sqrt{3} \)
(c) 4
(d) 1
Answer: (b) \( 2 - \sqrt{3} \)
Sol. Let other zero be \( \alpha \),

\( \therefore \) \( (2 + \sqrt{3}) + \alpha = -b/a = -(-4/1) \)

\( \implies \) \( \alpha = 4 - 2 - \sqrt{3} = 2 - \sqrt{3} \)

Question. The zeroes of the polynomial \( (x - 2)^2 + 4 \) is
(a) \( \pm 1 \)
(b) \( \pm 2 \)
(c) \( \pm \sqrt{2} \)
(d) no zero
Answer: (d) no zero
Sol. For zeroes \( (x - 2)^2 + 4 = 0 \)
\( (x - 2)^2 + 2^2 = 0 \)
Sum of two perfect squares is zero if each of them is zero.

\( \therefore \) No zero.

Question. Find the zeroes of \( \sqrt{3}x^2 + 10x + 7\sqrt{3} \).
Answer: Sol. \( \sqrt{3}x^2 + 10x + 7\sqrt{3} \)
\( = \sqrt{3}x^2 + 3x + 7x + 7\sqrt{3} \)
\( = \sqrt{3}x(x + \sqrt{3}) + 7(x + \sqrt{3}) = (\sqrt{3}x + 7)(x + \sqrt{3}) \)
For zeroes of the polynomial,
\( (\sqrt{3}x + 7)(x + \sqrt{3}) = 0 \)

\( \implies \) \( \sqrt{3}x + 7 = 0 \) or \( x + \sqrt{3} = 0 \)

\( \implies \) \( \sqrt{3}x = - 7 \) or \( x = - \sqrt{3} \) \( \implies \) \( x = \frac{-7}{\sqrt{3}}, -\sqrt{3} \)

Question. 100% Scoring Tips
If question is “Find zeroes of polynomial \( x^2 - 9 \).”
Then, answer is given as “\( x^2 = 9 \implies x = \pm 3 \). There are two zeroes, i.e. \( x = 3, x = - 3 \).”
Note: But don’t answer as “\( x^2 - 9 = 0 \implies x^2 = 9 \implies x = 3 \). There is only one zero, \( x = 3 \).” This is not the correct answer.

Question. Find a quadratic polynomial whose zeroes are –9 and \( -1/9 \).
Answer: Sol. Sum of zeroes \( = -9 + (-1/9) = \frac{-81-1}{9} = \frac{-82}{9} \)
Product of zeroes \( = (-9) \times (-1/9) = 1 \)
\( \therefore \) Required polynomial is given by,
\( p(x) = k[x^2 - (\text{sum of zeroes})x + \text{product of zeroes}] \), where \( k \) is a non-zero real number
\( = k[x^2 - (-82/9)x + 1] = \frac{k}{9}[9x^2 + 82x + 9] \)
\( = 9x^2 + 82x + 9 \) [taking \( k = 9 \)]

Question. If the sum of the zeroes of the quadratic polynomial \( ky^2 + 2y - 3k \) is equal to twice their product, find the value of \( k \).
Answer: Sol. Consider, \( p(y) = ky^2 + 2y - 3k \)
\( a = k \), \( b = 2 \), \( c = -3k \)
A.T.Q., Sum of zeroes = 2 \( \times \) product of zeroes

\( \implies \) \( -b/a = 2 \times c/a \)

\( \implies \) \( -2/k = 2 \times (-3k/k) \)

\( \implies \) \( -2/k = -6 \)

\( \implies \) \( k = 1/3 \)

Question. If zeroes of \( p(x) = ax^2 + bx + c \) are negative reciprocal of each other, find the relationship between \( a \) and \( c \).
Answer: Sol. \( p(x) = ax^2 + bx + c \)
Let one zero = \( \alpha \)
\( \therefore \) Other zero = \( -1/\alpha \)
Now, product of zeroes \( = c/a \)

\( \implies \) \( \alpha \times (-1/\alpha) = c/a \)

\( \implies \) \( c/a = -1 \)

\( \implies \) \( c = -a \) or \( a + c = 0 \)

Short Answer Type Questions

Question. Find the value of k such that the polynomial \( x^2 - (k + 6)x + 2(2k - 1) \) has sum of its zeroes equal to half of their product. 
Answer:
Sum of zeroes \( = k + 6 \)
Product of zeroes \( = 2(2k - 1) \)
Hence \( k + 6 = \frac{1}{2} \times 2(2k - 1) \)

\( \implies \) \( k = 7 \)

Question. If one root of the quadratic polynomial \( 2x^2 - 3x + p \) is 3, find the other root. Also, find the value of p.
Answer:
\( \because \) 3 is a root (zero) of \( p(x) \)

\( \implies \) \( 2(3)^2 - 3 \times 3 + p = 0 \)

\( \implies \) \( 18 - 9 + p = 0 \)
\( \implies \) \( p = - 9 \)
Now \( p(x) = 2x^2 - 3x - 9 = 2x^2 - 6x + 3x - 9 \)
\( = 2x(x - 3) + 3(x - 3) \)
\( = (x - 3)(2x + 3) \)
For roots of polynomial, \( p(x) = 0 \)

\( \implies \) \( (x - 3)(2x + 3) = 0 \)

\( \implies \) \( x = 3 \) or \( x = -\frac{3}{2} \), Other root \( = -\frac{3}{2} \)

Question. If \( \alpha \) and \( \beta \) are zeroes of the quadratic polynomial \( 4x^2 + 4x + 1 \), then form a quadratic polynomial whose zeroes are \( 2\alpha \) and \( 2\beta \). 
Answer:
\( p(x) = 4x^2 + 4x + 1 \)
\( \because \alpha, \beta \) are zeroes of \( p(x) \)
\( \therefore \alpha + \beta = \text{sum of zeroes} = -\frac{b}{a} \)

\( \implies \) \( \alpha + \beta = -\frac{4}{4} = - 1 \) ...(i)
Also \( \alpha \cdot \beta = \text{Product of zeroes} = \frac{c}{a} \)

\( \implies \) \( \alpha \cdot \beta = \frac{1}{4} \) ...(ii)
Now a quadratic polynomial whose zeroes are \( 2\alpha \) and \( 2\beta \).
\( x^2 - (\text{sum of zeroes})x + \text{Product of zeroes} \)
\( = x^2 - (2\alpha + 2\beta)x + 2\alpha \times 2\beta \)
\( = x^2 - 2(\alpha + \beta)x + 4(\alpha\beta) \)
\( = x^2 - 2 \times (-1)x + 4 \times \frac{1}{4} \)
[Using eq.(i) and (ii)]
\( = x^2 + 2x + 1 \)

Question. Find the zeroes of the quadratic polynomial \( 7y^2 - \frac{11}{3}y - \frac{2}{3} \) and verify the relationship between the zeroes and the coefficients.
Answer:
Here \( p(y) = 7y^2 - \frac{11}{3}y - \frac{2}{3} \)
For zeroes of \( p(y) \), \( p(y) = 0 \)

\( \implies \) \( 7y^2 - \frac{11}{3}y - \frac{2}{3} = 0 \)

\( \implies \) \( 21y^2 - 11y - 2 = 0 \)

\( \implies \) \( 21y^2 - 14y + 3y - 2 = 0 \)

\( \implies \) \( 7y(3y - 2) + 1(3y - 2) = 0 \)

\( \implies \) \( (7y + 1)(3y - 2) = 0 \)

\( \implies \) \( y = -\frac{1}{7}, \frac{2}{3} \)
\( \therefore \) zeroes are \( -\frac{1}{7} \) and \( \frac{2}{3} \)
Also \( a = 7, b = -\frac{11}{3}, c = -\frac{2}{3} \)
Sum of zeroes \( = -\frac{1}{7} + \frac{2}{3} = \frac{-3 + 14}{21} = \frac{11}{21} \)
Also Sum of zeroes \( = -\frac{b}{a} = \frac{-(-11/3)}{7} = \frac{11}{21} \) (verified)
and product of zeroes \( = -\frac{1}{7} \times \frac{2}{3} = -\frac{2}{21} \)
Also product of zeroes \( = \frac{c}{a} = \frac{-2/3}{7} = \frac{-2}{21} \) (verified)

Question. If the zeroes of \( x^2 - px + 6 \) are in the ratio 2 : 3, find p. 
Answer:
\( p(x) = x^2 - px + 6 \)
Let zeroes are \( 2m \) and \( 3m \)
Sum of zeroes \( = -\frac{b}{a} \)

\( \implies \) \( 2m + 3m = \frac{-(-p)}{1} \)

\( \implies \) \( 5m = p \) ...(i)
Product of zeroes \( = \frac{c}{a} \)

\( \implies \) \( 2m \times 3m = \frac{6}{1} \)

\( \implies \) \( 6m^2 = 6 \)

\( \implies \) \( m^2 = 1 \)

\( \implies \) \( m = \pm 1 \)
When \( m = 1 \), eq (i) becomes
\( 5 \times 1 = p \)

\( \implies \) \( p = 5 \)
When \( m = -1 \), eq (i) becomes
\( 5 \times -1 = p \)

\( \implies \) \( p = -5 \)

\( \therefore \) \( p = \pm 5 \)

Question. If \( \alpha, \beta \) are the zeroes of polynomial \( p(x) = x^2 - k(x + 1) - p \) such that \( (\alpha + 1)(\beta + 1) = 0 \), find p. 
Answer:
\( p(x) = x^2 - kx - k - p = 0 \)
\( a = 1, b = -k, c = -k - p \)
\( \because \alpha, \beta \) are zeroes of \( p(x) \)
\( \therefore \alpha + \beta = -\frac{b}{a} \)

\( \implies \) \( \alpha + \beta = k \)
and \( \alpha\beta = \frac{c}{a} \)

\( \implies \) \( \alpha\beta = -k - p \)
Also \( (\alpha + 1)(\beta + 1) = 0 \)

\( \implies \) \( \alpha\beta + \alpha + \beta + 1 = 0 \)

\( \implies \) \( (-k - p) + k + 1 = 0 \)
\( -p + 1 = 0 \)

\( \implies \) \( p = 1 \)

Question. a, b, c are co-prime \( a \neq 1 \) such that \( 2b = a + c \). If \( ax^2 - 2bx + c \) and \( 2x^3 - 5x^2 + kx + 4 \) has one integral root common, then find the value of k.
Answer:
\( p(x) = ax^2 - 2bx + c \)
\( p(1) = a(1)^2 - 2b \times 1 + c \)
\( = a - 2b + c \)
\( = a + c - 2b \)
Given: \( a + c = 2b \)
\( \therefore p(1) = 2b - 2b = 0 \)

\( \implies \) \( x = 1 \) is a zero of \( p(x) \)
Now, product of zeroes of \( p(x) = \frac{c}{a} \)
Other root \( = \frac{c/a}{1} = \frac{c}{a} \)
Roots are 1 and \( \frac{c}{a} \)
\( \because \frac{c}{a} \) are co-prime
\( \therefore \) integral root of \( p(x) = 1 \)
A.T.Q., 1 is a root of \( f(x) = 2x^3 - 5x^2 + kx + 4 \)

\( \implies \) \( f(1) = 0 \)

\( \implies \) \( 2(1)^3 - 5(1)^2 + k \times 1 + 4 = 0 \)
\( k = - 1 \)

Long Answer Type Question

Question. If one zero of the quadratic polynomial \( f(x) = 4x^2 - 8kx + 8x - 9 \) is negative of the other, then find zeroes of \( kx^2 + 3kx + 2 \). 
Answer:
\( f(x) = 4x^2 - 8kx + 8x - 9 = 4x^2 + (8 - 8k)x - 9 \)
Let one zero \( = \alpha \)
\( \therefore \) other zero \( = -\alpha \) [A.T.Q.]
Now Sum of zeroes \( = -\frac{b}{a} \)

\( \implies \) \( \alpha + (-\alpha) = \frac{-(8 - 8k)}{4} \)

\( \implies \) \( 0 = \frac{-8 + 8k}{4} \)

\( \implies \) \( -8 + 8k = 0 \implies k = 1 \)
Polynomial \( p(x) = kx^2 + 3kx + 2 \)
becomes \( p(x) = 1 \times x^2 + 3 \times x + 2 \) [Using \( k = 1 \)]
\( = x^2 + 3x + 2 \)
For zeroes of \( p(x) \), \( x^2 + 3x + 2 = 0 \)

\( \implies \) \( (x + 2)(x + 1) = 0 \)

\( \implies \) \( x = -2, x = -1 \)