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NCERT Exemplar · Class 10 Mathematics Polynomials

29 questions · 29 still being checked

EXERCISE 2.3 1–10 (part 3 of 4)

  1. Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:

    Exercise 1

    4x2−3x−1\displaystyle 4 x^2-3 x-1

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    NCERT’s answer
    \(\displaystyle 1,-\frac{1}{4}\)
    \[4x^2-3x-1 = 4x^2-4x+x-1 \] \[= 4x(x-1)+1(x-1) = (4x+1)(x-1) \] \[x = -\frac{1}{4}, \quad x = 1 \] \[\alpha+\beta = -\frac{1}{4}+1 = \frac{3}{4} = -\frac{b}{a} \] \[\alpha\beta = \left(-\frac{1}{4}\right)(1) = -\frac{1}{4} = \frac{c}{a} \] Answer: zeroes \(\displaystyle -\frac{1}{4}, 1\); relations verified.

  2. Exercise 2

    3x2+4x−4\displaystyle 3 x^2+4 x-4

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    NCERT’s answer
    \(\displaystyle \frac{2}{3},-2\)
    \[3x^2+4x-4 = 3x^2+6x-2x-4 \] \[= 3x(x+2)-2(x+2) = (3x-2)(x+2) \] \[x = \frac{2}{3}, \quad x = -2 \] \[\alpha+\beta = \frac{2}{3}-2 = -\frac{4}{3} = -\frac{b}{a} \] \[\alpha\beta = \left(\frac{2}{3}\right)(-2) = -\frac{4}{3} = \frac{c}{a} \] Answer: zeroes \(\displaystyle \frac{2}{3}, -2\); relations verified.

  3. Exercise 3

    5t2+12t+7\displaystyle 5 t^2+12 t+7

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    NCERT’s answer
    \(\displaystyle -1, \frac{-7}{5}\)
    \[5t^2+12t+7 = 5t^2+5t+7t+7 \] \[= 5t(t+1)+7(t+1) = (5t+7)(t+1) \] \[t = -\frac{7}{5}, \quad t = -1 \] \[\alpha+\beta = -\frac{7}{5}-1 = -\frac{12}{5} = -\frac{b}{a} \] \[\alpha\beta = \left(-\frac{7}{5}\right)(-1) = \frac{7}{5} = \frac{c}{a} \] Answer: zeroes \(\displaystyle -\frac{7}{5}, -1\); relations verified.

  4. Exercise 4

    t3−2t2−15t\displaystyle t^3-2 t^2-15 t

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    NCERT’s answer
    $\displaystyle 0$, -$\displaystyle 3$, $\displaystyle 5$
    \[t^3-2t^2-15t = t(t^2-2t-15) \] \[= t(t-5)(t+3) \] \[t = 0, \quad t = 5, \quad t = -3 \] \[\alpha+\beta+\gamma = 0+5-3 = 2 = -\frac{b}{a} \] \[\alpha\beta+\beta\gamma+\gamma\alpha = (0)(5)+(5)(-3)+(-3)(0) = -15 = \frac{c}{a} \] \[\alpha\beta\gamma = (0)(5)(-3) = 0 = -\frac{d}{a} \] Answer: zeroes \(\displaystyle 0, 5, -3\); relations verified.

  5. Exercise 5

    2x2+72x+34\displaystyle 2 x^2+\frac{7}{2} x+\frac{3}{4}

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    NCERT’s answer
    \(\displaystyle \frac{-3}{2}, \frac{-1}{4}\)
    \[2x^2+\frac{7}{2}x+\frac{3}{4} = \frac{1}{4}\left(8x^2+14x+3\right) \] \[8x^2+14x+3 = 8x^2+12x+2x+3 = 4x(2x+3)+1(2x+3) \] \[= (4x+1)(2x+3) \] \[x = -\frac{1}{4}, \quad x = -\frac{3}{2} \] \[\alpha+\beta = -\frac{1}{4}-\frac{3}{2} = -\frac{7}{4} = -\frac{b}{a} \] \[\alpha\beta = \left(-\frac{1}{4}\right)\left(-\frac{3}{2}\right) = \frac{3}{8} = \frac{c}{a} \] Answer: zeroes \(\displaystyle -\frac{1}{4}, -\frac{3}{2}\); relations verified.

  6. Exercise 6

    4x2+52x−3\displaystyle 4 x^2+5 \sqrt{2} x-3

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    NCERT’s answer
    \(\displaystyle \frac{\sqrt{2}}{4}, \frac{-3 \sqrt{2}}{2}\)
    Split the middle term: \[4x^2+5\sqrt2\,x-3 = 4x^2+6\sqrt2\,x-\sqrt2\,x-3 \] \[= 2\sqrt2\,x(\sqrt2\,x+3)-1(\sqrt2\,x+3) \] \[= (\sqrt2\,x+3)(2\sqrt2\,x-1) \] Zeroes: \[x=-\frac{3}{\sqrt2}=-\frac{3\sqrt2}{2},\qquad x=\frac{1}{2\sqrt2}=\frac{\sqrt2}{4} \] Check, with \(\displaystyle a=4,\ b=5\sqrt2,\ c=-3\): \[\alpha+\beta=-\frac{3\sqrt2}{2}+\frac{\sqrt2}{4}=-\frac{5\sqrt2}{4}=-\frac{b}{a} \] \[\alpha\beta=\left(-\frac{3\sqrt2}{2}\right)\!\left(\frac{\sqrt2}{4}\right)=-\frac34=\frac{c}{a} \] Answer: zeroes \(\displaystyle -\dfrac{3\sqrt2}{2}\) and \(\displaystyle \dfrac{\sqrt2}{4}\); both relations hold.

  7. Exercise 7

    2s2−(1+22)s+2\displaystyle 2 s^2-(1+2 \sqrt{2}) s+\sqrt{2}

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    NCERT’s answer
    \(\displaystyle \frac{1}{2}, \sqrt{2}\)
    Split the middle term: \[2s^2-(1+2\sqrt2)s+\sqrt2 = 2s^2-s-2\sqrt2\,s+\sqrt2 \] \[= s(2s-1)-\sqrt2(2s-1) \] \[= (2s-1)(s-\sqrt2) \] Zeroes: \[s=\frac12,\qquad s=\sqrt2 \] Check, with \(\displaystyle a=2,\ b=-(1+2\sqrt2),\ c=\sqrt2\): \[\alpha+\beta=\frac12+\sqrt2=\frac{1+2\sqrt2}{2}=-\frac{b}{a} \] \[\alpha\beta=\frac12\cdot\sqrt2=\frac{\sqrt2}{2}=\frac{c}{a} \] Answer: zeroes \(\displaystyle \dfrac12\) and \(\displaystyle \sqrt2\); both relations hold.

  8. Exercise 8

    v2+43v−15\displaystyle v^2+4 \sqrt{3} v-15

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    NCERT’s answer
    \(\displaystyle \sqrt{3},-5 \sqrt{3}\)
    Split the middle term: \[v^2+4\sqrt3\,v-15 = v^2+5\sqrt3\,v-\sqrt3\,v-15 \] \[= v(v+5\sqrt3)-\sqrt3(v+5\sqrt3) \] \[= (v+5\sqrt3)(v-\sqrt3) \] Zeroes: \[v=-5\sqrt3,\qquad v=\sqrt3 \] Check, with \(\displaystyle a=1,\ b=4\sqrt3,\ c=-15\): \[\alpha+\beta=-5\sqrt3+\sqrt3=-4\sqrt3=-\frac{b}{a} \] \[\alpha\beta=(-5\sqrt3)(\sqrt3)=-15=\frac{c}{a} \] Answer: zeroes \(\displaystyle -5\sqrt3\) and \(\displaystyle \sqrt3\); both relations hold.

  9. Exercise 9

    y2+325y−5\displaystyle y^2+\frac{3}{2} \sqrt{5} y-5

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    NCERT’s answer
    \(\displaystyle -2 \sqrt{5}, \frac{\sqrt{5}}{2}\)
    Split the middle term: \[y^2+\frac{3\sqrt5}{2}y-5 = y^2+2\sqrt5\,y-\frac{\sqrt5}{2}y-5 \] \[= y(y+2\sqrt5)-\frac{\sqrt5}{2}(y+2\sqrt5) \] \[= \left(y+2\sqrt5\right)\!\left(y-\frac{\sqrt5}{2}\right) \] Zeroes: \[y=-2\sqrt5,\qquad y=\frac{\sqrt5}{2} \] Check, with \(\displaystyle a=1,\ b=\frac{3\sqrt5}{2},\ c=-5\): \[\alpha+\beta=-2\sqrt5+\frac{\sqrt5}{2}=-\frac{3\sqrt5}{2}=-\frac{b}{a} \] \[\alpha\beta=(-2\sqrt5)\!\left(\frac{\sqrt5}{2}\right)=-5=\frac{c}{a} \] Answer: zeroes \(\displaystyle -2\sqrt5\) and \(\displaystyle \dfrac{\sqrt5}{2}\); both relations hold.

  10. Exercise 10

    7y2−113y−23\displaystyle 7 y^2-\frac{11}{3} y-\frac{2}{3}

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    NCERT’s answer
    \(\displaystyle \frac{2}{3},-\frac{1}{7}\)
    Clear the fractions (scaling by $\displaystyle 3$ leaves the zeroes unchanged), then split the middle term: \[7y^2-\frac{11}{3}y-\frac23=\frac13\left(21y^2-11y-2\right) \] \[21y^2-11y-2 = 21y^2-14y+3y-2 \] \[= 7y(3y-2)+1(3y-2)=(3y-2)(7y+1) \] Zeroes: \[y=\frac23,\qquad y=-\frac17 \] Check, with \(\displaystyle a=7,\ b=-\frac{11}{3},\ c=-\frac23\): \[\alpha+\beta=\frac23-\frac17=\frac{11}{21}=-\frac{b}{a} \] \[\alpha\beta=\frac23\cdot\left(-\frac17\right)=-\frac{2}{21}=\frac{c}{a} \] Answer: zeroes \(\displaystyle \dfrac23\) and \(\displaystyle -\dfrac17\); both relations hold.