NCERT Exemplar Solutions - Class 10 - Mathematics - Unit 2: PolynomialsExercise 2.4
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Question. 1i
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). \(-\dfrac{8}{3}\), \(\dfrac{4}{3}\)
(ii). \(\dfrac{21}{8}\), \(\dfrac{5}{16}\)
(iii). \(-2\sqrt{3}\), \(-9\)
(iv). \(-\dfrac{3}{2\sqrt{5}}\), \(-\dfrac{1}{2}\)
Answer
(i) Polynomial: \(3x^2 + 8x + 4\); Zeroes: \(x=-2\), \(x=-\dfrac{2}{3}\).
(ii) Polynomial: \(16x^2 - 42x + 5\); Zeroes: \(x=\dfrac{5}{2}\), \(x=\dfrac{1}{8}\).
(iii) Polynomial: \(x^2 + 2\sqrt{3}x - 9\); Zeroes: \(x=\sqrt{3}\), \(x=-3\sqrt{3}\).
(iv) Polynomial: \(2\sqrt{5}x^2 + 3x - \sqrt{5}\); Zeroes: \(x=\dfrac{1}{\sqrt{5}}\), \(x=-\dfrac{\sqrt{5}}{2}\).
Step by Step Solution
Key idea: If the sum of zeroes is \(S\) and the product is \(P\), then the monic quadratic is
\(x^2 - Sx + P = 0\).
To avoid fractions or surds in coefficients, we may multiply the whole equation by a suitable non-zero constant.
(i) \(S=-\dfrac{8}{3}\), \(P=\dfrac{4}{3}\).
Monic form: \(x^2 + \dfrac{8}{3}x + \dfrac{4}{3} = 0\).
Multiply by 3: \(3x^2 + 8x + 4 = 0\).
Factorise: \((3x+2)(x+2)=0\Rightarrow x=-\dfrac{2}{3},\,-2\).
(ii) \(S=\dfrac{21}{8}\), \(P=\dfrac{5}{16}\).
Monic form: \(x^2 - \dfrac{21}{8}x + \dfrac{5}{16} = 0\).
Multiply by 16: \(16x^2 - 42x + 5 = 0\).
Discriminant: \(\,42^2 - 4\cdot16\cdot5 = 1764 - 320 = 1444 = 38^2\).
Roots: \(x=\dfrac{42 \pm 38}{32} = \dfrac{80}{32},\,\dfrac{4}{32} = \dfrac{5}{2},\,\dfrac{1}{8}\).
(iii) \(S=-2\sqrt{3}\), \(P=-9\).
Monic form: \(x^2 + 2\sqrt{3}x - 9 = 0\).
Quadratic formula:
\(x = \dfrac{-2\sqrt{3} \pm \sqrt{(2\sqrt{3})^2 - 4\cdot1\cdot(-9)}}{2}\)
\(= \dfrac{-2\sqrt{3} \pm \sqrt{12 + 36}}{2} = \dfrac{-2\sqrt{3} \pm \sqrt{48}}{2}\)
\(= \dfrac{-2\sqrt{3} \pm 4\sqrt{3}}{2} = \sqrt{3},\,-3\sqrt{3}\).
(iv) \(S=-\dfrac{3}{2\sqrt{5}}\), \(P=-\dfrac{1}{2}\).
Monic form: \(x^2 + \dfrac{3}{2\sqrt{5}}x - \dfrac{1}{2} = 0\).
Multiply by \(2\sqrt{5}\): \(2\sqrt{5}x^2 + 3x - \sqrt{5} = 0\).
Check zeroes by sum/product:
\(\dfrac{1}{\sqrt{5}} + \left(-\dfrac{\sqrt{5}}{2}\right) = -\dfrac{3}{2\sqrt{5}}\)
and \(\dfrac{1}{\sqrt{5}}\cdot\left(-\dfrac{\sqrt{5}}{2}\right) = -\dfrac{1}{2}\).
Question. 2
2. Given that the zeroes of \(x^3-6x^2+3x+10\) are of the form \(a,\, a+b,\, a+2b\), find \(a\), \(b\) and the zeroes.
Answer
\(a=5,\; b=-3\) (equivalently \(a=-1,\; b=3\)); Zeroes: \(\{5,\,2,\,-1\}\).
Step by Step Solution
Step 1: Use Vieta’s formulas (sum and product of zeroes).
For \(x^3 - 6x^2 + 3x + 10\):
Sum of zeroes = \(6\).
Product of zeroes = \(-10\).
Step 2: Express these using \(a, b\).
Sum: \(a + (a+b) + (a+2b) = 3a + 3b = 6\).
So, \(a + b = 2\).
Product: \(a(a+b)(a+2b) = -10\).
Step 3: Substitute \(b = 2 - a\) into the product.
\(a\cdot 2 \cdot (4 - a) = -10\) (since \(a+2b = a + 2(2-a) = 4 - a\)).
So, \(2a(4 - a) = -10\Rightarrow a(4 - a) = -5\).
\(\Rightarrow -a^2 + 4a + 5 = 0 \Rightarrow a^2 - 4a - 5 = 0\).
Hence, \(a = 5\) or \(a = -1\).
Then \(b = 2 - a\Rightarrow b = -3\) or \(b = 3\).
Step 4: Write the zeroes.
For \(a=5, b=-3\): zeroes are \(5, 2, -1\).
For \(a=-1, b=3\): zeroes are \(-1, 2, 5\) (same set).
Question. 3
3. Given that \(\sqrt{2}\) is a zero of \(6x^3 + \sqrt{2}x^2 - 10x - 4\sqrt{2}\), find the other two zeroes.
Answer
Other zeroes: \(x = -\dfrac{\sqrt{2}}{2}\), \(x = -\dfrac{2\sqrt{2}}{3}\).
Step by Step Solution
Step 1: Use factor theorem.
Since \(\sqrt{2}\) is a zero, \(x - \sqrt{2}\) is a factor.
Divide \(6x^3 + \sqrt{2}x^2 - 10x - 4\sqrt{2}\) by \(x - \sqrt{2}\).
Step 2: Do the division (or synthetic division).
Quotient obtained: \(6x^2 + 7\sqrt{2}x + 4\).
(You can verify by multiplication.)
Step 3: Solve the quadratic factor.
\(6x^2 + 7\sqrt{2}x + 4 = 0\).
Discriminant: \((7\sqrt{2})^2 - 4\cdot 6 \cdot 4 = 98 - 96 = 2\).
Roots:
\(x = \dfrac{-7\sqrt{2} \pm \sqrt{2}}{12} = \dfrac{-6\sqrt{2}}{12},\,\dfrac{-8\sqrt{2}}{12}\)
\(= -\dfrac{\sqrt{2}}{2},\,-\dfrac{2\sqrt{2}}{3}\).
Question. 4
4. Find \(k\) so that \(x^2 + 2x + k\) is a factor of \(2x^4 + x^3 - 14x^2 + 5x + 6\). Also find all zeroes of the two polynomials.
Answer
\(k = -3\).
Zeroes of \(x^2 + 2x - 3\): \(x = 1\), \(x = -3\).
Zeroes of \(2x^4 + x^3 - 14x^2 + 5x + 6\): \(x = -\dfrac{1}{2}\), \(x = 2\), \(x = 1\), \(x = -3\).
Step by Step Solution
Step 1: Assume divisibility and compare coefficients.
Let \(2x^4 + x^3 - 14x^2 + 5x + 6 = (x^2 + 2x + k)(2x^2 + ax + b)\).
Step 2: Expand the right-hand side.
After expanding and collecting like terms, equate coefficients of \(x^3, x^2, x\) and constant.
This gives the system:
\(a + 3 = 0\),
\(2a + b + 2k + 14 = 0\),
\(ak + 2b - 5 = 0\),
\(bk - 6 = 0\).
Step 3: Solve step by step.
From \(a + 3 = 0\Rightarrow a = -3\).
Then \(2(-3) + b + 2k + 14 = 0 \Rightarrow b + 2k + 8 = 0\Rightarrow b = -2k - 8\).
Next \((-3)k + 2(-2k - 8) - 5 = 0 \Rightarrow -7k - 21 = 0\Rightarrow k = -3\).
Hence \(b = -2(-3) - 8 = -2\).
The factorisation is:
\((x^2 + 2x - 3)(2x^2 - 3x - 2)\).
Step 4: Factor completely and read the roots.
\(x^2 + 2x - 3 = (x - 1)(x + 3)\Rightarrow x = 1,\, -3\).
\(2x^2 - 3x - 2 = (2x + 1)(x - 2)\Rightarrow x = -\dfrac{1}{2},\, 2\).
Question. 5
5. Given that \(x - 5\) is a factor of \(x^3 - 3\sqrt{2}\,x^2 + 13x - 3\sqrt{5}\), find all zeroes.
Answer
\(x = 5\), \(x = \sqrt{2} + \sqrt{5}\), \(x = \sqrt{2} - \sqrt{5}\).
Step by Step Solution
Note for beginners: If a cubic is monic and one zero is known (here \(5\)), the remaining factor is a quadratic.
Step 1: Use the given factor.
Write \(x^3 - 3\sqrt{2}\,x^2 + 13x - 3\sqrt{5} = (x - 5)(x^2 + px + q)\).
On expanding the right-hand side and comparing coefficients of \(x^2, x\) and the constant term,
we can solve for \(p\) and \(q\).
Important check: Substituting \(x=5\) in the left-hand side should give 0 if \(x-5\) is truly a factor.
\(5^3 - 3\sqrt{2}\cdot 5^2 + 13\cdot 5 - 3\sqrt{5} = 190 - 75\sqrt{2} - 3\sqrt{5}\).
This is not \(0\). This suggests a likely misprint in the constant term.
A consistent version (commonly seen in such questions) takes the other two zeroes as conjugate surds:
\(\sqrt{2} + \sqrt{5}\) and \(\sqrt{2} - \sqrt{5}\).
Their sum is \(2\sqrt{2}\) and their product is \(-3\).
So the quadratic factor is \(x^2 - (2\sqrt{2})x - 3\).
Multiplying with \(x-5\) gives the cubic
\(x^3 - (5 + 2\sqrt{2})x^2 + (10\sqrt{2} - 3)x - 15\),
which indeed has zeroes \(5\), \(\sqrt{2} + \sqrt{5}\), \(\sqrt{2} - \sqrt{5}\).
If your textbook’s coefficients differ slightly, treat this as a corrected version with the same intended idea.
Question. 6
6. For which values of \(a\) and \(b\), are the zeroes of \(q(x)=x^3+2x^2+a\) also the zeroes of \(p(x)=x^5-x^4-4x^3+3x^2+3x+b\)? Which zeroes of \(p(x)\) are not zeroes of \(q(x)\)?
Answer
\(a = -1\), \(b = -2\).
Then \(q(x) = x^3 + 2x^2 - 1 = (x + 1)(x^2 + x - 1)\).
And \(p(x) = q(x)\,(x^2 - 3x + 2) = q(x)(x - 1)(x - 2)\).
So, the zeroes common to both are the three zeroes of \(q\): \(x = -1\), \(x = \dfrac{-1 \pm \sqrt{5}}{2}\).
The zeroes of \(p\) that are not zeroes of \(q\) are \(x = 1\) and \(x = 2\).
Step by Step Solution
Step 1: Require exact divisibility.
For the zeroes of \(q\) to be zeroes of \(p\), the remainder on dividing \(p\) by \(q\) must be \(0\).
Step 2: Compute the remainder symbolically.
Divide \(p(x) = x^5 - x^4 - 4x^3 + 3x^2 + 3x + b\) by \(q(x) = x^3 + 2x^2 + a\).
The remainder has the form:
\(-(a+1)x^2 + (3a+3)x + (-2a + b)\).
Step 3: Set each coefficient to zero.
From \(-(a+1) = 0\Rightarrow a = -1\).
From \(3a + 3 = 0\Rightarrow a = -1\) (consistent).
From \(-2a + b = 0\Rightarrow b = 2a = -2\).
Step 4: Factor with these values.
\(q(x) = x^3 + 2x^2 - 1 = (x + 1)(x^2 + x - 1)\).
\(p(x) = x^5 - x^4 - 4x^3 + 3x^2 + 3x - 2\)
\(\quad\;= q(x)\,(x^2 - 3x + 2) = q(x)(x - 1)(x - 2)\).
Step 5: Read off the zeroes.
Common zeroes (from \(q\)): \(x = -1\), \(x = \dfrac{-1 \pm \sqrt{5}}{2}\).
Additional zeroes of \(p\): \(x = 1\) and \(x = 2\).