NCERT solution for class 10 maths polynomials ( Chapter 2)

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Solution for Exercise 2.1

Q1. The graphs of y=p(x) are given to us, for some polynomials p(x). Find the number of zeroes of p(x), in each case.
(i) Graph 1

(ii) Graph 2

(iii) Graph 3

(iv) Graph 4

(v) Graph 5

(vi) Graph 6

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Answer :

(i) Graph does not meets x-axis at all, so it does not have any zero.
(ii) Graph meets x-axis 1 time, so it has one zero.
(iii) Graph meets x-axis 3 times, so it has three zero.
(iv) Graph meets x-axis 2 times, so it has two zero.
(v) Graph meets x-axis 4 times, so it has four zero.
(vi) Graph meets x-axis 3 times, so it has three zero.

Solution for Exercise 2.2

Q1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeros and the coefficients.
(i) \(x^2 - 2x - 8\)
(ii) \(4s^2 - 4s + 1\)
(iii) \(6x^2 - 3 - 7x\)
(iv) \(4u^2 + 8u\)
(v) \(t^2 - 15 \)
(vi) \(3x^2 - x - 4\)

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Answer :

General form of equation is \(ax^2 + bx + c\)
where, a = Coefficient of \(x^2\)
b = Coefficient of x
c = Constant term
Once we'll find the zeroes of the polynomial we can verify it using the following rules
Sum of zeroes = \({{-b}\over {a}}\)
Product of zeroes = \({{c}\over {a}}\)

(i) \(x^2 - 2x - 8\)
=>\(x^2 - 4x + 2x - 8\)
=> \(x(x - 4) + 2(x - 4)\)
=> \((x + 2)(x - 4)\)
So, zeroes are -2 and 4
Now, Verification:
Sum of the zeroes = 2 = \({{-b}\over {a}}\) = \({{-(-2)}\over {1}}\) = 2
Product of the zeroes = -8 = \({{c}\over {a}}\) = \({{-8}\over {1}}\) = -8

(ii) \(4s^2 - 4s + 1\)
=>\(4s^2 - 2s - 2s + 18\)
=> \(2s(2s - 1) - 1(2s - 1)\)
=> \((2s - 1)(2s - 1)\)
So, zeroes are \({{1}\over {2}}\) and \({{1}\over {2}}\)
Now, Verification:
Sum of the zeroes = 1 = \({{-b}\over {a}}\) = \({{-(-4)}\over {4}}\) = 1
Product of the zeroes = \({{1}\over {4}}\) = \({{c}\over {a}}\) = \({{1}\over {4}}\)

(iii) \(6x^2 - 7x - 3\)
=>\(6x^2 - 9x + 2x -3\)
=> \(3x(2x - 3) + 1(2x - 3)\)
=> \((3x + 1)(2x - 3)\)
So, zeroes are \({{-1}\over {3}}\) and \({{3}\over {2}}\)
Now, Verification:
Sum of the zeroes = \({{-1}\over {3}} + {{3}\over {2}}\) = \({{7}\over {6}}\) = \({{-b}\over {a}}\) = \({{-(-7)}\over {6}}\)
Product of the zeroes = \({{-1}\over {3}} × {{3}\over {2}}\) = \({{-1}\over {2}}\) = \({{c}\over {a}}\) = \({{-3}\over {6}}\) = \({{-1}\over {2}}\)

(iv) \(4u^2 + 8u\)
=>\(4u(u + 2)\)
=>\(4u = 0\)
=>\(u = 0\)
=>\(u = -2\)
So, zeroes are 0 and -2
Now, Verification:
Sum of the zeroes = -2 = \({{-b}\over {a}}\) = \({{-(8)}\over {4}}\) = -2
Product of the zeroes = 0 = \({{c}\over {a}}\) = \({{0}\over {4}}\) = 0

(v) \(t^2 - 15 \)
=>\(t^2 - (\sqrt{15})^2\)
=> \((t + \sqrt{15})(t -\sqrt{15})\)
So, zeroes are \(-\sqrt{15}\) and \(\sqrt{15}\)
Now, Verification:
Sum of the zeroes = 0 = \({{-b}\over {a}}\) = \({{-(0)}\over {1}}\) = 0
Product of the zeroes = -15 = \({{c}\over {a}}\) = \({{-15}\over {1}}\) = -15

(vi) \(3x^2 - x - 4\)
=>\(3x^2 - 4x + 3x -4\)
=> \(x(3x - 4) + 1(3x - 4)\)
=> \((3x - 4)(x + 1)\)
So, zeroes are \({{4}\over {3}}\) and -1
Now, Verification:
Sum of the zeroes = \({{4}\over {3}} + (-1)\) = \({{1}\over {3}}\) = \({{-b}\over {a}}\) = \({{-(-1)}\over {3}}\) = \({{1}\over {3}}\)
Product of the zeroes = \({{4}\over {3}} × (-1)\) = \({{-4}\over {3}}\) = \({{c}\over {a}}\) = \({{-4}\over {3}}\)

Q2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) \({{1} \over {4}} , -1\)
(ii) \(\sqrt{2}, {{1}\over{3}} \)
(iii) \(0, \sqrt{5}\)
(iv) \(1,1\)
(v) \({{-1} \over {4}},{{1} \over {4}} \)
(vi) \(4,1\)

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Answer :

(i) \({{1} \over {4}} , -1\)
=>Sum of the zeroes = \({{1} \over {4}}\) = \({{-b}\over {a}}\)
=> Product of the zeroes = -1 = \({{c}\over {a}}\)
Polynomial can be formed by : \(x^2\) - (sum of zeroes)x + (product of zeroes)
=>\(x^2 - {{1} \over {4}}x + (-1)\) => \(4x^2 - x - 4\)

(ii) \( \sqrt{2} , {{1}\over{3}}\)
=>Sum of the zeroes = \( \sqrt{2}\) = \({{-b}\over {a}}\)
=> Product of the zeroes = \({{1} \over {3}}\) = \({{c}\over {a}}\)
Polynomial can be formed by : \(x^2\) - (sum of zeroes)x + (product of zeroes)
=>\(x^2 - \sqrt{2} x + {{1} \over {3}}\) => \(3x^2 - 3\sqrt{2}x + 1\)

(iii) \(0 , \sqrt{5}\)
=>Sum of the zeroes = 0 = \({{-b}\over {a}}\)
=> Product of the zeroes = \(\sqrt{5}\) = \({{c}\over {a}}\)
Polynomial can be formed by : \(x^2\) - (sum of zeroes)x + (product of zeroes)
=>\(x^2 - 0x + ( \sqrt{5})\) => \(x^2 + \sqrt{5}\)

(iv) \(1 , 1\)
=>Sum of the zeroes = 1 = \({{-b}\over {a}}\)
=> Product of the zeroes = 1 = \({{c}\over {a}}\)
Polynomial can be formed by : \(x^2\) - (sum of zeroes)x + (product of zeroes)
=>\(x^2 - 1x + 1\)

(v) \({{-1} \over {4}} , {{1} \over {4}}\)
=>Sum of the zeroes = \({{-1} \over {4}}\) = \({{-b}\over {a}}\)
=> Product of the zeroes = \({{1} \over {4}}\) = \({{c}\over {a}}\)
Polynomial can be formed by : \(x^2\) - (sum of zeroes)x + (product of zeroes)
=>\(x^2 - {{-1} \over {4}}x + {{1} \over {4}}\) => \(4x^2 + x + 1\)

(vi) \(4 , 1\)
=>Sum of the zeroes = 4 = \({{-b}\over {a}}\)
=> Product of the zeroes = 1 = \({{c}\over {a}}\)
Polynomial can be formed by : \(x^2\) - (sum of zeroes)x + (product of zeroes)
=>\(x^2 - 4x + 1\)

Solution for Exercise 2.3

Q1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following.
(i) \(p(x) = x^3 - 3x^2 + 5x - 3 , g(x)= x^2 - 2\)
(ii) \(p(x) = x^4 - 3x^2 + 4x + 5 , g(x)= x^2 - x + 1\)
(iii) \(p(x) = x^4 - 5x + 6 , g(x)= 2 - x^2\)

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Answer :

(i) Solving it using long division:
\(\begin{array}{rrrr|ll} x^3 & - 3x^2 & + 5x & - 3 & x^2 - 2\\ -x^3 & & + 2x & & x - 3 \\ \hline & -3x^2 & +7x & -3\\ & \phantom{-}3x^2 & & -6 & & & & \\ \hline & & +7x & -9 \\ \hline \end{array}\)
Here, Quotient = \(x - 3\) and Remainder = \(7x - 9\).

(ii) Solving it using long division:
\(\begin{array}{rrrr|ll} x^4 & - 3x^2 & + 4x & + 5 & x^2 - x + 1\\ x^3 -x^4 & -x^2 & & & x^2 + x - 3 \\ \hline x^3 & - 4x^2 & + 4x & + 5\\ \phantom{-}x^3 & + x^2 & -x & & & \\ \hline & -3x^2 & + 3x & +5 \\ & +3x^2 & -3x & +3 \\ \hline & & & 8 \\ \hline \end{array}\)
Here, Quotient = \(x^2 + x - 3\) and Remainder = \( 8\).

(iii) Solving it using long division:
\(\begin{array}{rrrr|ll} &x^4 & - 5x &+ 6 &-x^2 + 2\\ &2x^2 - x^4 & & & -x^2 - 2 \\ \hline & 2x^2 & - 5x & + 6\\ & \phantom{-}-2x^2 & & + 4 & & & & \\ \hline & & -5x &+10 \\ \hline \end{array}\)
Here, Quotient = \(-x^2 - 2\) and Remainder = \(-5x + 10\).

Q2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial.
(i) \(t^2 - 3 , 2t^4 + 3t^3 -2t^2 - 9t -12\)
(ii) \( x^2 + 3x + 1 , 3x^4 + 5x^3 - 7x^2 + 2x + 2\)
(iii) \(x^2 - 3x + 1 , x^5 - 4x^3 + x^2 + 3x + 1\)

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Answer :

(i) Solving it using long division:
\(\begin{array}{rrrr|ll} 3t^3 + 2t^4 & - 2t^2 & - 9t & -12 & t^2 - 3 \\-2t^4 & +6t^2 & & & 2t^2 + 3t + 4 \\ \hline 3t^3 & + 4t^2 & -9t & -12\\ \phantom{-}-3t^3 & & + 9t & & & \\ \hline & 4t^2 & & -12 \\ & -4t^2 & & +12 \\ \hline & & & 0\\ \hline \end{array}\)
Here, Remainder = 0 so the first polynomial \(t^2 - 3\) is a factor of second polynomial \(2t^4 + 3t^3 -2t^2 - 9t -12\).

(ii) Solving it using long division:
\(\begin{array}{rrrr|ll} 3x^4 + 5x^3 & - 7x^2 & + 2x & +2 & x^2 + 3x + 1 \\-3x^4 - 9x^3 & -3x^2 & & & 3x^2 - 4x + 2 \\ \hline -4x^3 & - 10x^2 & +2x & +2\\ \phantom{-}+4x^3 & +12x^2 & + 4x & & & \\ \hline & 2x^2 & +6x & +2 \\ & -2x^2 & -6x & -2 \\ \hline & & & 0\\ \hline \end{array}\)
Here, Remainder = 0 so the first polynomial \(x^2 + 3x + 1\) is a factor of second polynomial \(3x^4 + 5x^3 - 7x^2 + 2x + 2\).

(iii) Solving it using long division:
\(\begin{array}{rrrr|ll} x^5 - 4x^3 & + x^2 & + 3x & +1 & x^3 - 3x + 1\\ -x^5 + 3x^3 & -x^2 & & & x^2 - 1 \\ \hline & -x^3 & +3x & + 1\\ & \phantom{-} +x^3 & -3x & + 1 & & & & \\ \hline & & & 2 \\ \hline \end{array}\)
Here, Remainder = 2 so the first polynomial \(x^2 - 3x + 1\) is not a factor of second polynomial \(x^5 - 4x^3 + x^2 + 3x + 1\).

Q3.Obtain all other zeroes of \(3x^4 + 6x^3 - 2x^2 -10x - 5\) if two of its zeroes are \(\sqrt{{5}\over{3}}\) and \(-\sqrt{{5}\over{3}}\).

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Answer :

Two zeroes are \(\sqrt{{5}\over{3}}\) and \(-\sqrt{{5}\over{3}}\).
\(=> (x -\sqrt{{5}\over{3}}) ( x + \sqrt{{5}\over{3}}) = x ^2 - {{5}\over {3}}\)
=> \({{1}\over {3}} × (3x^2 - 5)\)
Both \({{1}\over {3}}\) and \((3x^2 - 5)\) are the factors of \(3x^4 + 6x^3 - 2x^2 -10x - 5\).
So, we'll divide the given polynomial by \(3x^2 - 5\)
Solving it using long division:
\(\begin{array}{rrrr|ll} 6x^3 + 3x^4 & -2x^2 & -10x & -5 & 3x^2 - 5 \\ -3x^4 & +5x^2 & & & x^2 + 2x + 1 \\ \hline 6x^3 & +3x^2 & -10x & -5\\ \phantom{-}-6x^3 & & + 10x & & & \\ \hline & 3x^2 & & -5 \\ & -3x^2 & & +5 \\ \hline & & & 0\\ \hline \end{array}\)
\(3x^4 + 6x^3 - 2x^2 -10x - 5\)
=> \((3x^2 - 5) (x^2 + 2x + 1)\)
=> \((3x^2 - 5) (x(x + 1) + 1(x + 1))\)
=> \((3x^2 - 5) (x + 1)(x + 1)\)
So the other two zeroes are x = -1 and -1.

Q4.On dividing \(x^3 - 3x^2 + x + 2\) by a polynomial g(x), the quotient and remainder were \(x-2\) and \(-2x+4\) respectively. Find g(x).

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Answer :

Here, p(x) = \(x^3 - 3x^2 + x + 2\), q(x) = \(x-2\), r(x) = \(-2x+4\)
=>According to polynomial division algorithm,
=>p(x) = g(x).q(x) + r(x)
=>\(x^3 - 3x^2 + x + 2\) = g(x).\((x-2)\) + \((-2x+4)\)
=> \(x^3 - 3x^2 + x + 2 + 2x - 4\) = g(x).\((x-2)\)
=> \(x^3 - 3x^2 + 3x - 2\) = g(x).\((x-2)\)
=> \(g(x) = {{x^3 - 3x^2 + 3x - 2}\over{(x-2)}}\)
Solving it using long division:
\(\begin{array}{rrrr|ll} x^3 & -3x^2 & +3x & -2 & x - 2 \\ -x^3 & +2x^2 & & & x^2 - x + 1 \\ \hline & -x^2 & +3x & -2\\ \phantom{-} & +x^2 & -2x & & & \\ \hline & & x & -2 \\ & & -x & + 2 \\ \hline & & & 0\\ \hline \end{array}\)
=>g(x) = \(x^2 - x + 1\)

Q5. Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
(i) deg p(x) = deg q(x)
(ii) deg q(x) = deg r(x)
(iii) deg r(x) = 0

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Answer :

(i) p(x) = \(2x^2 + 2x + 8\)
=>q(x) = \(x^2 + x + 4\)
=> g(x) = \(2\) and r(x) = \(0\)

(ii) p(x) = \(x^3 + x^2 + x + 1\)
=>q(x) = \(x + 1\)
=> g(x) = \(x^2 - 1\) and r(x) = \(2x + 2\)

(iii) p(x) = \(x^3 - x^2 + 2x + 3\)
=>q(x) = \(x^2 + 2\)
=> g(x) = \(x - 1\) and r(x) = \(5\)

Solution for Exercise 2.4

Q1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:
(i) \(2x^3 + x^2 - 5x + 2 ; {{1}\over{2}},1,-2\)
(ii) \(x^3 - 4x^2 + 5x - 2 ; 2,1,1\)

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Answer :

(i)The general form of cubic equation is \(ax^3 + bx^2 + cx + d\).
Here, \(a = 2,b = 1,c = -5,d = 2\)
On substituting the value of zeroes in the given equation, p(x)= \(2x^3 + x^2 - 5x + 2\)
p\(({{1}\over{2}})\)= \(2({{1}\over{2}})^3 + ({{1}\over{2}})^2 + -5({{1}\over{2}}) + 2\)
=>\({{1}\over{4}} + {{1}\over{4}} - {{5}\over{2}} + 2 = 0 \)
p\((1)\)= \(2(1)^3 + (1)^2 + -5(1) + 2\)
=>\(2 + 1 - 5 + 2 = 0 \)
p\((-2)\)= \(2(-2)^3 + (-2)^2 + -5(-2) + 2\)
=>\(-16 + 4 + 10 + 2 = 0 \)
Hence, these zeroes satisfies the the given equation.
Now, we'll check whether the zeroes satisfies the following equations:
Let the three zeroes be p,q and r
\(p + q + r\) = \({{-b}\over{a}}\)............(i)
\(pq + qr + rp\) = \({{c}\over{a}}\)..........(ii)
\(pqr\) = \({{-d}\over{a}}\).................(iii)
On checking equation (i):=> \({{1}\over{2}} + 1 + (-2)\)
=>\({{3}\over{2}} - 2 \)
=> \({{-1}\over{2}} = {{-b}\over{a}}\)
On checking equation (ii):=> \({{1}\over{2}} × 1 + 1 × (-2) + (-2) × {{1}\over{2}}\)
=>\({{1}\over{2}} - 2 - 1\)
=> \({{-5}\over{2}} = {{c}\over{a}}\)
On checking equation (iii):=> \({{1}\over{2}} × 1 × (-2)\)
=>\({{-2}\over{2}}\)
=> \( -1 = {{-d}\over{a}}\)
Hence it is verified the numbers given alongside of the cubic polynomials are their zeroes.

(ii) The general form of cubic equation is \(ax^3 + bx^2 + cx + d\).
Here, \(a = 1,b = -4,c = 5,d = -2\)
On substituting the value of zeroes in the given equation, p(x)= \(x^3 - 4x^2 + 5x - 2 \)
p\((2)\)= \(2^3 - 4(2)^2 + 5(2) - 2 \)
=>\(8 - 16 + 10 - 2 = 0 \)
p\((1)\)= \(1^3 - 4(1)^2 + 5(1) - 2\)
=>\(1 - 4 + 5 - 2 = 0\)
Hence, these zeroes satisfies the the given equation.
Now, we'll check whether the zeroes satisfies the following equations:
Let the three zeroes be p,q and r
\(p + q + r\) = \({{-b}\over{a}}\)............(i)
\(pq + qr + rp\) = \({{c}\over{a}}\)..........(ii)
\(pqr\) = \({{-d}\over{a}}\).................(iii)
On checking equation (i):=> \(2 + 1 + 1\)
=>\(4\)
=> \({{-(-4)}\over{1}} = {{-b}\over{a}}\)
On checking equation (ii):=> \(2 × 1 + 1 × 1 + 1 × 2\)
=>\(2 + 1 + 2 = 5\)
=> \({{5}\over{1}} = {{c}\over{a}}\)
On checking equation (iii):=> \(2 × 1 × 1\)
=>\(2\)
=> \( {{-(-2)}\over{1}}= {{-d}\over{a}}\)
Hence it is verified the numbers given alongside of the cubic polynomials are their zeroes.

Q2. Find a cubic polynomial with the sum, the sum of the product of its zeroes taken two at a time and the product of its zeroes are \(2, -7,-14\) respectively.

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Answer :

The general form of cubic equation is \(ax^3 + bx^2 + cx + d\)
Let the three zeroes be p,q and r
Here:
\(p + q + r\) = \({{-b}\over{a}} = 2\) ............(i)
\(pq + qr + rp\) = \({{c}\over{a}} = -7\)..........(ii)
\(pqr\) = \({{-d}\over{a}} = -14\).................(iii)
On equating (i), (ii) and (iii) => \(a = 1,b = -2,c = -7,d = 14\)
Putting the values of a,b,c and d in general form
=> \(x^3 - 2x^2 - 7x + 14\)

Q3.If the zeroes of the polynomial \(x^3 - 3x^2 + x + 1\) are \(a-b , a , a+b\) find \(a\) and \(b\).

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Answer :

Let the three zeroes be p,q and r
We know:
\(p + q + r\) = \({{-b}\over{a}} = \) ............(i)
\(pq + qr + rp\) = \({{c}\over{a}} \)..........(ii)
\(pqr\) = \({{-d}\over{a}} \).................(iii)
On checking equation (i):=> \(a - b + a + a + b = 3a\)
=>\(3a = {{-b}\over{a}} = {{-(-3)}\over{1}}\)
=> \(3a = 3\)
=> \( a = 1\)
On checking equation (ii):=> \((a - b) × a + a × (a + b) + (a + b) × (a - b)\)
=>\(a^2 - ab + a^2 + ab + a^2 - b^2 = {{c}\over{a}} = {{1}\over{1}} = 1\)
=> \(3a^2 - b^2 = 1 \)
=> \(3(1)^2 - b^2 = 1 (a = 1)\)
=>\(3 - b^2 = 1\)
=> \(b^2 = 4\)
=> \(b = +2, -2\)
Therefore, \(a = 1, b = +2, -2\)

Q4. If the two zeroes of the polynomial \(x^4- 6x^3 - 26x^2 -138x - 35\) are \(2 + \sqrt{3}\) and \(2 - \sqrt{3}\) find other zeroes.

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Answer :

Two zeroes are\(2 + \sqrt{3}\) and \(2 - \sqrt{3}\).
\(=> (x -(2 + \sqrt{3})) (x - (2 - \sqrt{3})) = (x - 2 - \sqrt{3}) (x - 2 + \sqrt{3})\)
\(=> (x - 2)^2 - ( \sqrt{3})^2 = x^2 - 4x + 1\)
\(x^2 - 4x + 1\) the factors of \(x^4- 6x^3 - 26x^2 -138x - 35\).
So, we'll divide the given polynomial by \(x^2 - 4x + 1\) Solving it using long division:
\(\begin{array}{rrrr|ll} x^4 -6x^3 & -26x^2 & +138x & -35 & x^2 -4x +1 \\ -x^4 + 4x^3 & -x^2 & & & x^2 -2x - 35 \\ \hline -2x^3 & -27x^2 & +138x & -35\\ \phantom{-}+2x^3 & -8x^2 & + 2x & & & \\ \hline & -35x^2 & +140x & -35 \\ & +35x^2 & -140x & +35 \\ \hline & & & 0\\ \hline \end{array}\)
\(x^4- 6x^3 - 26x^2 -138x - 35\)
=> \(( x^2 -4x +1) (x^2 -2x - 35)\)
=> \(( x^2 -4x +1) (x^2 -7x + 5x - 35)\)
=> \((x^2 -4x +1) (x(x - 7) + 5(x - 7))\)
=> \((x^2 -4x +1) (x + 5)(x - 7)\)

Thus, \(x + 5 = 0\) and \(x - 7 = 0\) are the solutions.
So, the other two zeroes are \(x = -5 , 7\).

Q5.If the polynomial \(x^4- 6x^3 + 16x^2 -25x + 10\) is divided by another polynomial \(x^2 -2x + k \) the remainder comes out to be \(x +a\) find \(k\) and \(a\).

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Answer :

Solving it using long division:
\(\begin{array}{rrrr|ll} x^4 -6x^3 & +16x^2 & -25x & +10 & x^2 -2x + k\\ -x^4 + 2x^3 & -kx^2 & & & x^2 -4x + (8-k) \\ \hline -4x^3 & +(16-k)x^2 & -25x & +10 \\ \phantom{-}+4x^3 & -8x^2 & + 4kx & & & \\ \hline & (8-k)x^2 & +(4k-25)x & +10 \\ & -(8-k)x^2 & +2(8-k)x & -(8-k)k \\ \hline & (2k-9)x & -(8-k)k & +10\\ \hline \end{array}\)
Here remainder = \((2k-9)x -(8-k)k +10\)...........(i)
Given remainder = \(x + a\)..........................(ii)
On comparing the coefficients x in equation (i) and (ii):
=>\(2k - 9 = 1\)
=>\(2k = 10\)
=>\(k = 5\)
On comparing the a in equation (i) and (ii):
=>\(-(8-k)k + 10 = a\)
=>\(-(8-5)5 + 10 = a\)
=>\(a = -15 + 10 = -5\)