EXAMPLES
Quadratic Equations • 8 Questions
Question 1
Hint available
Represent the following situations mathematically: (i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with. (ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ` 750. We would like to find out the number of toys produced on that day.
Key Idea
Translate the verbal information into algebraic equations. Use the sum and product of two numbers for part (i) and the relation ‘cost per unit × number of units = total cost’ for part (ii). Both lead to quadratic equations of the form $ax^2+bx+c=0$, which are solved using factorisation or the quadratic formula.
Step-by-Step Solution
### Part (i)
1. Let $x$ = number of marbles John originally had, $y$ = number of marbles Jivanti originally had.
2. From the statement "together have 45 marbles":
$$x + y = 45 \tag{1}$$
3. After each loses 5 marbles, the numbers become $x-5$ and $y-5$. Their product is 124:
$$(x-5)(y-5) = 124 \tag{2}$$
4. Using (1), express $y = 45 - x$ and substitute in (2):
$$(x-5)(45 - x - 5) = 124$$
$$ (x-5)(40 - x) = 124$$
5. Expand and bring all terms to one side:
$$ -x^{2} + 45x - 200 = 124$$
$$ -x^{2} + 45x - 324 = 0$$
Multiply by $-1$:
$$ x^{2} - 45x + 324 = 0$$
6. Solve the quadratic. Discriminant $D = 45^{2} - 4\times1\times324 = 2025 - 1296 = 729 = 27^{2}$.
7. Roots:
$$ x = \frac{45 \pm 27}{2}$$
Hence $x = 36$ or $x = 9$.
8. Using (1), the corresponding $y$ values are $9$ or $36$ respectively.
9. Result: John and Jivanti originally had $36$ and $9$ marbles (order can be interchanged).
### Part (ii)
1. Let $n$ = number of toys produced in a day.
2. Cost of production of each toy = $55 - n$ rupees.
3. Total cost = $750$ rupees, so:
$$ n\,(55 - n) = 750 \tag{3}$$
4. Expand and rearrange:
$$ -n^{2} + 55n - 750 = 0$$
Multiply by $-1$:
$$ n^{2} - 55n + 750 = 0$$
5. Solve the quadratic. Discriminant $D = 55^{2} - 4\times1\times750 = 3025 - 3000 = 25 = 5^{2}$.
6. Roots:
$$ n = \frac{55 \pm 5}{2}$$
Hence $n = 30$ or $n = 25$.
7. Both values give a positive cost per toy ( $55-30 = 25$ rupees and $55-25 = 30$ rupees ), so both are admissible.
8. Result: The cottage industry could have produced either $30$ toys (cost per toy = Rs. 25) or $25$ toys (cost per toy = Rs. 30) on that day.
1. Let $x$ = number of marbles John originally had, $y$ = number of marbles Jivanti originally had.
2. From the statement "together have 45 marbles":
$$x + y = 45 \tag{1}$$
3. After each loses 5 marbles, the numbers become $x-5$ and $y-5$. Their product is 124:
$$(x-5)(y-5) = 124 \tag{2}$$
4. Using (1), express $y = 45 - x$ and substitute in (2):
$$(x-5)(45 - x - 5) = 124$$
$$ (x-5)(40 - x) = 124$$
5. Expand and bring all terms to one side:
$$ -x^{2} + 45x - 200 = 124$$
$$ -x^{2} + 45x - 324 = 0$$
Multiply by $-1$:
$$ x^{2} - 45x + 324 = 0$$
6. Solve the quadratic. Discriminant $D = 45^{2} - 4\times1\times324 = 2025 - 1296 = 729 = 27^{2}$.
7. Roots:
$$ x = \frac{45 \pm 27}{2}$$
Hence $x = 36$ or $x = 9$.
8. Using (1), the corresponding $y$ values are $9$ or $36$ respectively.
9. Result: John and Jivanti originally had $36$ and $9$ marbles (order can be interchanged).
### Part (ii)
1. Let $n$ = number of toys produced in a day.
2. Cost of production of each toy = $55 - n$ rupees.
3. Total cost = $750$ rupees, so:
$$ n\,(55 - n) = 750 \tag{3}$$
4. Expand and rearrange:
$$ -n^{2} + 55n - 750 = 0$$
Multiply by $-1$:
$$ n^{2} - 55n + 750 = 0$$
5. Solve the quadratic. Discriminant $D = 55^{2} - 4\times1\times750 = 3025 - 3000 = 25 = 5^{2}$.
6. Roots:
$$ n = \frac{55 \pm 5}{2}$$
Hence $n = 30$ or $n = 25$.
7. Both values give a positive cost per toy ( $55-30 = 25$ rupees and $55-25 = 30$ rupees ), so both are admissible.
8. Result: The cottage industry could have produced either $30$ toys (cost per toy = Rs. 25) or $25$ toys (cost per toy = Rs. 30) on that day.
Question 2
Hint available
Check whether the following are : (i) (x – 2)2 + 1 = 2x – 3 (ii) x(x + 1) + 8 = (x + 2) (x – 2) (iii) x (2x + 3) = x2 + 1 (iv) (x + 2)3 = x3 – 4
Key Idea
To verify whether an equation is an identity, expand both sides, bring all terms to one side and simplify. If the resulting expression reduces to 0 (or a true statement) for all values of \(x\), the equation is an identity. If it holds only for particular values of \(x\) or never, it is not an identity.
Step-by-Step Solution
1. Equation (i)\
\[ (x-2)^2 + 1 = 2x-3 \]
Expand the left side: \( (x-2)^2 = x^2-4x+4 \).\
Hence \( LHS = x^2-4x+4+1 = x^2-4x+5 \).
Bring all terms to one side:\
\[ x^2-4x+5-(2x-3)=0 \Rightarrow x^2-6x+8=0 \]
This quadratic factors as \((x-2)(x-4)=0\), giving \(x=2\) or \(x=4\). Since it is true only for these two values, the given relation is not an identity.
2. Equation (ii)\
\[ x(x+1)+8 = (x+2)(x-2) \]
Expand both sides: \( LHS = x^2 + x + 8 \), \( RHS = x^2 -4 \).
Subtract RHS from LHS:\
\[ x^2 + x + 8 - (x^2 -4)=0 \Rightarrow x + 12 = 0 \]
Hence \( x = -12 \). The equality holds only for \(x=-12\); therefore it is not an identity.
3. Equation (iii)\
\[ x(2x+3) = x^2 + 1 \]
Expand the left side: \( LHS = 2x^2 + 3x \).
Bring all terms to one side:\
\[ 2x^2 + 3x - (x^2 + 1)=0 \Rightarrow x^2 + 3x - 1 = 0 \]
Solving the quadratic gives \( x = \frac{-3 \pm \sqrt{13}}{2} \). The relation is true only for these two values, so it is not an identity.
4. Equation (iv)\
\[ (x+2)^3 = x^3 - 4 \]
Expand the left side using \((a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\):\
\[ (x+2)^3 = x^3 + 6x^2 + 12x + 8 \]
Bring all terms to one side:\
\[ x^3 + 6x^2 + 12x + 8 - (x^3 - 4)=0 \Rightarrow 6x^2 + 12x + 12 = 0 \]
Divide by 6: \( x^2 + 2x + 2 = 0 \).
The discriminant \( D = 2^2 - 4\cdot1\cdot2 = -4 < 0 \); there is no real solution. Hence the given equation is not an identity (it is never true for real \(x\)).
Summary: None of the four relations are identities; each holds only for specific values of \(x\) (or none in the real domain for (iv)).
\[ (x-2)^2 + 1 = 2x-3 \]
Expand the left side: \( (x-2)^2 = x^2-4x+4 \).\
Hence \( LHS = x^2-4x+4+1 = x^2-4x+5 \).
Bring all terms to one side:\
\[ x^2-4x+5-(2x-3)=0 \Rightarrow x^2-6x+8=0 \]
This quadratic factors as \((x-2)(x-4)=0\), giving \(x=2\) or \(x=4\). Since it is true only for these two values, the given relation is not an identity.
2. Equation (ii)\
\[ x(x+1)+8 = (x+2)(x-2) \]
Expand both sides: \( LHS = x^2 + x + 8 \), \( RHS = x^2 -4 \).
Subtract RHS from LHS:\
\[ x^2 + x + 8 - (x^2 -4)=0 \Rightarrow x + 12 = 0 \]
Hence \( x = -12 \). The equality holds only for \(x=-12\); therefore it is not an identity.
3. Equation (iii)\
\[ x(2x+3) = x^2 + 1 \]
Expand the left side: \( LHS = 2x^2 + 3x \).
Bring all terms to one side:\
\[ 2x^2 + 3x - (x^2 + 1)=0 \Rightarrow x^2 + 3x - 1 = 0 \]
Solving the quadratic gives \( x = \frac{-3 \pm \sqrt{13}}{2} \). The relation is true only for these two values, so it is not an identity.
4. Equation (iv)\
\[ (x+2)^3 = x^3 - 4 \]
Expand the left side using \((a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\):\
\[ (x+2)^3 = x^3 + 6x^2 + 12x + 8 \]
Bring all terms to one side:\
\[ x^3 + 6x^2 + 12x + 8 - (x^3 - 4)=0 \Rightarrow 6x^2 + 12x + 12 = 0 \]
Divide by 6: \( x^2 + 2x + 2 = 0 \).
The discriminant \( D = 2^2 - 4\cdot1\cdot2 = -4 < 0 \); there is no real solution. Hence the given equation is not an identity (it is never true for real \(x\)).
Summary: None of the four relations are identities; each holds only for specific values of \(x\) (or none in the real domain for (iv)).
Question 3
Hint available
Find the roots of the equation 2x2 – 5x + 3 = 0, by factorisation.
Key Idea
Factorise the quadratic expression into two linear factors of the form \((ax + b)(cx + d) = 0\). The values of \(x\) that make each factor zero are the roots of the equation.
Step-by-Step Solution
1. Write the quadratic in standard form
$$2x^{2} - 5x + 3 = 0.$$
2. Assume a factorisation
Since the leading coefficient is 2, try factors of the form \((2x + p)(x + q)\) where \(p\) and \(q\) are integers. Expanding gives
$$ (2x + p)(x + q) = 2x^{2} + (2q + p)x + pq.$$
Compare with \(2x^{2} - 5x + 3\).
3. Match the coefficients
- Coefficient of \(x\): \(2q + p = -5\).
- Constant term: \(pq = 3\).
4. Find integer pairs \((p,q)\) satisfying \(pq = 3\)
The possibilities are \((1,3), (3,1), (-1,-3), (-3,-1)\).
5. Test each pair in the linear‑coefficient equation
- For \(p = -3, q = -1\): \(2(-1) + (-3) = -2 -3 = -5\) ✔️
Hence \(p = -3\) and \(q = -1\) satisfy both conditions.
6. Write the factorised form
$$2x^{2} - 5x + 3 = (2x - 3)(x - 1).$$
7. Apply the zero‑product property
Set each factor to zero:
\[2x - 3 = 0 \quad \text{or} \quad x - 1 = 0.\]
Solving gives
\[x = \frac{3}{2} \quad \text{or} \quad x = 1.\]
8. State the roots
The roots of the quadratic equation are \(x = 1\) and \(x = \dfrac{3}{2}\).
$$2x^{2} - 5x + 3 = 0.$$
2. Assume a factorisation
Since the leading coefficient is 2, try factors of the form \((2x + p)(x + q)\) where \(p\) and \(q\) are integers. Expanding gives
$$ (2x + p)(x + q) = 2x^{2} + (2q + p)x + pq.$$
Compare with \(2x^{2} - 5x + 3\).
3. Match the coefficients
- Coefficient of \(x\): \(2q + p = -5\).
- Constant term: \(pq = 3\).
4. Find integer pairs \((p,q)\) satisfying \(pq = 3\)
The possibilities are \((1,3), (3,1), (-1,-3), (-3,-1)\).
5. Test each pair in the linear‑coefficient equation
- For \(p = -3, q = -1\): \(2(-1) + (-3) = -2 -3 = -5\) ✔️
Hence \(p = -3\) and \(q = -1\) satisfy both conditions.
6. Write the factorised form
$$2x^{2} - 5x + 3 = (2x - 3)(x - 1).$$
7. Apply the zero‑product property
Set each factor to zero:
\[2x - 3 = 0 \quad \text{or} \quad x - 1 = 0.\]
Solving gives
\[x = \frac{3}{2} \quad \text{or} \quad x = 1.\]
8. State the roots
The roots of the quadratic equation are \(x = 1\) and \(x = \dfrac{3}{2}\).
Question 4
Hint available
Find the roots of the quadratic equation 6x2 – x – 2 = 0.
Key Idea
Apply the quadratic formula $x = \dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ for a quadratic equation $ax^{2}+bx+c=0$.
Step-by-Step Solution
Given equation: $$6x^{2} - x - 2 = 0$$
1. Identify coefficients: $a = 6$, $b = -1$, $c = -2$.
2. Compute the discriminant $D$:
$$D = b^{2} - 4ac = (-1)^{2} - 4\times 6 \times (-2) = 1 + 48 = 49.$$
Since $D > 0$, the equation has two distinct real roots.
3. Apply the quadratic formula:
$$x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-(-1) \pm \sqrt{49}}{2\times 6} = \frac{1 \pm 7}{12}.$$
4. Evaluate the two possibilities:
- For the ‘+’ sign:
$$x_{1} = \frac{1 + 7}{12} = \frac{8}{12} = \frac{2}{3}.$$
- For the ‘–’ sign:
$$x_{2} = \frac{1 - 7}{12} = \frac{-6}{12} = -\frac{1}{2}.$$
Thus, the roots of the quadratic equation are $x = \dfrac{2}{3}$ and $x = -\dfrac{1}{2}$.
1. Identify coefficients: $a = 6$, $b = -1$, $c = -2$.
2. Compute the discriminant $D$:
$$D = b^{2} - 4ac = (-1)^{2} - 4\times 6 \times (-2) = 1 + 48 = 49.$$
Since $D > 0$, the equation has two distinct real roots.
3. Apply the quadratic formula:
$$x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-(-1) \pm \sqrt{49}}{2\times 6} = \frac{1 \pm 7}{12}.$$
4. Evaluate the two possibilities:
- For the ‘+’ sign:
$$x_{1} = \frac{1 + 7}{12} = \frac{8}{12} = \frac{2}{3}.$$
- For the ‘–’ sign:
$$x_{2} = \frac{1 - 7}{12} = \frac{-6}{12} = -\frac{1}{2}.$$
Thus, the roots of the quadratic equation are $x = \dfrac{2}{3}$ and $x = -\dfrac{1}{2}$.
Question 5
Hint available
Find the roots of the quadratic equation 2 6 x x .
Key Idea
Use the quadratic formula $x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ for a quadratic equation $ax^{2}+bx+c=0$.
Step-by-Step Solution
1. Write the given equation in standard form $ax^{2}+bx+c=0$:\\
$$2x^{2}-6x+2=0$$\
Hence, $a=2$, $b=-6$, $c=2$.\\
2. Compute the discriminant $D=b^{2}-4ac$:\\
$$D = (-6)^{2}-4\times2\times2 = 36-16 = 20.$$\\
3. Since $D>0$, the equation has two distinct real roots.\\
4. Apply the quadratic formula:\\
$$x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-(-6) \pm \sqrt{20}}{2\times2} = \frac{6 \pm \sqrt{20}}{4}.$$\\
5. Simplify $\sqrt{20}=\sqrt{4\times5}=2\sqrt{5}$ and reduce the fraction:\\
$$x = \frac{6 \pm 2\sqrt{5}}{4}=\frac{2(3 \pm \sqrt{5})}{4}=\frac{3 \pm \sqrt{5}}{2}.$$\\
6. Hence the two roots are:\\
$$x_{1}=\frac{3+\sqrt{5}}{2}, \qquad x_{2}=\frac{3-\sqrt{5}}{2}.$$
$$2x^{2}-6x+2=0$$\
Hence, $a=2$, $b=-6$, $c=2$.\\
2. Compute the discriminant $D=b^{2}-4ac$:\\
$$D = (-6)^{2}-4\times2\times2 = 36-16 = 20.$$\\
3. Since $D>0$, the equation has two distinct real roots.\\
4. Apply the quadratic formula:\\
$$x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-(-6) \pm \sqrt{20}}{2\times2} = \frac{6 \pm \sqrt{20}}{4}.$$\\
5. Simplify $\sqrt{20}=\sqrt{4\times5}=2\sqrt{5}$ and reduce the fraction:\\
$$x = \frac{6 \pm 2\sqrt{5}}{4}=\frac{2(3 \pm \sqrt{5})}{4}=\frac{3 \pm \sqrt{5}}{2}.$$\\
6. Hence the two roots are:\\
$$x_{1}=\frac{3+\sqrt{5}}{2}, \qquad x_{2}=\frac{3-\sqrt{5}}{2}.$$
Question 6
Hint available
Find the dimensions of the prayer hall discussed in Section 4.1.
Key Idea
Model the length and breadth of the rectangular hall with variables, use the relation Area = length × breadth, and the given condition that length exceeds breadth by 5 m. This leads to a quadratic equation whose positive root gives the required dimensions.
Step-by-Step Solution
1. Let the breadth of the hall be $x$ metres.
2. Then the length is $x+5$ metres (5 m more than the breadth).
3. Given that the area of the hall is $1200\,\text{m}^2$, we have
$$x(x+5)=1200.$$
4. Expand and bring all terms to one side:
$$x^{2}+5x-1200=0.$$
5. Solve the quadratic equation using the quadratic formula:
$$x=\frac{-5\pm\sqrt{5^{2}-4\cdot1\cdot(-1200)}}{2\cdot1}
=\frac{-5\pm\sqrt{25+4800}}{2}
=\frac{-5\pm\sqrt{4825}}{2}.$$
6. Since a length cannot be negative, take the positive root:
$$x=\frac{-5+\sqrt{4825}}{2}\approx\frac{-5+69.46}{2}\approx32.23\text{ m.}$$
7. Hence the breadth $=x\approx32.23\,\text{m}$.
8. The length $=x+5\approx32.23+5=37.23\,\text{m}.$
9. For practical purposes, the dimensions are taken as approximately $32\,\text{m}$ (breadth) and $37\,\text{m}$ (length).
2. Then the length is $x+5$ metres (5 m more than the breadth).
3. Given that the area of the hall is $1200\,\text{m}^2$, we have
$$x(x+5)=1200.$$
4. Expand and bring all terms to one side:
$$x^{2}+5x-1200=0.$$
5. Solve the quadratic equation using the quadratic formula:
$$x=\frac{-5\pm\sqrt{5^{2}-4\cdot1\cdot(-1200)}}{2\cdot1}
=\frac{-5\pm\sqrt{25+4800}}{2}
=\frac{-5\pm\sqrt{4825}}{2}.$$
6. Since a length cannot be negative, take the positive root:
$$x=\frac{-5+\sqrt{4825}}{2}\approx\frac{-5+69.46}{2}\approx32.23\text{ m.}$$
7. Hence the breadth $=x\approx32.23\,\text{m}$.
8. The length $=x+5\approx32.23+5=37.23\,\text{m}.$
9. For practical purposes, the dimensions are taken as approximately $32\,\text{m}$ (breadth) and $37\,\text{m}$ (length).
Question 7
Hint available
A pole has to be erected at a point on the boundary of a circular park of diameter 13 metres in such a way that the differences of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected?
Key Idea
Use Thales' theorem: the angle subtended by a diameter at any point on the circle is a right angle. Hence triangle formed by the two gates and the required point is a right‑angled triangle with hypotenuse equal to the diameter. Apply the Pythagorean theorem together with the given difference of the two sides, leading to a quadratic equation.
Step-by-Step Solution
1. Let the circular park have centre O and diameter AB = 13 m. Points A and B are diametrically opposite gates.
2. Let P be the required point on the circumference. Denote PA = x m and PB = y m.
3. Since AB is a diameter, by Thales' theorem \(\angle APB = 90^{\circ}\). Therefore triangle APB is right‑angled with hypotenuse AB.
4. Apply the Pythagorean theorem:
$$x^{2}+y^{2}=AB^{2}=13^{2}=169.$$
5. The condition on the distances is \(|x-y|=7|\). Assume \(x>y\) (the other case gives the symmetric position), so
$$x-y=7 \quad\Rightarrow\quad x = y+7.$$
6. Substitute \(x = y+7\) into the Pythagorean relation:
$$(y+7)^{2}+y^{2}=169$$
$$y^{2}+14y+49+y^{2}=169$$
$$2y^{2}+14y-120=0$$
Divide by 2:
$$y^{2}+7y-60=0.$$
7. Factor the quadratic:
$$(y+12)(y-5)=0$$
Hence \(y=5\) (positive) or \(y=-12\) (reject as distance cannot be negative).
8. Find \(x\):
$$x = y+7 = 5+7 = 12\;\text{metres}.$$
9. Thus the required distances are 12 m from one gate and 5 m from the other. Because the figure is symmetric about the line AB, there are two possible positions of P on the circle, each giving the same pair of distances.
10. Since both distances are less than the diameter, the construction is feasible.
2. Let P be the required point on the circumference. Denote PA = x m and PB = y m.
3. Since AB is a diameter, by Thales' theorem \(\angle APB = 90^{\circ}\). Therefore triangle APB is right‑angled with hypotenuse AB.
4. Apply the Pythagorean theorem:
$$x^{2}+y^{2}=AB^{2}=13^{2}=169.$$
5. The condition on the distances is \(|x-y|=7|\). Assume \(x>y\) (the other case gives the symmetric position), so
$$x-y=7 \quad\Rightarrow\quad x = y+7.$$
6. Substitute \(x = y+7\) into the Pythagorean relation:
$$(y+7)^{2}+y^{2}=169$$
$$y^{2}+14y+49+y^{2}=169$$
$$2y^{2}+14y-120=0$$
Divide by 2:
$$y^{2}+7y-60=0.$$
7. Factor the quadratic:
$$(y+12)(y-5)=0$$
Hence \(y=5\) (positive) or \(y=-12\) (reject as distance cannot be negative).
8. Find \(x\):
$$x = y+7 = 5+7 = 12\;\text{metres}.$$
9. Thus the required distances are 12 m from one gate and 5 m from the other. Because the figure is symmetric about the line AB, there are two possible positions of P on the circle, each giving the same pair of distances.
10. Since both distances are less than the diameter, the construction is feasible.
Question 8
Hint available
Find the discriminant of the equation 3x2 – 2x + 1 3 = 0 and hence find the nature of its roots. Find them, if they are real.
Key Idea
For a quadratic equation $ax^2+bx+c=0$, the discriminant $D=b^2-4ac$ determines the nature of its roots: $D>0$ ⇒ two distinct real roots, $D=0$ ⇒ equal real roots, $D<0$ ⇒ two imaginary (complex conjugate) roots. The roots are given by $x=\frac{-b\pm\sqrt{D}}{2a}$.
Step-by-Step Solution
1. Identify the coefficients\
The given quadratic is $3x^2-2x+13=0$, so\
\[ a=3,\quad b=-2,\quad c=13. \]
2. Compute the discriminant\
\[ D = b^2-4ac = (-2)^2-4\cdot3\cdot13 = 4-156 = -152. \]
3. Interpret the discriminant\
Since $D=-152<0$, the equation has no real roots; the roots are a pair of complex conjugates.
4. Find the roots (for completeness)\
Using the quadratic formula:\
\[ x = \frac{-b\pm\sqrt{D}}{2a} = \frac{-(-2)\pm\sqrt{-152}}{2\cdot3} = \frac{2\pm\sqrt{-152}}{6}. \]
Write $\sqrt{-152}=i\sqrt{152}=i\,2\sqrt{38}$, then\
\[ x = \frac{2\pm i\,2\sqrt{38}}{6}=\frac{1\pm i\sqrt{38}}{3}. \]
5. Conclusion\
The discriminant is $-152$, indicating two imaginary (complex) roots. The roots are $\displaystyle x = \frac{1}{3} \pm \frac{i\sqrt{38}}{3}$.
Hence, the equation has no real roots.
The given quadratic is $3x^2-2x+13=0$, so\
\[ a=3,\quad b=-2,\quad c=13. \]
2. Compute the discriminant\
\[ D = b^2-4ac = (-2)^2-4\cdot3\cdot13 = 4-156 = -152. \]
3. Interpret the discriminant\
Since $D=-152<0$, the equation has no real roots; the roots are a pair of complex conjugates.
4. Find the roots (for completeness)\
Using the quadratic formula:\
\[ x = \frac{-b\pm\sqrt{D}}{2a} = \frac{-(-2)\pm\sqrt{-152}}{2\cdot3} = \frac{2\pm\sqrt{-152}}{6}. \]
Write $\sqrt{-152}=i\sqrt{152}=i\,2\sqrt{38}$, then\
\[ x = \frac{2\pm i\,2\sqrt{38}}{6}=\frac{1\pm i\sqrt{38}}{3}. \]
5. Conclusion\
The discriminant is $-152$, indicating two imaginary (complex) roots. The roots are $\displaystyle x = \frac{1}{3} \pm \frac{i\sqrt{38}}{3}$.
Hence, the equation has no real roots.