NCERT Exemplar Solutions - Class 10 - Mathematics - Unit 2: PolynomialsExercise 2.3
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Question. 1
1. Find the zeroes of \(4x^2-3x-1\) by factorisation and verify relations.
Answer
\(x=1,\; x=-\dfrac{1}{4}\)
Step by Step Solution
Step 1: Split the middle term.
Product = \(4 \times -1 = -4\)
Sum = \(-3\)
Numbers = \(-4\) and \(+1\)
Step 2: Write and group.
\(4x^2 - 3x - 1 = 4x^2 - 4x + x - 1\)
= \((4x^2 - 4x) + (x - 1)\)
= \(4x(x - 1) + 1(x - 1)\)
= \((4x + 1)(x - 1)\)
Step 3: Zeroes.
\(4x + 1 = 0 \Rightarrow x = -\dfrac{1}{4}\)
\(x - 1 = 0 \Rightarrow x = 1\)
Step 4: Verify.
Sum = \(1 - \dfrac{1}{4} = \dfrac{3}{4}\)
= \(-b/a = -(-3)/4\)
Product = \(1 × -\dfrac{1}{4} = -\dfrac{1}{4}\)
= \(c/a = -1/4\)
Question. 2
2. Find the zeroes of \(3x^2+4x-4\) and verify relations.
Answer
\(x=\dfrac{2}{3},\; x=-2\)
Step by Step Solution
Step 1: Split the middle term.
Product = \(3 \times -4 = -12\)
Sum = \(4\)
Numbers = \(6\) and \(-2\)
Step 2: Write and group.
\(3x^2 + 4x - 4 = 3x^2 + 6x - 2x - 4\)
= \((3x^2 + 6x) + (-2x - 4)\)
= \(3x(x + 2) - 2(x + 2)\)
= \((x + 2)(3x - 2)\)
Step 3: Zeroes.
\(x + 2 = 0 \Rightarrow x = -2\)
\(3x - 2 = 0 \Rightarrow x = \dfrac{2}{3}\)
Step 4: Verify.
Sum = \(-2 + \dfrac{2}{3} = -\dfrac{4}{3}\)
= \(-b/a = -4/3\)
Product = \(-2)(\dfrac{2}{3}) = -\dfrac{4}{3}\)
= \(c/a = -4/3\)
Question. 3
3. Find the zeroes of \(5t^2+12t+7\) and verify relations.
Answer
\(t=-1,\; t=-\dfrac{7}{5}\)
Step by Step Solution
Step 1: Factorise directly.
\((5t + 7)(t + 1) = 5t^2 + 12t + 7\)
Step 2: Zeroes.
\(5t + 7 = 0 \Rightarrow t = -\dfrac{7}{5}\)
\(t + 1 = 0 \Rightarrow t = -1\)
Step 3: Verify.
Sum = \(-1 - \dfrac{7}{5} = -\dfrac{12}{5}\)
= \(-b/a = -12/5\)
Product = \((-1)(-\dfrac{7}{5}) = 7/5\)
= \(c/a = 7/5\)
Question. 4
4. Find the zeroes of \(t^3-2t^2-15t\) and verify relations.
Answer
\(t=0,\; t=5,\; t=-3\)
Step by Step Solution
Step 1: Factorise.
Take common factor: \(t(t^2 - 2t - 15)\)
Quadratic: \(t^2 - 2t - 15 = (t - 5)(t + 3)\)
So, \(t(t - 5)(t + 3)\)
Step 2: Zeroes.
\(t = 0,\; t = 5,\; t = -3\)
Step 3: Verify.
Sum = \(0 + 5 + (-3) = 2\)
= \(-a = -(-2)\)
Pairwise sum = \(0×5 + 5×(-3) + (-3)×0 = -15\)
= \(b = -15\)
Product = \(0×5×(-3) = 0\)
= \(-c = 0\)
Question. 5
5. Find the zeroes of \(2x^2+\dfrac{7}{2}x+\dfrac{3}{4}\) and verify relations.
Answer
\(x=-\dfrac{1}{4},\; x=-\dfrac{3}{2}\)
Step by Step Solution
Step 1: Remove fractions.
Multiply whole equation by 4:
\(8x^2 + 14x + 3\)
Step 2: Factorise.
\(8x^2 + 14x + 3 = (4x + 1)(2x + 3)\)
Step 3: Zeroes.
\(4x + 1 = 0 \Rightarrow x = -\dfrac{1}{4}\)
\(2x + 3 = 0 \Rightarrow x = -\dfrac{3}{2}\)
Step 4: Verify.
Sum = \(-\dfrac{1}{4} - \dfrac{3}{2} = -\dfrac{7}{4}\)
= \(-b/a = - (7/2)/2 = -7/4\)
Product = \((-\dfrac{1}{4})(-\dfrac{3}{2}) = 3/8\)
= \(c/a = (3/4)/2 = 3/8\)
Question. 6
6. Find the zeroes of \(4x^2+5\sqrt{2}x-3\) and verify relations.
Answer
\(x=\dfrac{\sqrt{2}}{4},\; x=-\dfrac{3\sqrt{2}}{2}\)
Step by Step Solution
Step 1: Quadratic formula.
\(a = 4, b = 5\sqrt{2}, c = -3\)
Discriminant = \((5\sqrt{2})^2 - 4(4)(-3)\)
= 50 + 48 = 98
\(\sqrt{98} = 7\sqrt{2}\)
Step 2: Roots.
\(x = \dfrac{-5\sqrt{2} + 7\sqrt{2}}{8} = \dfrac{\sqrt{2}}{4}\)
\(x = \dfrac{-5\sqrt{2} - 7\sqrt{2}}{8} = -\dfrac{3\sqrt{2}}{2}\)
Step 3: Verify.
Sum = \(\dfrac{\sqrt{2}}{4} - \dfrac{3\sqrt{2}}{2} = -5\sqrt{2}/4\)
= \(-b/a = -(5\sqrt{2})/4\)
Product = \((\dfrac{\sqrt{2}}{4})(-\dfrac{3\sqrt{2}}{2}) = -3/4\)
= \(c/a = -3/4\)
Question. 7
7. Find the zeroes of \(2s^2-(1+2\sqrt{2})s+\sqrt{2}\) and verify relations.
Answer
\(s=\dfrac{1}{2},\; s=\sqrt{2}\)
Step by Step Solution
Step 1: Check zeroes.
Substitute \(s = \sqrt{2}\)
\(2(2) - (1 + 2\sqrt{2})(\sqrt{2}) + \sqrt{2}\)
= 4 - \sqrt{2} - 4 + \sqrt{2} = 0
So, \(s = \sqrt{2}\) is a root.
Substitute \(s = 1/2\)
\(2(1/4) - (1 + 2\sqrt{2})(1/2) + \sqrt{2}\)
= 1/2 - 1/2 - \sqrt{2} + \sqrt{2} = 0
So, \(s = 1/2\) is a root.
Step 2: Verify.
Sum = \(1/2 + \sqrt{2}\)
= \((1 + 2\sqrt{2})/2 = -b/a\)
Product = \((1/2)(\sqrt{2}) = \sqrt{2}/2\)
= c/a
Question. 8
8. Find the zeroes of \(v^2+4\sqrt{3}v-15\) and verify relations.
Answer
\(v=\sqrt{3},\; v=-5\sqrt{3}\)
Step by Step Solution
Step 1: Quadratic formula.
\(a=1, b=4\sqrt{3}, c=-15\)
Discriminant = \((4\sqrt{3})^2 - 4(1)(-15)\)
= 48 + 60 = 108
\(\sqrt{108} = 6\sqrt{3}\)
Step 2: Roots.
\(v = \dfrac{-4\sqrt{3} + 6\sqrt{3}}{2} = \sqrt{3}\)
\(v = \dfrac{-4\sqrt{3} - 6\sqrt{3}}{2} = -5\sqrt{3}\)
Step 3: Verify.
Sum = \(\sqrt{3} - 5\sqrt{3} = -4\sqrt{3}\)
= -b/a = -4\sqrt{3}
Product = \(\sqrt{3} × -5\sqrt{3} = -15\)
= c/a = -15
Question. 9
9. Find the zeroes of \(y^2+\dfrac{\sqrt{3}}{2}y-5\) and verify relations.
Answer
\(y=\dfrac{-\sqrt{3}+\sqrt{83}}{4},\; y=\dfrac{-\sqrt{3}-\sqrt{83}}{4}\)
Step by Step Solution
Step 1: Identify coefficients.
\(a=1, b=\sqrt{3}/2, c=-5\)
Step 2: Discriminant.
\(b^2 - 4ac = (\sqrt{3}/2)^2 - 4(1)(-5)\)
= 3/4 + 20 = 83/4
\(\sqrt{83/4} = \sqrt{83}/2\)
Step 3: Roots.
\(y = \dfrac{-\sqrt{3}/2 ± \sqrt{83}/2}{2}\)
= \(\dfrac{-\sqrt{3} ± \sqrt{83}}{4}\)
Step 4: Verify.
Sum = \((-\sqrt{3} + \sqrt{83})/4 + (-\sqrt{3} - \sqrt{83})/4\)
= -2\sqrt{3}/4 = -\sqrt{3}/2 = -b/a
Product = \((-\sqrt{3} + \sqrt{83})(-\sqrt{3} - \sqrt{83})/16\)
= (83 - 3)/16 = 80/16 = 5 = c/a
Question. 10
10. Find the zeroes of \(7y^2-\dfrac{11}{3}y-23\) and verify relations.
Answer
\(y=\dfrac{11+\sqrt{5917}}{42},\; y=\dfrac{11-\sqrt{5917}}{42}\)
Step by Step Solution
Step 1: Identify coefficients.
\(a = 7,\; b = -\dfrac{11}{3},\; c = -23\)
Step 2: Discriminant.
\(b^2 - 4ac = \left(-\dfrac{11}{3}\right)^2 - 4(7)(-23)\)
= \(\dfrac{121}{9} + 644\)
= \(\dfrac{5917}{9}\)
So, \(\sqrt{\Delta} = \dfrac{\sqrt{5917}}{3}\)
Step 3: Roots.
\(y = \dfrac{ -b \pm \sqrt{\Delta} }{ 2a }\)
= \(\dfrac{ \dfrac{11}{3} \pm \dfrac{\sqrt{5917}}{3} }{ 14 }\)
= \(\dfrac{ 11 \pm \sqrt{5917} }{ 42 }\)
Step 4: Verify.
Sum = \(\dfrac{11 + \sqrt{5917}}{42} + \dfrac{11 - \sqrt{5917}}{42}\)
= \(\dfrac{22}{42} = \dfrac{11}{21}\)
= \(-\dfrac{b}{a} = -\dfrac{-11/3}{7} = \dfrac{11}{21}\)
Product = \(\dfrac{(11 + \sqrt{5917})(11 - \sqrt{5917})}{42^2}\)
= \(\dfrac{121 - 5917}{1764}\)
= \(\dfrac{-5796}{1764}\)
= \(-\dfrac{23}{7}\)
= \(\dfrac{c}{a}\)