SolveIt is under development
SolveItNCERT · CBSE · NEET

NCERT Exemplar · Class 10 Mathematics Polynomials

29 questions · 29 still being checked

EXERCISE 2.4 1–6 (part 4 of 4)

  1. Exercise 1

    For each of the following, find a quadratic polynomial whose sum and product respectively of the zeroes are as given. Also find the zeroes of these polynomials by factorisation.
    (i)
    −83,43\displaystyle \frac{-8}{3}, \frac{4}{3}
    (ii)
    218,516\displaystyle \frac{21}{8}, \frac{5}{16}
    (iii)
    −23,−9\displaystyle -2 \sqrt{3},-9
    (iv)
    −325,−12\displaystyle \frac{-3}{2 \sqrt{5}},-\frac{1}{2}

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    (i)
    \(\displaystyle -2, -\frac{2}{3}\)
    (ii)
    \(\displaystyle \frac{5}{2}, \frac{1}{8}\)
    (iii)
    \(\displaystyle -3 \sqrt{3}, \sqrt{3}\)
    (iv)
    \(\displaystyle \frac{\sqrt{5}}{5}, \frac{-\sqrt{5}}{2}\)
    (i)
    \[x^2-(\text{Sum})x+\text{Product}=x^2+\tfrac{8}{3}x+\tfrac{4}{3} \]
    \[3x^2+8x+4=0 \]
    \[3x^2+6x+2x+4=0 \]
    \[3x(x+2)+2(x+2)=0 \]
    \[(3x+2)(x+2)=0 \]
    \[x=-\tfrac{2}{3},\ -2 \]
    (ii)
    \[x^2-\tfrac{21}{8}x+\tfrac{5}{16}=0 \]
    \[16x^2-42x+5=0 \]
    \[16x^2-40x-2x+5=0 \]
    \[8x(2x-5)-1(2x-5)=0 \]
    \[(8x-1)(2x-5)=0 \]
    \[x=\tfrac{1}{8},\ \tfrac{5}{2} \]
    (iii)
    \[x^2+2\sqrt3\,x-9=0 \]
    \[(x+3\sqrt3)(x-\sqrt3)=x^2+2\sqrt3\,x-9 \]
    \[x=-3\sqrt3,\ \sqrt3 \]
    (iv)
    \[x^2+\tfrac{3}{2\sqrt5}x-\tfrac12=0 \]
    \[2\sqrt5\,x^2+3x-\sqrt5=0 \]
    \[2\sqrt5\,x^2+5x-2x-\sqrt5=0 \]
    \[\sqrt5\,x(2x+\sqrt5)-1(2x+\sqrt5)=0 \]
    \[(2x+\sqrt5)(\sqrt5\,x-1)=0 \]
    \[x=-\tfrac{\sqrt5}{2},\ \tfrac{1}{\sqrt5} \]
    Answer: (i) \(\displaystyle 3x^2+8x+4\), zeroes \(\displaystyle -\tfrac23,-2\). (ii) \(\displaystyle 16x^2-42x+5\), zeroes \(\displaystyle \tfrac18,\tfrac52\). (iii) \(\displaystyle x^2+2\sqrt3x-9\), zeroes \(\displaystyle -3\sqrt3,\sqrt3\). (iv) \(\displaystyle 2\sqrt5x^2+3x-\sqrt5\), zeroes \(\displaystyle -\tfrac{\sqrt5}{2},\tfrac{1}{\sqrt5}\).

  2. Exercise 2

    Given that the zeroes of the cubic polynomial x3−6x2+3x+10\displaystyle x^3-6 x^2+3 x+10 are of the form a\displaystyle a, a+b,a+2b\displaystyle a+b, a+2 b for some real numbers a\displaystyle a and b\displaystyle b, find the values of a\displaystyle a and b\displaystyle b as well as the zeroes of the given polynomial.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle a=-1\) and \(\displaystyle b=3\) or \(\displaystyle a=5, b=-3\). Zeroes are -$\displaystyle 1$, $\displaystyle 2$, $\displaystyle 5$
    \[a+(a+b)+(a+2b)=6 \quad\Rightarrow\quad a+b=2 \] confirming \(\displaystyle x=2\) is a zero: \[(2)^3-6(2)^2+3(2)+10=0 \] \[a(a+b)(a+2b)=-10 \quad\Rightarrow\quad a(a+2b)=-5 \] \[a=2-b,\quad a+2b=2+b \] \[(2-b)(2+b)=-5 \quad\Rightarrow\quad b^2=9 \quad\Rightarrow\quad b=\pm3 \] \[b=3:\ a=-1,\ a+b=2,\ a+2b=5 \] \[b=-3:\ a=5,\ a+b=2,\ a+2b=-1 \]Answer: \(\displaystyle a=-1,\,b=3\) or \(\displaystyle a=5,\,b=-3\); zeroes \(\displaystyle -1,\,2,\,5\).

  3. Exercise 3

    Given that 2\displaystyle \sqrt{2} is a zero of the cubic polynomial 6x3+2x2−10x−42\displaystyle 6 x^3+\sqrt{2} x^2-10 x-4 \sqrt{2}, find its other two zeroes.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle \frac{-\sqrt{2}}{2}, \frac{-2 \sqrt{2}}{3}\)
    \[6x^3+\sqrt2x^2-10x-4\sqrt2\ \div\ (x-\sqrt2) \] \[6,\ \sqrt2,\ -10,\ -4\sqrt2 \] \[6\sqrt2+\sqrt2=7\sqrt2,\quad 7\sqrt2\cdot\sqrt2=14,\quad -10+14=4,\quad 4\sqrt2-4\sqrt2=0 \] \[\text{Quotient: } 6x^2+7\sqrt2x+4 \] \[x=\dfrac{-7\sqrt2\pm\sqrt{(7\sqrt2)^2-4(6)(4)}}{2(6)}=\dfrac{-7\sqrt2\pm\sqrt2}{12} \] \[x=-\dfrac{\sqrt2}{2},\ -\dfrac{2\sqrt2}{3} \]Answer: \(\displaystyle -\dfrac{\sqrt2}{2}\) and \(\displaystyle -\dfrac{2\sqrt2}{3}\).

  4. Exercise 4

    Find k\displaystyle k so that x2+2x+k\displaystyle x^2+2 x+k is a factor of 2x4+x3−14x2+5x+6\displaystyle 2 x^4+x^3-14 x^2+5 x+6. Also find all the zeroes of the two polynomials.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle k=-3\) Zeroes of \(\displaystyle 2 x^4+x^3-14 x^2+5 x+6\) are $\displaystyle 1$, -$\displaystyle 3$, $\displaystyle 2$, \(\displaystyle -\frac{1}{2}\) Zeroes of \(\displaystyle x^2+2 x-3\) are $\displaystyle 1$, -$\displaystyle 3$
    \[2x^4+x^3-14x^2+5x+6=2x^2(x^2+2x+k)+\big(-3x^3+(-14-2k)x^2+5x+6\big) \] \[-3x^3+(-14-2k)x^2+5x+6=-3x(x^2+2x+k)+\big((-8-2k)x^2+(5+3k)x+6\big) \] \[(-8-2k)x^2+(5+3k)x+6=(-8-2k)(x^2+2x+k)+\big((21+7k)x+(2k^2+8k+6)\big) \] \[21+7k=0,\quad 2k^2+8k+6=0\ \Rightarrow\ k=-3 \] \[x^2+2x-3=(x+3)(x-1)\ \Rightarrow\ x=-3,\ 1 \] \[2x^4+x^3-14x^2+5x+6=(x^2+2x-3)(2x^2-3x-2) \] \[2x^2-3x-2=(2x+1)(x-2)\ \Rightarrow\ x=-\tfrac12,\ 2 \]Answer: \(\displaystyle k=-3\); zeroes of \(\displaystyle x^2+2x+k\): \(\displaystyle -3,1\); zeroes of \(\displaystyle 2x^4+x^3-14x^2+5x+6\): \(\displaystyle -3,1,-\tfrac12,2\).

  5. Exercise 5

    Given that x−5\displaystyle x-\sqrt{5} is a factor of the cubic polynomial x3−35x2+13x−35\displaystyle x^3-3 \sqrt{5} x^2+13 x-3 \sqrt{5}, find all the zeroes of the polynomial.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle \sqrt{5}, \sqrt{5}+\sqrt{2}, \sqrt{5}-\sqrt{2}\)
    \[x^3-3\sqrt5x^2+13x-3\sqrt5\ \div\ (x-\sqrt5) \] \[1,\ -3\sqrt5,\ 13,\ -3\sqrt5 \] \[\sqrt5-3\sqrt5=-2\sqrt5,\quad -2\sqrt5\cdot\sqrt5=-10,\quad 13-10=3,\quad 3\sqrt5-3\sqrt5=0 \] \[\text{Quotient: } x^2-2\sqrt5x+3 \] \[x=\dfrac{2\sqrt5\pm\sqrt{(2\sqrt5)^2-4(3)}}{2}=\dfrac{2\sqrt5\pm\sqrt8}{2}=\sqrt5\pm\sqrt2 \]Answer: \(\displaystyle \sqrt5,\ \sqrt5+\sqrt2,\ \sqrt5-\sqrt2\).

  6. Exercise 6

    For which values of a\displaystyle a and b\displaystyle b, are the zeroes of q(x)=x3+2x2+a\displaystyle q(x)=x^3+2 x^2+a also the zeroes of the polynomial p(x)=x5−x4−4x3+3x2+3x+b\displaystyle p(x)=x^5-x^4-4 x^3+3 x^2+3 x+b? Which zeroes of p(x)\displaystyle p(x) are not the zeroes of q(x)\displaystyle q(x)?

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle a=-1, b=-2\) $\displaystyle 1$ and $\displaystyle 2$ are the zeroes of \(\displaystyle q(x)\) which are not the zeroes of \(\displaystyle p(x)\).
    \[x^5-x^4-4x^3+3x^2+3x+b=x^2(x^3+2x^2+a)+\big(-3x^4-4x^3+(3-a)x^2+3x+b\big) \] \[-3x^4-4x^3+(3-a)x^2+3x+b=-3x(x^3+2x^2+a)+\big(2x^3+(3-a)x^2+(3+3a)x+b\big) \] \[2x^3+(3-a)x^2+(3+3a)x+b=2(x^3+2x^2+a)+\big((-1-a)x^2+(3+3a)x+(b-2a)\big) \] \[\text{quotient } x^2-3x+2,\quad \text{remainder } (-1-a)x^2+(3+3a)x+(b-2a) \] \[-1-a=0,\ 3+3a=0\ \Rightarrow\ a=-1 \] \[b-2a=0\ \Rightarrow\ b=2a=-2 \] \[x^2-3x+2=(x-1)(x-2) \]Answer: \(\displaystyle a=-1,\ b=-2\); zeroes of \(\displaystyle p(x)\) not zeroes of \(\displaystyle q(x)\): \(\displaystyle 1,\ 2\).