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NCERT Exemplar · Class 10 Mathematics Quadratic Equations

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EXERCISE 4.2 1–7 (part 2 of 4)

  1. Exercise 1

    State whether the following quadratic equations have two distinct real roots. Justify your answer.
    (i)
    x2−3x+4=0\displaystyle x^2-3 x+4=0
    (ii)
    2x2+x−1=0\displaystyle 2 x^2+x-1=0
    (iii)
    2x2−6x+92=0\displaystyle 2 x^2-6 x+\frac{9}{2}=0
    (iv)
    3x2−4x+1=0\displaystyle 3 x^2-4 x+1=0
    (v)
    (x+4)2−8x=0\displaystyle (x+4)^2-8 x=0
    (vi)
    (x−2)2−2(x+1)=0\displaystyle (x-\sqrt{2})^2-2(x+1)=0
    (vii)
    2x2−32x+12=0\displaystyle \sqrt{2} x^2-\frac{3}{\sqrt{2}} x+\frac{1}{\sqrt{2}}=0
    (viii)
    x(1−x)−2=0\displaystyle x(1-x)-2=0
    (ix)
    (x−1)(x+2)+2=0\displaystyle (x-1)(x+2)+2=0
    (x)
    (x+1)(x−2)+x=0\displaystyle (x+1)(x-2)+x=0

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    NCERT’s answer
    (i)
    No, because discriminant \(\displaystyle =-7<0\).
    (ii)
    Yes, because discriminant \(\displaystyle =9>0\).
    (iii)
    No, because discriminant = 0.
    (iv)
    Yes, because discriminant \(\displaystyle =4>0\).
    (v)
    No, because discriminant \(\displaystyle =-64<0\).
    (vi)
    Yes, because discriminant \(\displaystyle =(2+2 \sqrt{2})^2>0\).
    (vii)
    Yes, because discriminant \(\displaystyle =1>0\).
    (viii)
    No, because discriminant \(\displaystyle =-7<0\).
    (ix)
    Yes, because discriminant \(\displaystyle =1>0\).
    (x)
    Yes, because discriminant \(\displaystyle =8>0\).
    (i) No. \[D=(-3)^2-4(1)(4)=9-16=-7<0 \](ii) Yes. \[D=1^2-4(2)(-1)=1+8=9>0 \](iii) No. \[D=(-6)^2-4(2)\left(\frac{9}{2}\right)=36-36=0 \] Equal roots.(iv) Yes. \[D=(-4)^2-4(3)(1)=16-12=4>0 \](v) No. \[(x+4)^2-8x=x^2+16=0 \] \[D=0^2-4(1)(16)=-64<0 \](vi) Yes. \[(x-\sqrt2)^2-2(x+1)=x^2-2(\sqrt2+1)x=0 \] \[D=[2(\sqrt2+1)]^2>0 \](vii) Yes. \[D=\left(-\frac{3}{\sqrt2}\right)^2-4(\sqrt2)\left(\frac{1}{\sqrt2}\right)=\frac{9}{2}-4=\frac12>0 \](viii) No. \[x(1-x)-2=0\Rightarrow x^2-x+2=0 \] \[D=(-1)^2-4(1)(2)=-7<0 \](ix) Yes. \[(x-1)(x+2)+2=0\Rightarrow x^2+x=0 \] \[D=1^2-4(1)(0)=1>0 \](x) Yes. \[(x+1)(x-2)+x=0\Rightarrow x^2-2=0 \] \[D=0^2-4(1)(-2)=8>0 \]Answer: Yes: (ii), (iv), (vi), (vii), (ix), (x). No: (i), (iii), (v), (viii).

  2. Exercise 2

    Write whether the following statements are true or false. Justify your answers.
    (i)
    Every quadratic equation has exactly one root.
    (ii)
    Every quadratic equation has at least one real root.
    (iii)
    Every quadratic equation has at least two roots.
    (iv)
    Every quadratic equations has at most two roots.
    (v)
    If the coefficient of x2\displaystyle x^2 and the constant term of a quadratic equation have opposite signs, then the quadratic equation has real roots.
    (vi)
    If the coefficient of x2\displaystyle x^2 and the constant term have the same sign and if the coefficient of x\displaystyle x term is zero, then the quadratic equation has no real roots.

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    NCERT’s answer
    (i)
    False, for example : \(\displaystyle x^2=1\) is a quadratic equation with two roots.
    (ii)
    False, for example \(\displaystyle x^2+1=0\) has no real root.
    (iii)
    False, for example : \(\displaystyle x^2+1=0\) is a quadratic equation which has no real roots.
    (iv)
    True, because every quadratic polynomial has almost two zeroes.
    (v)
    True, because if in \(\displaystyle a x^2+b x+c=0, a\) and \(\displaystyle c\) have opposite signs, then \(\displaystyle a c<0\) and so \(\displaystyle b^2-4 a c>0\).
    (vi)
    True, because if in \(\displaystyle a x^2+b x+c=0, a\) and \(\displaystyle c\) have same sign and \(\displaystyle b=0\), then \(\displaystyle b^2-4 a c=-4 a c<0\).
    (i) False. \(\displaystyle x^2-4=0\) has two roots: \[x=\pm2 \](ii) False. \[x^2+1=0\Rightarrow D=0-4=-4<0 \] No real root.(iii) False. The equation in (ii), \(\displaystyle x^2+1=0\), has no real root, so it does not have two.(iv) True. A degree-$\displaystyle 2$ equation has at most two roots.(v) True. \(\displaystyle a,c\) opposite in sign \(\displaystyle \Rightarrow ac<0\). \[D=b^2-4ac\ge-4ac>0 \] Two distinct real roots.(vi) True. \(\displaystyle b=0,\ ac>0\). \[D=-4ac<0 \] No real roots.

  3. Exercise 3

    A quadratic equation with integral coefficient has integral roots. Justify your answer.

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    NCERT’s answer
    \(\displaystyle x^2-3 x+1=0\) is an equation with integral coefficients but its roots are not integers.
    No. Counter-example: \[2x^2+x-1=0 \] has integral coefficients, but by the quadratic formula \[x=\frac{-1\pm\sqrt{1+8}}{4}=\frac{-1\pm3}{4} \] giving \(\displaystyle x=\frac12,-1\); the root \(\displaystyle \frac12\) is not an integer.

  4. Exercise 4

    Does there exist a quadratic equation whose coefficients are rational but both of its roots are irrational? Justify your answer.

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    NCERT’s answer
    \(\displaystyle x^2-6 x+7=0\), which has roots \(\displaystyle 3+\sqrt{2}, 3-\sqrt{2}\)
    Yes. \[x^2-2=0 \] has rational coefficients \(\displaystyle (1,0,-2)\) and \[D=0^2-4(1)(-2)=8 \] Since $\displaystyle 8$ is not a perfect square, \(\displaystyle \sqrt8\) is irrational, so \[x=\pm\sqrt2 \] both roots are irrational.

  5. Exercise 5

    Does there exist a quadratic equation whose coefficients are all distinct irrationals but both the roots are rationals? Why?

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    NCERT’s answer
    Yes. \(\displaystyle \sqrt{3} x^2-7 \sqrt{3} x+12 \sqrt{3}=0\), which has roots $\displaystyle 3$, $\displaystyle 4$
    Yes. \[\sqrt2\,x^2-3\sqrt2\,x+2\sqrt2=0 \] has coefficients \(\displaystyle \sqrt2,-3\sqrt2,2\sqrt2\) — all irrational, all distinct. Dividing by \(\displaystyle \sqrt2\;(\ne0)\): \[x^2-3x+2=0\Rightarrow(x-1)(x-2)=0 \] roots \(\displaystyle x=1,2\), both rational.

  6. Exercise 6

    Is 0.2\displaystyle 0.2 a root of the equation x2−0.4=0\displaystyle x^2-0.4=0? Justify.

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    NCERT’s answer
    No.
    No. \[(0.2)^2-0.4=0.04-0.4=-0.36\ne0 \] $\displaystyle 0.2$ does not satisfy the equation.

  7. Exercise 7

    If b=0,c<0\displaystyle b=0, c<0, is it true that the roots of x2+bx+c=0\displaystyle x^2+b x+c=0 are numerically equal and opposite in sign? Justify.

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    NCERT’s answer
    Yes
    True. With \(\displaystyle b=0\): \[x^2+c=0\Rightarrow x^2=-c \] Since \(\displaystyle c<0\), \(\displaystyle -c>0\), so \[x=\pm\sqrt{-c} \] Roots \(\displaystyle \sqrt{-c},-\sqrt{-c}\) are numerically equal, opposite in sign.