CBSE Class 10 Mathematics Polynomials VBQs Set 05

Question. 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) has no linear term and the constant term is negative.

Question. If one zero of the quadratic polynomial \( x^2 + 3x + k \) is 2, then the value of k is
(a) 10
(b) -10
(c) 5
(d) -5
Answer: (b) -10

Question. 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) \( a = 0, b = -6 \)

Question. What should be added to the polynomial \( x^2 - 5x + 4 \), so that 3 is the zero of the resulting polynomial:
(a) 1
(b) 2
(c) 4
(d) 5
Answer: (b) 2

Question. Zeroes of \( p(x) = \sqrt{2}x^2 + 7x + 5\sqrt{2} \) are
(a) \( -\frac{5}{\sqrt{2}}, -\sqrt{2} \)
(b) \( \frac{5}{\sqrt{2}}, \sqrt{2} \)
(c) \( -\frac{5}{\sqrt{2}}, \sqrt{2} \)
(d) \( \frac{5}{\sqrt{2}}, -\sqrt{2} \)
Answer: (a) \( -\frac{5}{\sqrt{2}}, -\sqrt{2} \)

Question. If \( \alpha \) and \( \beta \) are the zeroes of the quadratic polynomial \( f(x) = x^2 - x - 4 \), then the value of \( \frac{1}{\alpha} + \frac{1}{\beta} - \alpha\beta \) is
(a) \( \frac{15}{4} \)
(b) \( -\frac{15}{4} \)
(c) 4
(d) 15
Answer: (a) \( \frac{15}{4} \)

Question. If the zeroes of the quadratic polynomial \( x^2 + (a + 1)x + b \) are 4 and -3, then \( a - b \) is
(a) 12
(b) 10
(c) 7
(d) 1
Answer: (b) 10

Question. If zeroes of \( p(x) = 2x^2 - 7x + k \) are reciprocal of each other, then value of k is
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (b) 2

Question. If \( \alpha, \beta \) are the zeroes of a polynomial, such that \( \alpha + \beta = 6 \) and \( \alpha\beta = 4 \), then write the polynomial.
Answer: \( x^2 - 6x + 4 \)

Question. Find a quadratic polynomial, the sum of whose zeroes is 0 and one zero is 5.
Answer: One zero is 5, sum is 0 \( \implies \) other zero is -5. Product \( = 5 \times (-5) = -25 \). Polynomial \( = x^2 - 25 \).

Question. Find the zeroes of the quadratic polynomial \( 6x^2 - 3 - 7x \) and verify the relationship between the zeroes and the coefficients of the polynomial.
Answer: \( 6x^2 - 7x - 3 = 6x^2 - 9x + 2x - 3 = 3x(2x - 3) + 1(2x - 3) = (3x + 1)(2x - 3) \). Zeroes are \( -1/3, 3/2 \).

Question. Find a quadratic polynomial whose zeroes are reciprocals of the zeroes of the polynomial \( f(x) = ax^2 + bx + c, a \neq 0, c \neq 0 \). 
Answer: \( cx^2 + bx + a \)

Question. Find a quadratic polynomial whose one zero is 5 and product of zeroes is 30.
Answer: \( x^2 - 11x + 30 \)

Question. Find a quadratic polynomial whose one zero is 7 and sum of zeroes is -18.
Answer: \( x^2 + 18x - 175 \)

Question. Form a quadratic polynomial, one of whose zero is \( \sqrt{5} \) and the product of the zeroes is \( -2\sqrt{5} \).
Answer: \( x^2 - (\sqrt{5} - 2)x - 2\sqrt{5} \)

Question. Find a quadratic polynomial whose zeroes are \( 5 + \sqrt{2} \) and \( 5 - \sqrt{2} \).
Answer: \( x^2 - 10x + 23 \)

Question. Show that \( 1/2 \) and \( -3/2 \) are the zeroes of the polynomial \( 4x^2 + 4x - 3 \) and verify the relationship between zeroes and co-efficients of polynomial.
Answer: Verification involves substitution and checking sum/product formulas.

Question. If one zero of the quadratic polynomial \( 2x^2 + px + 4 \) is 2, find the other zero. Also, find the value of p.
Answer: Other zero \( = 1 \), \( p = -6 \)

Question. If \( \alpha, \beta \) are zeroes/roots of polynomial \( x^2 - 6x + k \), find the value of k such that \( (\alpha + \beta)^2 - 2\alpha\beta = 40 \).
Answer: \( k = -2 \)

Question. If 2 and 3 are zeroes of polynomial \( 3x^2 - 2kx + 2m \), find the values of k and m.
Answer: \( k = 7.5, m = 9 \)

Question. If \( \alpha, \beta \) are zeroes of polynomial \( p(x) = 5x^2 + 5x + 1 \), then find the value of (i) \( \alpha^2 + \beta^2 \) (ii) \( \alpha^{-1} + \beta^{-1} \).
Answer: (i) \( 3/5 \), (ii) \( -5 \)

Question. If both the zeroes of the quadratic polynomial \( p(x) = (k + 2)x^2 - (k - 2)x - 5 \) are equal in magnitude but opposite in sign, then find the value of k.
Answer: \( k = 2 \)

Question. If \( \alpha \) and \( \beta \) are the zeroes of the polynomial \( x^2 + 6x + 9 \), then form a polynomial whose zeroes are \( -\alpha \) and \( -\beta \). 
Answer: \( x^2 - 6x + 9 \)

Question. If \( \alpha, \beta \) are zeroes of the polynomial \( p(x) = 3x^2 - 12x + k \) such that \( \alpha - \beta = 2 \), find the value of k.
Answer: \( k = 9 \)

Question. Find the value of p and q if they are the zeroes of \( p(x) = 2x^2 + px + q \).
Answer: \( p = 1/2, q = -3/4 \) (Assuming zeroes are not zero)

Question. If one zero of the polynomial \( p(x) = 6x^2 + 37x - (k - 2) \) is reciprocal of the other, then find the value of k.
Answer: \( k = -4 \)

Question. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) \( x^2 - 2x - 8 \)
(ii) \( 4s^2 - 4s + 1 \)
(iii) \( 6x^2 - 3 - 7x \)
(iv) \( 4u^2 + 8u \)
(v) \( t^2 - 15 \)
(vi) \( 3x^2 - x - 4 \)
Answer: (i) Quadratic polynomial is
\( x^2 - 2x - 8 = x^2 - 4x + 2x - 8 \)
\( = x(x - 4) + 2(x - 4) \)
\( = (x + 2) (x - 4) \)
So, zeroes are – 2 and 4 because value of \( x^2 - 2x - 8 \) is zero, when \( x + 2 = 0 \)
\( \implies \) \( x = -2 \) and \( x - 4 = 0 \)
\( \implies \) \( x = 4 \).
Verification:
zeroes are –2 and 4.
\( \alpha = -2 \) and \( \beta = 4 \)
\( \alpha + \beta = \frac{-\text{coefficient of } x}{\text{coefficient of } x^2} \)
\( \therefore -2 + 4 = \frac{-(-2)}{1} \)
\( \implies \) \( 2 = 2 \) (verified)
\( \alpha\beta = \frac{\text{constant term}}{\text{coefficient of } x^2} \)
\( (-2)(4) = \frac{-8}{1} = -8 \)
\( -8 = -8 \) (verified)

(ii) \( 4s^2 - 4s + 1 = 4s^2 - 2s - 2s + 1 \)
\( = 2s(2s - 1) - 1(2s - 1) \)
\( = (2s - 1)(2s - 1) \)
zeroes are \( \frac{1}{2} \) and \( \frac{1}{2} \). So, the value of \( 4s^2 - 4s + 1 \) is zero,
when \( 2s - 1 = 0 \)
\( \implies \) \( 2s = 1 \)
\( \implies \) \( s = \frac{1}{2} \)
or \( 2s - 1 = 0 \)
\( \implies \) \( s = \frac{1}{2} \)
Verification:
\( \alpha = \frac{1}{2}, \beta = \frac{1}{2} \)
\( \therefore \alpha + \beta = \frac{-\text{coefficient of } s}{\text{coefficient of } s^2} \)
\( \frac{1}{2} + \frac{1}{2} = \frac{-(-4)}{4} \)
\( 1 = -(-1) = 1 \) (verified)
\( \alpha\beta = \frac{\text{constant term}}{\text{coefficient of } s^2} \)
\( (\frac{1}{2})(\frac{1}{2}) = \frac{1}{4} \)
\( \frac{1}{4} = \frac{1}{4} \) (verified)

(iii) \( 6x^2 - 3 - 7x = 6x^2 - 7x - 3 \)
\( = 6x^2 - 9x + 2x - 3 \)
\( = 3x(2x - 3) + 1 (2x - 3) \)
\( = (3x + 1) (2x - 3) \)
zeroes are \( -\frac{1}{3} \) and \( \frac{3}{2} \)
So, the value of \( 6x^2 - 7x - 3 \) is zero, when
\( 3x + 1 = 0 \)
\( \implies \) \( x = -\frac{1}{3} \)
\( 2x - 3 = 0 \)
\( \implies \) \( x = \frac{3}{2} \)
Verification:
\( \alpha = -\frac{1}{3}, \beta = \frac{3}{2} \)
\( \therefore \alpha + \beta = \frac{-\text{coefficient of } x}{\text{coefficient of } x^2} \)
\( = -\frac{1}{3} + \frac{3}{2} = \frac{-2 + 9}{6} \)
\( = \frac{7}{6} = \frac{-(-7)}{6} \)
and \( \alpha\beta = \frac{\text{constant term}}{\text{coefficient of } x^2} \)
\( = -\frac{1}{3} \times \frac{3}{2} = \frac{-1}{2} = \frac{-3}{6} \)

(iv) \( 4u^2 + 8u = 4u(u + 2) \)
zeroes are \( 0 \) and \( -2 \), so, value of \( 4u^2 + 8u \) is zero,
when \( 4u = 0 \)
\( \implies \) \( u = 0 \)
or \( u + 2 = 0 \)
\( \implies \) \( u = -2 \)
Verification:
\( \alpha = 0, \beta = -2 \)
\( \therefore \alpha + \beta = 0 + (-2) = -2 = -\frac{8}{4} \)
\( = \frac{-\text{coefficient of } u}{\text{coefficient of } u^2} \) (verified)
and \( \alpha\beta = 0(-2) = \frac{0}{4} \)
\( = \frac{\text{constant term}}{\text{coefficient of } u^2} = 0 \).

(v) \( t^2 - 15 = t^2 - (\sqrt{15})^2 \)
Using \( a^2 - b^2 = (a + b)(a - b) \)
\( = (t - \sqrt{15})(t + \sqrt{15}) \)
zeroes are \( \sqrt{15} \) and \( -\sqrt{15} \), so, value of \( t^2 - 15 \) is zero
when \( t - \sqrt{15} = 0 \)
\( \implies \) \( t = \sqrt{15} \)
or \( t + \sqrt{15} = 0 \)
\( \implies \) \( t = -\sqrt{15} \)
Verification:
\( \alpha = -\sqrt{15}, \beta = \sqrt{15} \)
\( \therefore \alpha + \beta = -\sqrt{15} + \sqrt{15} = 0 = \frac{-0}{1} \)
\( = -\frac{\text{coefficient of } t}{\text{coefficient of } t^2} \) (verified)
and \( \alpha\beta = (-\sqrt{15})(\sqrt{15}) = -15 \)
\( = \frac{-15}{1} = \frac{\text{constant term}}{\text{coefficient of } t^2} \) (verified)

(vi) \( 3x^2 - x - 4 = 3x^2 - 4x + 3x - 4 \)
\( = x(3x - 4) + 1 (3x - 4) \)
\( = (3x - 4)(x + 1) \) is zero,
when \( 3x - 4 = 0 \)
\( \implies \) \( x = 4/3 \)
or \( x + 1 = 0 \)
\( \implies \) \( x = -1 \)
Verification:
\( \alpha = \frac{4}{3}, \beta = -1 \)
\( \therefore \alpha + \beta = \frac{4}{3} + (-1) \)
\( = \frac{1}{3} = \frac{-(-1)}{3} \)
\( = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2} \) (verified)
\( \alpha\beta = \frac{4}{3}(-1) = -\frac{4}{3} \)
\( = \frac{\text{constant term}}{\text{coefficient of } x^2} \) (verified)

Question. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) \( \frac{1}{4}, -1 \)
(ii) \( \sqrt{2}, \frac{1}{3} \)
(iii) \( 0, \sqrt{5} \)
(iv) \( 1, 1 \)
(v) \( -\frac{1}{4}, \frac{1}{4} \)
(vi) \( 4, 1 \)
Answer: (i) Zeroes of polynomial are not given,
sum of zeroes = \( \frac{1}{4} \)
and product of zeroes = \( -1 \)
If \( ax^2 + bx + c \) is a quadratic polynomial, then
\( \alpha + \beta = \text{sum of zeroes} = \frac{-b}{a} = \frac{1}{4} \)
and \( \alpha\beta = \text{product of zeroes} = \frac{c}{a} = -1 \)
Quadratic polynomial is \( ax^2+bx+c \)
Let \( a = k \), \( \therefore b = -\frac{k}{4} \) and \( c = -k \)
\( \therefore \) equation is \( k[x^2 - \frac{1}{4}x - 1] = k[x^2 - \frac{x}{4} - 1] \)
\( = \frac{k}{4}(4x^2 - x - 4) = k(4x^2 - x - 4) \)
For different values of \( k \), we can have quadratic polynomials all having sum of zeroes as \( \frac{1}{4} \) and product of zeroes as –1.

(ii) Sum of zeroes = \( \alpha + \beta = \sqrt{2} = \frac{-b}{a} \)
and product of zeroes = \( \alpha\beta = \frac{1}{3} = \frac{c}{a} \)
quadratic polynomial is \( ax^2+bx+c \)
let \( a = k, b = -\sqrt{2}k \) and \( c = \frac{k}{3} \).
putting values we get
\( k[x^2 - \sqrt{2}x + \frac{1}{3}] = k \frac{(3x^2 - 3\sqrt{2}x + 1)}{3} \)
\( = \frac{k}{3}[3x^2 - 3\sqrt{2}x + 1] = k[3x^2 - 3\sqrt{2}x + 1] \)
For all different real values of \( k \), we can have different quadratic polynomials of the form \( 3x^2 - 3\sqrt{2}x + 1 \) having sum of zeroes = \( \sqrt{2} \) and product of zeroes = \( \frac{1}{3} \).

(iii) Sum of zeroes = \( \alpha + \beta = 0 = -\frac{b}{a} \)
Product of zeroes = \( \alpha\beta = \sqrt{5} = \frac{c}{a} \)
let quadratic polynomial is \( ax^2 + bx + c \)
Here \( a = k, b = 0, c = \sqrt{5}k \)
Putting these values, we get
\( k[x^2 - 0x + \sqrt{5}] = k(x^2 + \sqrt{5}) \).
For different real values of \( k \), we can have different quadratic polynomials of the form \( x^2 + \sqrt{5} \), having sum of zeroes = 0 and product of zeroes = \( \sqrt{5} \).

(iv) Sum of zeroes = \( \alpha + \beta = 1 = -\frac{b}{a} \)
Product of zeroes = \( \alpha\beta = 1 = \frac{c}{a} \)
Let quadratic polynomial is \( ax^2 + bx + c \).
Here \( a = k, c = k, b = -k \)
Putting values, we get
\( = k[x^2 - x + 1] \)
Quadratic polynomial is \( x^2 - x + 1 \) for different values of \( k \).

(v) Sum of zeroes = \( \alpha + \beta = -\frac{1}{4} = -\frac{b}{a} \)
Product of zeroes = \( \alpha\beta = \frac{1}{4} = \frac{c}{a} \)
Let quadratic polynomial is \( ax^2 + bx + c \)
Let \( a = k, b = \frac{k}{4}, c = \frac{k}{4} \)
Putting values, we get
\( = k[x^2 - (-\frac{1}{4})x + \frac{1}{4}] \)
\( = k[x^2 + \frac{x}{4} + \frac{1}{4}] = \frac{k}{4}(4x^2 + x + 1) \)
Quadratic polynomial is \( 4x^2 + x + 1 \).

(vi) Sum of zeroes = \( 4 = \alpha + \beta = -\frac{b}{a} \)
Product of zeroes = \( \alpha\beta = 1 = \frac{c}{a} \)
Let quadratic polynomial be \( ax^2 + bx + c \).
Here \( a = k, b = -4k \) and \( c = k \).
Putting values, we get
\( = k[x^2 - 4x + 1] \)
\( \therefore \) quadratic polynomial is \( x^2 - 4x + 1 \).

Question. If one of the zeroes of the quadratic polynomial \( (k - 1)x^2 + kx + 1 \) is –3, then the value of \( k \) is
(a) \( \frac{4}{3} \)
(b) \( -\frac{4}{3} \)
(c) \( \frac{2}{3} \)
(d) \( -\frac{2}{3} \)
Answer: (a) \( \frac{4}{3} \)
(k – 1)x² + kx + 1
One zero is – 3, so it must satisfy the equation and make it zero.
\( \therefore (k - 1) (-3)^2 + k(-3) + 1 = 0 \)
\( \implies \) \( 9k - 9 - 3k + 1 = 0 \)
\( \implies \) \( 6k - 8 = 0 \)
\( \implies \) \( k = \frac{8}{6} = \frac{4}{3} \)

Question. 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) \( a = 0, b = -6 \)
\( x^2 + (a + 1)x + b \)
\( \because x = 2 \) is a zero and \( x = -3 \) is another zero
\( \therefore (2)^2 + (a + 1) \cdot 2 + b = 0 \)
and \( (-3)^2 + (a + 1) (-3) + b = 0 \)
\( \implies \) \( 4 + 2a + 2 + b = 0 \) and \( 9 - 3a - 3 + b = 0 \)
\( \implies \) \( 2a + b = -6 \) ...(i) and \( -3a + b = -6 \) ...(ii)
Solving (i) and (ii), we get \( 5a = 0 \)
\( \implies \) \( a = 0 \) and \( b = -6 \).

Question. The number of polynomials having zeroes as –2 and 5 is
(a) 1
(b) 2
(c) 3
(d) more than 3
Answer: (d) more than 3
\( \because x^2 - 3x - 10, 2x^2 - 6x - 20, \frac{1}{2}x^2 - \frac{3}{2}x - 5, 3x^2 - 9x - 30 \) etc., have zeroes – 2 and 5.
[If \( \alpha, \beta \) are zeroes of \( ax^2 + bx + c \) then \( k(ax^2 + bx + c) (k \in R, k \neq 0) \) also has zeroes as \( \alpha \) and \( \beta \)]

Question. Find the zeroes of the following polynomials by factorisation method and verify the relationship between the zeroes and coefficients of the polynomials. \( 4x^2 + 5\sqrt{2}x - 3 \)
Answer: \( 4x^2 + 5\sqrt{2}x - 3 = 4x^2 + 6\sqrt{2}x - \sqrt{2}x - 3 = 0 \)
\( = 2\sqrt{2}x (\sqrt{2}x + 3) - 1 (\sqrt{2}x + 3) = 0 \)
\( = (2\sqrt{2}x - 1) (\sqrt{2}x + 3) = 0 \)
\( \implies \) \( x = \frac{1}{2\sqrt{2}} \) and \( x = -\frac{3}{\sqrt{2}} \)
are zeroes of the polynomial.
Now, sum of zeroes
\( \frac{1}{2\sqrt{2}} + \frac{-3}{\sqrt{2}} = \frac{1 - 6}{2\sqrt{2}} = \frac{-5}{2\sqrt{2}} \)
Also \( -\frac{b}{a} = \frac{-5\sqrt{2}}{4} = \frac{-5}{2\sqrt{2}} \)
\( \implies \) Sum of zeroes = \( -\frac{b}{a} \)
Product of zeroes = \( \frac{1}{2\sqrt{2}} \times \frac{-3}{\sqrt{2}} = \frac{-3}{4} \)
Also \( \frac{c}{a} = \frac{-3}{4} \)
\( \therefore \) Product of zeroes = \( \frac{c}{a} \)

Question. Find the zeroes of the following polynomials by factorisation method and verify the relationship between the zeroes and coefficients of the polynomials. \( 2s^2 - (1 + 2\sqrt{2})s + \sqrt{2} \)
Answer: \( 2s^2 - (1 + 2\sqrt{2})s + \sqrt{2} = 2s^2 - s - 2\sqrt{2}s + \sqrt{2} = 0 \)
\( = s(2s - 1) - \sqrt{2}(2s - 1) = (2s - 1) (s - \sqrt{2}) = 0 \)
\( \implies \) \( s = \frac{1}{2}, \sqrt{2} \) are zeroes of the polynomial.
Sum of zeroes = \( \frac{1}{2} + \sqrt{2} = \frac{1 + 2\sqrt{2}}{2} \)
Also \( \frac{-b}{a} = \frac{1 + 2\sqrt{2}}{2} \)
\( \implies \) Sum of zeroes = \( -\frac{b}{a} \)
Product of zeroes = \( \frac{1}{2} \times \sqrt{2} = \frac{\sqrt{2}}{2} \)
and \( \frac{c}{a} = \frac{\sqrt{2}}{2} \)
\( \implies \) Product of zeroes = \( \frac{c}{a} \)

Question. Find the zeroes of the following polynomials by factorisation method and verify the relationship between the zeroes and coefficients of the polynomials. \( v^2 + 4\sqrt{3}v - 15 \)
Answer: \( v^2 + 4\sqrt{3}v - 15 = v^2 + 5\sqrt{3}v - \sqrt{3}v - 15 = 0 \)
\( = v(v + 5\sqrt{3}) - \sqrt{3}(v + 5\sqrt{3}) = (v - \sqrt{3}) (v + 5\sqrt{3}) = 0 \)
\( \implies \) \( v = \sqrt{3}, -5\sqrt{3} \) are zeroes of the polynomial.
Sum of zeroes = \( \sqrt{3} + (-5\sqrt{3}) = -4\sqrt{3} \)
Also, \( -\frac{b}{a} = \frac{-4\sqrt{3}}{1} = -4\sqrt{3} \)
\( \therefore \) Sum of zeroes = \( -\frac{b}{a} \)
Product of zeroes = \( \sqrt{3} \times -5\sqrt{3} = -15 \)
Also, \( \frac{c}{a} = \frac{-15}{1} = -15 \)
\( \implies \) Product of zeroes = \( \frac{c}{a} \)

Question. Find the zeroes of the following polynomials by factorisation method and verify the relationship between the zeroes and coefficients of the polynomials. \( 7y^2 - \frac{11}{3}y - \frac{2}{3} \)
Answer: \( 7y^2 - \frac{11}{3}y - \frac{2}{3} = \frac{1}{3}(21y^2 - 11y - 2) = 0 \)
\( = \frac{1}{3}(21y^2 - 14y + 3y - 2) = \frac{1}{3}[7y(3y - 2) + 1(3y - 2)] = 0 \)
\( = \frac{1}{3}(3y - 2) (7y + 1) = 0 \)
\( \implies \) \( y = \frac{2}{3}, -\frac{1}{7} \) are zeroes of the polynomial.
Sum of zeroes = \( \frac{2}{3} + \frac{-1}{7} = \frac{14 - 3}{21} = \frac{11}{21} \)
Also \( -\frac{b}{a} = \frac{-(-\frac{11}{3})}{7} = \frac{11}{21} \)
\( \implies \) Sum of zeroes = \( -\frac{b}{a} \)
Now, product of zeroes = \( \frac{2}{3} \times -\frac{1}{7} = -\frac{2}{21} \)
Also, \( \frac{c}{a} = \frac{-\frac{2}{3}}{7} = -\frac{2}{21} \)
\( \implies \) Product of zeroes = \( \frac{c}{a} \)