Application of Determinants MCQ Quiz - Objective Question with Answer for Application of Determinants - Download Free PDF
Last updated on Apr 7, 2025
Latest Application of Determinants MCQ Objective Questions
Application of Determinants Question 1:
If the vertices of a triangle are (1, 2), (2, 5) and (4, 3) then the area of the triangle is :
Answer (Detailed Solution Below)
Application of Determinants Question 1 Detailed Solution
Concept
The area of a triangle given its three vertices is
\(A = \frac{1}{2} | x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2 ) | \)
Explanation:
\(A = \frac{1}{2} [|1(5-3) + 2(3-2) + 4(2-5)|] \)
\(A = \frac{1}{2} |1× 2 + 2 × 1 + 4 × (-3) | \)
\(A = \frac{1}{2} | 2 + 2 - 12| \)
\(A = \frac{1}{2} |-8| \)
\(A = \frac{1}{2} × 8 = 4 \)
Hence Option(2) is the correct answer.
Application of Determinants Question 2:
If the vertices of a triangle are (1, 2), (2, 5) and (4, 3) then the area of the triangle is :
Answer (Detailed Solution Below)
Application of Determinants Question 2 Detailed Solution
Concept
The area of a triangle given its three vertices is
\(A = \frac{1}{2} | x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2 ) | \)
Explanation:
\(A = \frac{1}{2} [|1(5-3) + 2(3-2) + 4(2-5)|] \)
\(A = \frac{1}{2} |1× 2 + 2 × 1 + 4 × (-3) | \)
\(A = \frac{1}{2} | 2 + 2 - 12| \)
\(A = \frac{1}{2} |-8| \)
\(A = \frac{1}{2} × 8 = 4 \)
Hence Option(2) is the correct answer.
Application of Determinants Question 3:
Consider a system equation
\(2x+y-z=0\),
\(4x - py + 4z = 4\) and
\(x-y+z=q\)
where \(p, q \in I\) and \(p, q \in [1, 10]\), then identify the correct statement(s).
List-I | List-II | |
---|---|---|
(I) | Number of ordered pairs \((p, q)\) for which system of equation has unique solution is | (P) 1 |
(II) | Number of ordered pairs \((p, q)\) for which system of equation has no solution is | (Q) 9 |
(III) | Number of ordered pairs \((p, q)\) for which system of equation has infinite solution is | (R) 91 |
(IV) | Number of ordered pairs \((p, q)\) for which system of equation has atleast one solution is | (S) 90 |
Answer (Detailed Solution Below)
I → S, II → Q, III → P, IV → R
Application of Determinants Question 3 Detailed Solution
Calculation:
Given:
System of linear equations:
\(2x + y - z = 0\)
\(4x - py + 4z = 4\)
\(x - y + z = q\)
\(p, q \in I\) and \(p, q \in [1, 10]\)
The coefficient matrix A is:
\(A = \begin{bmatrix} 2 & 1 & -1 \\ 1 & -\frac{p}{4} & 1 \\ 1 & -1 & 1 \end{bmatrix}\)
The augmented matrix [A|B] is:
\([A|B] = \begin{bmatrix} 2 & 1 & -1 & 0 \\ 1 & -\frac{p}{4} & 1 & 1 \\ 1 & -1 & 1 & q \end{bmatrix}\)
Calculate the determinant of A, |A|:
\(|A| = 2(-\frac{p}{4} + 1) - 1(1 - 1) - 1(-1 + \frac{p}{4})\)
\(|A| = -\frac{p}{2} + 2 + 0 + 1 - \frac{p}{4}\)
\(|A| = 3 - \frac{3p}{4}\)
⇒ For unique solution, \(|A| \ne 0\):
\(3 - \frac{3p}{4} \ne 0\)
\(3 \ne \frac{3p}{4}\)
\(p \ne 4\)
For no solution or infinite solutions, \(|A| = 0\), so \(p = 4\).
⇒ If \(p=4\), the system becomes:
\(2x + y - z = 0\)
\(x - y + z = 1\)
\(x - y + z = q\)
⇒ From the second and third equations, for a solution to exist, \(1 = q\).
⇒ If \(p = 4\) and \(q = 1\), infinite solutions exist.
⇒ If \(p = 4\) and \(q \ne 1\), no solution exists.
(I) Unique solution: \(p \ne 4\). \(p\) can take 9 values (1 to 10 except 4). \(q\) can take 10 values. Total pairs: 9 × 10 = 90.
(II) No solution: \(p = 4\) and \(q \ne 1\). 9 values for \(q\). 1 pair for \(p\). Total pairs: 1 × 9 = 9.
(III) Infinite solutions: \(p = 4\) and \(q = 1\). 1 pair only.
(IV) At least one solution: Total pairs - No solution pairs = 100 - 9 = 91.
∴ (I) - (S), (II) - (Q), (III) - (P), (IV) - (R)
Application of Determinants Question 4:
If the points (5, 2, 4), (6, -1, 2) and (8, -7, k) are collinear, then k =
Answer (Detailed Solution Below)
Application of Determinants Question 4 Detailed Solution
Given:
Points are (5, 2, 4), (6, -1, 2) & (8, -7, k)
Concept used:
Area of the triangle formed by using these points = 0
\(1\over2\) ×= 0
Calculation:
⇒ 5(- k + 14) - 2(6k - 16) + 4(- 42 + 8) = 0
⇒ - 5k + 70 - 12k + 32 + 4(- 34) = 0
⇒ -17k + 102 - 136 = 0
⇒ -17k - 34 = 0
⇒ -17k = 34
⇒ k = -2
Hence the correct answer is "-2".
Application of Determinants Question 5:
Let α, β (α ≠ β) be the values of m, for which the equations x + y + z = 1; x + 2y + 4z = m and x + 4y + 10z = m2 have infinitely many solutions. Then the value of is \(\rm \displaystyle \sum_{n=1}^{10}\left(n^{\alpha}+n^{\beta}\right)\) equal to :
Answer (Detailed Solution Below)
Application of Determinants Question 5 Detailed Solution
\(\Delta =\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 4 & 10 \end{array}\right|=1(20-16)-1(10-4)+1(4-2) \)
= 4 – 6 + 2 = 0
For infinite solutions
Δx = Δy = Δz = 0
m2 – 3x + 2 = 0
m = 1, 2
α = 1, β = 2
\(\rm \displaystyle \therefore \sum_{n=1}^{10}\left(n^{\alpha}+n^{\beta}\right)=\sum_{n=1}^{10} n^{1}+\sum_{n=1}^{10} n^{2} \)
\(=\frac{10(11)}{2}+\frac{10(11)(21)}{6} \)
= 55 + 385
= 440
Top Application of Determinants MCQ Objective Questions
The system of equations kx + y + z = 1, x + ky + z = k and x + y + kz = k2 has no solution if k equals
Answer (Detailed Solution Below)
Application of Determinants Question 6 Detailed Solution
Download Solution PDFConcept
Let the system of equations be,
a1x + b1y + c1z = d1
a2x + b2y + c2z = d2
a3x + b3y + c3z = d3
\(\; \Rightarrow \left[ {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}}\\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right]\;\left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right] = \;\left[ {\begin{array}{*{20}{c}} {{d_1}}\\ {{d_2}}\\ {{d_3}} \end{array}} \right]\)
⇒ AX = B
⇒ X = A-1 B = \(\frac{{{\rm{adj\;}}\left( {\rm{A}} \right)}}{{\det {\rm{\;}}({\rm{A}})}}\;B\)
⇒ If det (A) ≠ 0, the system is consistent having unique solution.
⇒ If det (A) = 0 and (adj A). B = 0, system is consistent, with infinitely many solutions.
⇒ If det (A) = 0 and (adj A). B ≠ 0, system is inconsistent (no solution)
Calculation:
Given:
kx + y + z = 1, x + ky + z = k and x + y + kz = k2
\( \Rightarrow {\rm{A}} = {\rm{}}\left[ {\begin{array}{*{20}{c}} {\rm{k}}&1&1\\ 1&{\rm{k}}&1\\ 1&1&{\rm{k}} \end{array}} \right],{\rm{B}} = {\rm{}}\left[ {\begin{array}{*{20}{c}} {\rm{x}}\\ {\rm{y}}\\ {\rm{z}} \end{array}} \right]{\rm{and\;C}} = \left[ {\begin{array}{*{20}{c}} 1\\ {\rm{k}}\\ {{{\rm{k}}^2}} \end{array}} \right]\)
⇒ For the given equations to have no solution, |A| = 0
\(\Rightarrow \left| {\begin{array}{*{20}{c}} {\rm{k}}&1&1\\ 1&{\rm{k}}&1\\ 1&1&{\rm{k}} \end{array}} \right| = 0\)
⇒ k(k2 – 1) - 1(k – 1) + 1(1 – k) = 0
⇒ k3 – k – k + 1 + 1 – k = 0
⇒ k3 -3k +2 = 0
⇒ (k – 1) (k – 1) (k + 2) = 0
⇒ k = 1, -2
If we put k = 1 in the above given equations, then all the equations will become the same.
Hence, the given equations have no solution if k = - 2.
If the area of a triangle with vertices (-3, 0), (3, 0) and (0, k) is 9 square units, then what is the value of k?
Answer (Detailed Solution Below)
Application of Determinants Question 7 Detailed Solution
Download Solution PDFConcept:
Area of a triangle with vertices (x1, y1) , (x2, y2), (x3, y3) is given by
Area = \(\rm \dfrac 12 \rm \begin{vmatrix} \rm x_1&\rm y_1 &1 \\ x_2& \rm y_2&1 \\ \rm\rm x_3 &\rm y_3&1 \end{vmatrix}\)
Calculations:
Given that, vertices of triangle are (-3, 0), (3, 0) and (0, k)
By using the above formula,
⇒ Area = \(\rm \dfrac 12 \rm \begin{vmatrix} \rm -3&\rm 0 &1 \\ 3& 0&1 \\ 0 &k&1 \end{vmatrix}\)
⇒ Area = \(\dfrac 12\)[-3(0 - k) - 0 + 1(3k)]
⇒ Area = 3k
According to the question, area of triangle is 9 square unit,
⇒ 3k = 9
⇒ k = 3
∴ Required value of k is 3 unit.
An equilateral triangle has each side equal to a. If the co-ordinates of its vertices are (x1, y1); (x2, y2): (x3, y3) then the square of the determinant \(\begin{vmatrix} x_1 & y_1 & 1 \\\ x_2 & y_2& 1 \\\ x_3 & y_3 & 1 \end{vmatrix}\) equals:
Answer (Detailed Solution Below)
Application of Determinants Question 8 Detailed Solution
Download Solution PDFConcept:
Area of equilateral triangle = \(\frac{\sqrt{3}}{4}\)×a2
Calculation:
Given: The co-ordinates of its vertices are (x1, y1); (x2, y2): (x3, y3) then the square of the determinant \(\begin{vmatrix} x_1 & y_1 & 1 \\\ x_2 & y_2& 1 \\\ x_3 & y_3 & 1 \end{vmatrix}\)
⇒ (△ ABC ) = \(\dfrac{1}{2}\) \(\begin{vmatrix} x_1 & y_1 & 1 \\\ x_2 & y_2& 1 \\\ x_3 & y_3 & 1 \end{vmatrix}\) = ( \(\frac{\sqrt{3}}{4}\)) ×a2
On squaring both side,
⇒ \(\dfrac{1}{4}\) \(\begin{vmatrix} x_1 & y_1 & 1 \\\ x_2 & y_2& 1 \\\ x_3 & y_3 & 1 \end{vmatrix}^2\) = \(\dfrac{3a^4}{16}\)
⇒ \(\begin{vmatrix} x_1 & y_1 & 1 \\\ x_2 & y_2& 1 \\\ x_3 & y_3 & 1 \end{vmatrix}^2\) = \(\dfrac{3a^4}{4}\)
If A is a 2 × 2 matrix and |A| = 5, what is |5A| ? (| | denotes determinant)
Answer (Detailed Solution Below)
Application of Determinants Question 9 Detailed Solution
Download Solution PDFConcept:
Properties of determinants:
For a n×n matrix A, det(kA) = kn det(A).
Calculation:
Given:
|A| = 5
k = 5
From the properties of the determinants, we know that |KA| = Kn |A|, where n is the order of the determinant.
Here, n = 2, therefore, the answer is K2 |A|.
|5A| = 52|A|
|5A| = 52 × 5 = 125
If \(\rm (x_1, y_1), (x_2, y_2)\) and \(\rm (x_3, y_3)\) are the vertices of a triangle whose area is ‘k’ square units, then \(\begin{vmatrix} \rm x_1 & \rm y_1 & 4 \\\ \rm x_2 & \rm y_2 & 4 \\\ \rm x_3 & \rm y_3 & 4 \end{vmatrix}^2\) is
Answer (Detailed Solution Below)
Application of Determinants Question 10 Detailed Solution
Download Solution PDFConcept:
If \(\rm (x_1, y_1), (x_2, y_2)\) and \(\rm (x_3, y_3)\) are the vertices of a triangle then,
Area = \(\rm \dfrac 12 \begin{vmatrix} \rm x_1 & \rm y_1 & 1 \\\ \rm x_2 & \rm y_2 & 1 \\\ \rm x_3 & \rm y_3 & 1 \end{vmatrix}\)
Properties of Determinants
\(\rm \begin{vmatrix} \rm x_1 & \rm y_1 & kx \\\ \rm x_2 & \rm y_2 & ky \\\ \rm x_3 & \rm y_3 & kz \end{vmatrix} = k \begin{vmatrix} \rm x_1 & \rm y_1 & x \\\ \rm x_2 & \rm y_2 & y \\\ \rm x_3 & \rm y_3 & z \end{vmatrix}\)
Calculations:
If \(\rm (x_1, y_1), (x_2, y_2)\) and \(\rm (x_3, y_3)\) are the vertices of a triangle then,
Area = \(\rm \dfrac 12 \begin{vmatrix} \rm x_1 & \rm y_1 & 1 \\\ \rm x_2 & \rm y_2 & 1 \\\ \rm x_3 & \rm y_3 & 1 \end{vmatrix}\)
Given: the area is ‘k’ square units
⇒ k = \(\rm \dfrac 12 \begin{vmatrix} \rm x_1 & \rm y_1 & 1 \\\ \rm x_2 & \rm y_2 & 1 \\\ \rm x_3 & \rm y_3 & 1 \end{vmatrix}\)
⇒ \(\rm \begin{vmatrix} \rm x_1 & \rm y_1 & 1 \\\ \rm x_2 & \rm y_2 & 1 \\\ \rm x_3 & \rm y_3 & 1 \end{vmatrix} = 2k\)
Now, \(\begin{vmatrix} \rm x_1 & \rm y_1 & 4 \\\ \rm x_2 & \rm y_2 & 4 \\\ \rm x_3 & \rm y_3 & 4 \end{vmatrix}^2\)
= \(\begin{vmatrix} \rm x_1 & \rm y_1 & 4 \\\ \rm x_2 & \rm y_2 & 4 \\\ \rm x_3 & \rm y_3 & 4 \end{vmatrix}\begin{vmatrix} \rm x_1 & \rm y_1 & 4 \\\ \rm x_2 & \rm y_2 & 4 \\\ \rm x_3 & \rm y_3 & 4 \end{vmatrix}\)
= \(4\begin{vmatrix} \rm x_1 & \rm y_1 & 1 \\\ \rm x_2 & \rm y_2 & 1 \\\ \rm x_3 & \rm y_3 & 1 \end{vmatrix}.4\begin{vmatrix} \rm x_1 & \rm y_1 & 1 \\\ \rm x_2 & \rm y_2 & 1 \\\ \rm x_3 & \rm y_3 & 1 \end{vmatrix}\)
= 16.(2k)(2k)
= 64k2
Find the area of triangle with vertices at points A (1, 1) ,B ( 6, 0) and C ( 3, 2).
Answer (Detailed Solution Below)
Application of Determinants Question 11 Detailed Solution
Download Solution PDFConcept:
If \(\rm (x_1, y_1), (x_2, y_2)\) and \(\rm (x_3, y_3)\) are the vertices of a triangle then,
Area = \(\rm \dfrac 12 \begin{vmatrix} \rm x_1 & \rm y_1 & 1 \\\ \rm x_2 & \rm y_2 & 1 \\\ \rm x_3 & \rm y_3 & 1 \end{vmatrix}\)
Note: Area is always a positive quantity, therefore we always take the absolute value of the determinant for the area.
Calculation:
Given vertices are A (1, 1) ,B ( 6, 0) and C ( 3, 2).
We know that area of triangle ABC is given by,
Δ = \(\rm \dfrac 12 \begin{vmatrix} \rm x_1 & \rm y_1 & 1 \\\ \rm x_2 & \rm y_2 & 1 \\\ \rm x_3 & \rm y_3 & 1 \end{vmatrix}\)
⇒ Δ = \(\rm\frac{1}{2}\begin{vmatrix} 1& 1 & 1\\ 6& 0 & 1\\ 3& 2 & 1 \end{vmatrix}\)
⇒ Δ = \(\frac{1}{2}\left [ 1\left ( 0-2 \right )-1 (6-3)+1 (12-0)\right ]\)
⇒ Δ = \(\frac{7}{2}\)
The correct option is 1.
The area of triangle with vertices (K, 0), (4, 0), (0, 2) is 4 square units, then value of K is
Answer (Detailed Solution Below)
Application of Determinants Question 12 Detailed Solution
Download Solution PDFConcept:
The area of a triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3), is given by the expression
\(\rm Area = \frac{1}{2}\begin{vmatrix} x_{1} &x_{2} & 1\\ y_{1}& y_{2} &1 \\ z_{1}&z_{2} & 1 \end{vmatrix}\)
Calculation:
Given, Area of triangle, A = 4 sq. unit and vertices (K, 0), (4, 0), (0, 2)
since the area is always Positive But the determinant can be both positive and negative.
∴ Δ = ± 4 .
⇒ ± 4 = \(\rm = \frac{1}{2}\begin{vmatrix} k &0 & 1\\ 4& 0 &1 \\ 0&2 & 1 \end{vmatrix}\)
⇒ ± 4 \(\rm = \frac{1}{2} \left [ k(0-2) - 0 (4-0)+1(8-0)\right ]\)
⇒ ± 8 = -2k + 8
So, 8 = -2k + 8 or -8 = -2k +8
⇒ k = 0 or 8 .
The correct option is 2.
The points (5, -2), (8, -3) and (a, -12) are collinear if the value of a is
Answer (Detailed Solution Below)
Application of Determinants Question 13 Detailed Solution
Download Solution PDFConcept:
If the points A (x1, y1), B (x2, y2) and C (x3, y3) are collinear then area of ΔABC = 0.
Let A (x1, y1), B (x2, y2) and C (x3, y3) be the vertices of a Δ ABC, then area (A) of ΔABC is given as;
\(A=\frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right|\)
Calculation:
Here, we have to find the value of a for which the points (5, -2), (8, -3) and (a, -12) are collinear
Let,
x1 = 5, y1 = -2,
x2 = 8, y2 = -3,
x3 = a, y3 = -12.
As we know that, if A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of a Δ ABC then area of Δ ABC = \(A= \frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right|\)
\(⇒ A = \frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{5}}&{{-2}}&1\\ {{8}}&{{-3}}&1\\ {{a}}&{{-12}}&1 \end{array}} \right|\)
2A = 5 (-3 + 12) + 2(8 - a) + 1(-96 + 3a)
2A = 45 + 16 - 2a - 96 + 3a
2A = a - 35
⇒ A = (a - 35)/2
∵ The given points are collinear.
As we know that, if the points A (x1, y1), B (x2, y2) and C (x3, y3) are collinear then area of ΔABC = 0.
⇒ (a - 35)/2 = 0
⇒ a = 35
Hence, option D is the correct answer.
Alternate Method
Concept:
Three or more points are collinear if the slope of any two pairs of points is the same.
The slope of a line passing through the distinct points (x1, y1) and (x2, y2) is \(\rm \frac{y_2 -y_1 }{x_2-x_1}\)
Calculation:
Let, A = (5, -2), B = (8, -3), C = (a, -12)
Now, the slope of AB = Slope of BC = Slope of AC (∵ points are collinear)
\(\rm \frac{-3-(-2)}{8-5}=\frac{-12-(-3)}{a-8}\\ ⇒ \frac{-1}{3}=\frac{-9}{a-8}\)
⇒ a - 8= 27
⇒ a = 27 + 8 = 35
Hence, option (4) is correct.
Consider the determinant
Δ = \(\left|\begin{array}{lll}\text{a}_{11} & \text{a}_{12} & \text{a}_{13} \\ \text{a}_{21} & \text{a}_{22} & \text{a}_{23} \\ \text{a}_{31} & \text{a}_{32} & \text{a}_{33}\end{array}\right|\)
If a13 = yz, a23 = zx, a33 = xy and the minors of a13, a23, a33 are respectively (z − y), (z − x), (y − x) then what is the value of Δ ?
Answer (Detailed Solution Below)
Application of Determinants Question 14 Detailed Solution
Download Solution PDFExplanation:
Given:
Δ = \(\left|\begin{array}{lll}\text{a}_{11} & \text{a}_{12} & \text{a}_{13} \\ \text{a}_{21} & \text{a}_{22} & \text{a}_{23} \\ \text{a}_{31} & \text{a}_{32} & \text{a}_{33}\end{array}\right|\)
a13 = yz, a23 = zx, a33 = xy and M13 = (z − y), M23 = (z − x), M33 = (y − x)
Expanding the determinant along Column 3,
⇒ Δ = a13 (a21a32 - a22a31) - a23 (a11a32 - a12a31) + a33 (a11a2 - a12a1)
⇒ Δ = a13M13 - a23M23 + a33M33
where Mij denotes the minor of aij element.
⇒ Δ = yz(z - y) - zx(z - x) + xy(y - x)
⇒ Δ = yz(z - y) - z2x + zx2 + xy2 - x2y
⇒ Δ = yz(z - y) + (xy2 - z2x) + (zx2 - x2y)
⇒ Δ = yz(z - y) + x(y2 - z2) + x2(z - y)
⇒ Δ = (z - y)[yz - x(y + z) + x2]
⇒ Δ = (z - y)[yz - xy - xz + x2]
⇒ Δ = (z - y)[y(z - x) - x(z - x)]
⇒ Δ = (z - y)[(z - x)(y - x)]
For what values of k is the system of equations 2k2x + 3y - 1 = 0, 7x - 2y + 3 = 0, 6kx + y + 1 = 0 consistent?
Answer (Detailed Solution Below)
Application of Determinants Question 15 Detailed Solution
Download Solution PDFConcept:
Consider three linear eqaution in two variable:
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
a3x + b3y + c3 = 0
Condition for the consistency of three simultaneous linear equations in 2 variables:
\( \left| {\begin{array}{*{20}{c}} a_1&b_1&c_1\\ a_2&b_2&c_2\\ a_3&b_3&c_3 \end{array}} \right|=0\)
Formula for Quadratic equation:
ax2 + bx + c = 0
x = \(\rm \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\)
Calculation:
2k2x + 3y - 1 = 0 ....(1)
7x - 2y + 3 = 0 ....(2)
6kx + y + 1 = 0 ....(3)
For consistency of given simultaneous equation,
\( \left| {\begin{array}{*{20}{c}} 2k^2&3&-1\\ 7&-2&3\\ 6k&1&1 \end{array}} \right|=0\)
⇒ 2k2(-2- 3) - 3(7 - 18k) - 1(7 + 12k) = 0
⇒ -10k2 - 21 + 54k - 7 - 12k = 0
⇒ -10k2 + 42k - 28 = 0
⇒ 5k2 - 21k + 14 = 0
By using the formula,
\(x=\rm \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\)
\(\Rightarrow k=\rm \frac{-(-21) \pm \sqrt{(-21)^{2} - 4(5)(14)}}{2\times 5}\)
\(\therefore k=\rm \frac{21 \pm \sqrt{161}}{10}\)