Given below are the Class 10 Maths Extra questions for Polynomials
a. Finding Zero's Questions
b. Short Answers Questions
c. Word Problems
d. Graph Questions
Question 1
Find a quadratic polynomial whose zeroes are $5 + \sqrt {2}$ and $5 - \sqrt {2}$ Solution
If $\alpha$ and $\beta$ are the zeroes of the polynomials
$x^2 - (\alpha + \beta) + \alpha \beta$
Here $\alpha + \beta =5 + \sqrt {2} + 5 - \sqrt {2} =10$
$\alpha \beta =(5 + \sqrt {2})(5 - \sqrt {2}) = 25 -4 =21$
Therefore quadratic Polynomial is
$x^2 - 10x + 21$
Question 2
If a and b are zeroes of quadratic polynomial $kx^2 + 4x + 4$, find the value of k such that $(a+b )^2- 2ab= 24$. Solution
Here $a + b= -\frac {4}{k}$
$ab=\frac {4}{k}$
Now $(a+b )^2- 2ab= 24$
$(-\frac {4}{k})^2 -2 \frac {4}{k}=24$
$16-8k=24k^2$ or $3K^2 +k-2=0$
$k=-1$ or $k=\frac {2}{3}$
Question 3 If one zero of $3x^2 - 4x + p$ is reciprocal to the other, then find the value of p Solution
Let a be one of zero, then 1/a be the another zero
Now $a \times \frac {1}{a} =\frac {p}{3}$
or p=3
Question 4
If p and q are the zeroes of $p(x) = kx^2 - 3x + 2k$ and $p+q=pq$ then find the value of k. Solution
Here $ p + q=\frac {3}{k}$
$pq=\frac {2k}{k}$
Now
$p+q=pq$
$\frac {3}{k} = 2$
$k= \frac {3}{2}$
Question 5 If a and b are zeroes of $x^2 - 6x + k$. What is the value of k, if $3a+2b= 20$? Solution
Here $a + b= - \frac {-6}{1} = 6$ --(1)
$ab= k$ -(3)
Given $3a+2b= 20$ --(2)
Solving (1) and (2)
a=14 and b=-8
Now substituting these values in (3)
k=-112
Question 6 If one zero of the quadratic polynomial $x^2 + 3x + k$ is 2, then find the value of k. Solution
Since 2 is zero of the quadratic polynomial
$2^2 + 3 \times 2 + k=0$
k=-10
Question 7 Find the zeroes of the quadratic polynomial $x^2 + 8x + 16$ and verify the relationship between the zeroes and the coefficients. Solution
Now $x^2 + 8x + 16$
$= x^2 + 4x + 4x + 16 = x(x+4) + 4(x+ 4) = (x+4)^2$
So zeroes are -4 ,-4
Now if $k_1$ ,$k_2$ are the roots of quadratic polynomials $ax^2 + bx + c$, then relationship between them is given by
$k_1 + k_2 = -\frac {b}{a}$ and $k_1 k_2= \frac {c}{a}$
Now lets verify the relationship in the above polynomial
$-4 + (-4) = -\frac {8}{1}$
$-8=-8$
Also
$(-4) \times (-4) = \frac {16}{1}$
16=16
Question 8 Find the value of a and b, if they are the zeroes of polynomial x2 + ax + b. Solution
Here
$a + b= -a$ or $2a + b=0$ -(1)
Also $ab= b$ or $a=1$
Substituting the value of a in (1)
b=-2
Hence a=1 and b=-2
Question 9 If m and n are the zeroes of the polynomial $3x^2 + 11x - 4$, find the value of $ \frac {m}{n} + \frac {n}{m}$ Solution
Question 10 Show that 2, -1 and 1/2 are the zeroes of the cubic polynomial
$p(x) = 2x^3 - 3x^2 - 3x + 2$
and then verify that the sum of the zeroes =-(Coeff of x2/Coeff of x3)
Question 11 Find the zeroes of the polynomial $f(x) = 4 \sqrt {3} x^2 + 5x - 2 \sqrt {3}$ , and verify the relationship between the zeroes and its coefficients. Solution
Question 16 If two zeroes of the polynomial $f(x) =x^3 - 4x^2 - 3x + 12$ are √3 and -√3 then find its third zero. Solution
If √3 and -√3 are zeroes of the polynomial, then $(x- \sqrt {3})(x+ \sqrt {3}) = x^2 -3$ should be factor of it.
$x^3 - 4x^2 - 3x + 12= x^2(x-4) -3(x -4) = (x^2 -3)(x-4)$
Clearly the other zero is 4
Question 17 If l and m are zeroes of the polynomial $p(x) = 2x^2 - 5x + 7$, find a polynomial whose zeroes are $2l+ 3$ and $2m+ 3$. Solution
Here $l+m= \frac {5}{2}$
$lm= \frac {7}{2}$
Now the polynomial whose zeroes are $2l+ 3$ and $2m+ 3$ is given by
$x^2 - (2l+3 + 2m + 3)x +(2l+3)(2m+3)=x^2 -2(l+m)x -6x + 4lm + 6l + 6m +9$
$=x^2 -10x -6x + 14 + 6(l+m) +9$
$=x^2 -16x +14+15+9=x^2 -16x + 38$
Question 18 If p ,q and r are zeroes of polynomial $6x^3+ 3x^2 - 5x + 1$, then find the value of $\frac {1}{p} + \frac {1}{q} + \frac {1}{r}$. Solution
Here $pq + qr+ pr=\frac {-5}{6}=- \frac {5}{6}$
$pqr= -\frac {1}{6}$
Now
$\frac {1}{p} + \frac {1}{q} + \frac {1}{r}$
$= \frac {pq+ qr + pr}{pqr} =5$
Question 19 If the zeroes of the polynomial $f(x) = ax^3 + 3bx^2 + 3cx + d$ are in A. P., prove that $2b^3 - 3abc + a^2d = 0$. Solution
Given Roots of given polynomial is in AP
let p-q, p, and p+q are the roots of
$f( x) = ax^3 + 3bx^2 + 3cx + d$
Now we know that
$sum \; of \; roots = - \frac {coefficient \; of \; x^2} {coefficient \;of\; x^3}$
$p-q +p + p+q = - \frac {3b}{a}$
or $3p = - \frac {3b}{a}$
or $p = - \frac {b}{a}$ -----------(1)
Also,
$sum \; of \; products \; of \; two roots = \frac {coefficient \; of\; x}{coefficient \; of \; x^3}$
$ (p -q)\times p + p \times (p+q) + (p-q)(p+q)=\frac {3c}{a}$
$3p^2 - q^2 = \frac {3c}{a}$ ------------------(2)
Also,
$products \; of \; all \; roots = -\frac {constant }{coefficient \; of \; x^3}$
$(p-q)\times p \times (p+q) = -\frac {d}{a}$
$p^3 - pq^2 = -\frac {d}{a}$ ----------(3)
Substituting the value of p from (1) in (2), we get the value of q as
$ 3 (-\frac {b}{a})^2 - q^2 = \frac {3c}{a}$
or
$q^2 = 3 \frac {b^2}{a^2} - 3 \frac {c}{a}$ ----------(4)
Now that we have got the values of p and q, substituting these in equation (3)
Question 20 Obtain all zeroes of the polynomial $f(x) = 2x^4 - 2x^3 - 7x^2 + 3x + 6$, if its two zeros are √3/2 and -√3/2 Solution
If √(3/2) and -√(3/2) are zeroes of the polynomial, then $(x- \sqrt {3/2})(x+ \sqrt {3/2}) = 2x^2 -3$ should be factor of it.
$f(x) = 2x^4 - 2x^3 - 7x^2 + 3x + 6$
Lets use division algorithm to find the other factors
So,
$2x^4 - 2x^3 - 7x^2 + 3x + 6= (2x^2 -3)(x^2 -x -2) = (2x^2 -3)(x^2 -2x + x-2) =(2x^2 -3)(x-2)(x+1)$
So other zeros are x=2 and x=-1
Question 21
The graphs of y = p(x) are given in below figure, for some polynomials p(x). Find the number of zeroes of p(x), in each case.
Solution
i. 1
ii. 2
iii. 3
iv 1
v. 4
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