CBSE Class 10 Quadratic Equations Sure Shot Questions Set 11

Solution of a Quadratic Equation by Quadratic Formula

The roots of a quadratic equation \( ax^2 + bx + c = 0 \) are given by \( \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), if \( b^2 - 4ac \geq 0 \).
This formula for finding the roots of a quadratic equation is often referred to as the quadratic formula.
The expression \( b^2 - 4ac \) is called the discriminant of the quadratic equation and generally denoted by D. If \( b^2 - 4ac < 0 \), then the equation will have no real roots.
As this formula was given by an ancient Indian mathematician Sridharacharya around AD 1025, it is known as Sridharacharya’s formula for determining the roots of the quadratic equations \( ax^2 + bx + c = 0 \).

Question. Find the solution of the quadratic equations by quadratic formula.
(i) \( \frac{1}{2}x^2 - \sqrt{11}x + 1 = 0 \) (ii) \( -x^2 + 7x - 10 = 0 \)
Answer: (i) \( \frac{1}{2}x^2 - \sqrt{11}x + 1 = 0 \Rightarrow x^2 - 2\sqrt{11}x + 2 = 0 \)
Here, \( a = 1, b = -2\sqrt{11}, c = 2 \)
Using the formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we get
\( x = \frac{+2\sqrt{11} \pm \sqrt{44 - 8}}{2} = \frac{2\sqrt{11} \pm 6}{2} \)
\( \Rightarrow x = \sqrt{11} + 3, \sqrt{11} - 3 \)
Hence, the roots of the given quadratic equation are \( \sqrt{11} + 3 \) and \( \sqrt{11} - 3 \).
(ii) \( -x^2 + 7x - 10 = 0 \)
Here, \( a = -1, b = 7, c = -10 \)
Using the formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we get
\( x = \frac{-7 \pm \sqrt{49 - 40}}{-2} = \frac{-7 \pm \sqrt{9}}{-2} \)
\( \Rightarrow x = \frac{-7 + 3}{-2} = \frac{-4}{-2} = 2 \), or \( \frac{-7 - 3}{-2} = \frac{-10}{-2} = 5 \)
Roots are 2 and 5.

Question. Using the quadratic formula, solve the equation \( a^2b^2x^2 - (4b^4 - 3a^4)x - 12a^2b^2 = 0 \)
Answer: Comparing given equation with \( Ax^2 + Bx + C = 0 \), we get
\( A = a^2b^2 \), \( B = -(4b^4 - 3a^4) \) and \( C = -12a^2b^2 \).
\( \therefore B^2 - 4AC = [-(4b^4 - 3a^4)]^2 - 4 \times a^2b^2 \times (-12a^2b^2) \)
\( = 16b^8 + 9a^8 - 24a^4b^4 + 48a^4b^4 \)
\( = 16b^8 + 9a^8 + 24a^4b^4 = (4b^4 + 3a^4)^2 \)
\( \Rightarrow \sqrt{B^2 - 4AC} = 4b^4 + 3a^4 \)
Now, \( x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} = \frac{+(4b^4 - 3a^4) \pm (4b^4 + 3a^4)}{2a^2b^2} \)
\( = \frac{4b^4 - 3a^4 + 4b^4 + 3a^4}{2a^2b^2} \) or \( \frac{4b^4 - 3a^4 - 4b^4 - 3a^4}{2a^2b^2} \)
\( = \frac{8b^4}{2a^2b^2} \) or \( \frac{-6a^4}{2a^2b^2} = \frac{4b^2}{a^2} \) or \( \frac{-3a^2}{b^2} \)

Question. Solve the quadratic equation \( 2x^2 + ax - a^2 = 0 \) for x using quadratic formula.
Answer: \( 2x^2 + ax - a^2 = 0 \)
Here, \( a = 2, b = a \) and \( c = -a^2 \).
Using the formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we get
\( x = \frac{-a \pm \sqrt{a^2 - 4 \times 2 \times (-a^2)}}{2 \times 2} = \frac{-a \pm \sqrt{9a^2}}{4} = \frac{-a \pm 3a}{4} \)
\( \Rightarrow x = \frac{-a + 3a}{4} = \frac{a}{2} \), \( x = \frac{-a - 3a}{4} = -a \)
Roots are \( \frac{a}{2}, -a \).

Question. Solve the following quadratic equation: \( 9x^2 - 9(a + b)x + [2a^2 + 5ab + 2b^2] = 0 \)
Answer: We have \( 9x^2 - 9(a + b)x + [2a^2 + 5ab + 2b^2] = 0 \)
Here, \( A = 9, B = -9(a + b) \) and \( C = [2a^2 + 5ab + 2b^2] \)
So, discriminant, \( D = B^2 - 4AC = \{-9(a + b)\}^2 - 4 \times 9(2a^2 + 5ab + 2b^2) \)
\( = 9^2(a + b)^2 - 4 \times 9(2a^2 + 5ab + 2b^2) \)
\( = 9\{9(a^2 + b^2 + 2ab) - 4(2a^2 + 5ab + 2b^2)\} \)
\( = 9\{9a^2 + 9b^2 + 18ab - 8a^2 - 20ab - 8b^2\} = 9(a^2 + b^2 - 2ab) = 9(a - b)^2 \)
Using the quadratic formula, \( x = \frac{-B \pm \sqrt{D}}{2A} \), we get
\( x = \frac{9(a + b) \pm \sqrt{9(a - b)^2}}{2 \times 9} \)
\( \Rightarrow x = \frac{9(a + b) \pm 3(a - b)}{2 \times 9} \Rightarrow x = \frac{3(a + b) \pm (a - b)}{6} \)
\( \Rightarrow x = \frac{(3a + 3b) + (a - b)}{6} \) or \( x = \frac{(3a + 3b) - (a - b)}{6} \)
\( \Rightarrow x = \frac{4a + 2b}{6} = \frac{2a + b}{3} \) or \( x = \frac{2a + 4b}{6} = \frac{a + 2b}{3} \)
\( x = \frac{2a + b}{3} \) or \( x = \frac{a + 2b}{3} \) are required solutions.

I. Very Short Answer Type Questions

Multiple Choice Questions 

Choose the correct answer from the given options:

Question. The discriminant of the equation \( 9x^2 + 6x + 1 = 0 \) is
(a) 0
(b) 1
(c) 2
(d) 3
Answer: (a)

Question. If D is the discriminant of the equation \( x^2 + 2x - 4 \), then 2D is:
(a) 20
(b) 40
(c) 60
(d) 80
Answer: (b)

Question.  The discriminant of the quadratic equation \( 4x^2 - 6x + 3 = 0 \) is:
(a) 12
(b) 84
(c) \( 2\sqrt{3} \)
(d) -12
Answer: (d)

Question. The roots of the quadratic equation \( ax^2 + bx + c = 0 \) are given by \( \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) if \( b^2 - 4ac \)
(a) < 0
(b) \( \leq 0 \)
(c) > 0
(d) \( \geq 0 \)
Answer: (d)

Question. The quadratic formula was given by an ancient Indian mathematician.
(a) Sridharacharya
(b) Aryabhata
(c) Brahmagupta
(d) None of these
Answer: (a)

Question. Assertion-Reason Type Question
Assertion (A): The values of x are \( -\frac{a}{2} \), \( a \) for a quadratic equation \( 2x^2 + ax - a^2 = 0 \).
Reason (R): For quadratic equation \( ax^2 + bx + c = 0 \), \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer: (d)

Question. Answer the following:
(1) Write the discriminant of the quadratic equation \( (x + 5)^2 = 2(5x - 3) \).
(2) Find the discriminant of the quadratic equation: \( 4x^2 - \frac{2}{3}x - \frac{1}{16} = 0 \).
Answer: (1) \( -124 \); (2) \( \frac{13}{9} \)

II. Short Answer Type Questions-I

Question. Find the roots of the equation \( ax^2 + a = a^2x + x \).
Answer: \( a, \frac{1}{a} \)

Question. Solve the following quadratic equation for x: \( 4x^2 - 4a^2x + (a^4 - b^4) = 0 \).
Answer: \( \frac{a^2 + b^2}{2}, \frac{a^2 - b^2}{2} \)

Question. Solve the following quadratic equation for x: \( 9x^2 - 6b^2x - (a^4 - b^4) = 0 \).
Answer: \( \frac{b^2 + a^2}{3}, \frac{b^2 - a^2}{3} \)

Question. Solve the following quadratic equation for x: \( 4x^2 + 4bx - (a^2 - b^2) = 0 \).
Answer: \( \frac{-b + a}{2}, \frac{-b - a}{2} \)

Question. Solve the following quadratic equation for x: \( x^2 - 2ax - (4b^2 - a^2) = 0 \).
Answer: \( a + 2b, a - 2b \)

III. Short Answer Type Questions-II

Question. Solve the following using quadratic formula: \( 2\sqrt{3}x^2 - 5x + \sqrt{3} = 0 \).
Answer: \( \frac{\sqrt{3}}{2}, \frac{1}{\sqrt{3}} \)

Question. Solve the following using quadratic formula: \( 3x^2 + 2\sqrt{5}x - 5 = 0 \).
Answer: \( \frac{\sqrt{5}}{3}, -\sqrt{5} \)

Question. Solve the following using quadratic formula: \( \frac{1}{x+4} - \frac{1}{x-7} = \frac{11}{30}, x \neq -4, 7 \).
Answer: 2, 1

Question. Find the roots of quadratic equation: \( x^2 - 3\sqrt{5}x + 10 = 0 \).
Answer: \( 2\sqrt{5}, \sqrt{5} \)

Question. Find the roots of quadratic equation: \( 5\sqrt{5}x^2 + 30x + 8\sqrt{5} = 0 \).
Answer: \( \frac{-4\sqrt{5}}{5}, \frac{-2\sqrt{5}}{5} \)

Question. Solve for x: \( 4x^2 - 4ax + (a^2 - b^2) = 0 \).
Answer: \( \frac{a+b}{2}, \frac{a-b}{2} \)

Question. Two water taps together can fill a tank in 9 hours 36 minutes. The tap of large diameter takes 8 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Answer: 16 hours, 24 hours

IV. Long Answer Type Questions

Question. A rectangular field is 20 m long and 14 m wide. There is a path of equal width all around it, having an area of 111 sq m. Find the width of the path.
Answer: 1.5 m

Question. At ‘t’ minutes past 2 pm, the time needed by the minute hand of a clock to show 3 pm was found to be 3 minutes less than \( \frac{t^2}{4} \) minutes. Find t.
Answer: 14 minutes

Question. Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number.
Answer: 12

Question. Solve for x: \( \frac{x-3}{x-4} + \frac{x-5}{x-6} = \frac{10}{6}, x \neq 4, 6 \).
Answer: \( 7, 2.8 \)

Question. Solve for x: \( \frac{x-2}{x-3} + \frac{x-4}{x-5} = \frac{10}{3}, x \neq 3, 5 \).
Answer: \( \frac{7}{2}, 6 \)

Question. Solve for x: \( 3\left(\frac{3x-1}{2x+3}\right) - 2\left(\frac{2x+3}{3x-1}\right) = 5, x \neq -\frac{3}{2}, \frac{1}{3} \).
Answer: \( 0, -7 \)

Question. The difference of squares of two numbers is 88. If the larger number is 5 less than twice the smaller number, then find the two numbers.
Answer: 9 and 13

Question. Sum of the areas of two squares is \( 544 \text{ m}^2 \). If the difference of their perimeters is 32 m, find the sides of the two squares.
Answer: 20 m and 12 m

Case Study Based Questions
I. Water Distribution System: Delhi Jal Board (DJB) is the main body of the Delhi Government which supplies drinking water in the National Capital Territory of Delhi. Distribution system is well knit and properly planned. Maintenance of underground pipe and hose system is also performed at regular interval of time... In our locality, DJB constructed two big reservoir labelled as Reservoir–A and Reservoir–B.
Refer to Reservoir-A

Question. Two pipes running together can fill the reservoir in \( 11\frac{1}{9} \) minutes. If one pipe takes 5 minutes more than the other to fill the reservoir, the time in which each pipe alone would fill the reservoir is
(a) 10 min, 12 min
(b) 25 min, 20 min
(c) 15 min, 18 min
(d) 22 min, 28 min
Answer: (b)

Question. Two pipes running together can fill a reservoir in 6 minutes. If one pipe takes 5 minutes more than the other to fill the reservoir, the time in which each pipe would fill the reservoir separately is
(a) 8 min, 6 min
(b) 10 min, 15 min
(c) 12 min, 16 min
(d) 16 min, 18 min
Answer: (b)

Refer to Reservoir-B

Question. Two water taps together can fill a reservoir in \( 9\frac{3}{8} \) hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the reservoir separately. The time in which each tap can separately fill the reservoir will be
(a) 15 hrs, 25 hrs
(b) 20 hrs, 22 hrs
(c) 14 hrs, 18 hrs
(d) 18 hrs, 16 hrs
Answer: (a)

Question. Two taps running together can fill the reservoir in \( 3\frac{1}{13} \) minutes. If one tap takes 3 minutes more than the other to fill it, how many minutes each tap would take to fill the reservoir?
(a) 12 min, 15 min
(b) 6 min, 9 min
(c) 18 min, 14 min
(d) 5 min, 8 min
Answer: (d)

Question. If two tapes function simultaneously, reservoir will be filled in 12 hours. One tap fills the reservoir 10 hours faster than the other. The time that the second tap takes to fill the reservoir is given by
(a) 25 hrs
(b) 28 hrs
(c) 30 hrs
(d) 32 hrs
Answer: (c)

4. Nature of Roots

We know that the roots of the equation \( ax^2 + bx + c = 0, a \neq 0 \) are given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( b^2 - 4ac = D \).
The quadratic equation \( ax^2 + bx + c = 0 \) has:
(i) two distinct real roots \( \alpha \) and \( \beta \), if \( b^2 - 4ac > 0 \).
(ii) two equal real roots (i.e., coincident roots), if \( b^2 - 4ac = 0 \). Roots are given by \( -\frac{b}{2a} \).
(iii) no real roots, if \( b^2 - 4ac < 0 \).

Formation of Quadratic Equation with given Roots

If \( \alpha \) and \( \beta \) are the two roots of a quadratic equation, then the formula to construct the quadratic equation is \( x^2 - (\alpha + \beta)x + \alpha\beta = 0 \).
That is \( x^2 - (\text{sum of roots})x + \text{product of roots} = 0 \).
Note: Let \( \alpha \) and \( \beta \) be the two roots of quadratic equation \( ax^2 + bx + c = 0 \). Then the formula to get sum and product of the roots are:
\( \alpha + \beta = -\frac{b}{a} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} \)
\( \alpha\beta = \frac{c}{a} = \frac{\text{Constant term}}{\text{Coefficient of } x^2} \)

Question. What will be the nature of roots of quadratic equation \( 2x^2 + 4x - 7 = 0 \)?
Answer: \( 2x^2 + 4x - 7 = 0 \)
Here, \( a = 2, b = 4, c = -7 \)
\( D = b^2 - 4ac = 16 - 4 \times 2 \times (-7) = 16 + 56 = 72 > 0 \)
Hence, roots of quadratic equation are real and unequal.

Question. If \( ax^2 + bx + c = 0 \) has equal roots, find the value of c.
Answer: For equal roots \( D = 0 \)
i.e., \( b^2 - 4ac = 0 \Rightarrow b^2 = 4ac \Rightarrow c = \frac{b^2}{4a} \)

Question. Find the value(s) of k for which the equation \( x^2 + 5kx + 16 = 0 \) has real and equal roots.
Answer: For roots to be real and equal, \( b^2 - 4ac = 0 \)
\( \Rightarrow (5k)^2 - 4 \times 1 \times 16 = 0 \Rightarrow 25k^2 = 64 \Rightarrow k = \pm \frac{8}{5} \)

Question. If the roots of the equation \( (a - b)x^2 + (b - c)x + (c - a) = 0 \) are equal, prove that \( 2a = b + c \).
Answer: Since the equation has equal roots, therefore discriminant \( D = (b - c)^2 - 4(a - b)(c - a) = 0 \)
\( \Rightarrow b^2 + c^2 - 2bc - 4(ac - a^2 - bc + ab) = 0 \)
\( \Rightarrow b^2 + c^2 - 2bc - 4ac + 4a^2 + 4bc - 4ab = 0 \)
\( \Rightarrow 4a^2 + b^2 + c^2 - 4ab + 2bc - 4ac = 0 \)
\( \Rightarrow (2a)^2 + (-b)^2 + (-c)^2 + 2(2a)(-b) + 2(-b)(-c) + 2(-c)(2a) = 0 \)
\( \Rightarrow (2a - b - c)^2 = 0 \Rightarrow 2a - b - c = 0 \Rightarrow 2a = b + c \). Hence Proved.

Question. If the equation \( (1 + m^2)x^2 + 2mcx + c^2 - a^2 = 0 \) has equal roots, show that \( c^2 = a^2(1 + m^2) \).
Answer: Here, \( A = 1 + m^2 \), \( B = 2mc \) and \( C = c^2 - a^2 \)
For equal roots, \( B^2 - 4AC = 0 \)
\( \Rightarrow (2mc)^2 - 4(1 + m^2)(c^2 - a^2) = 0 \)
\( \Rightarrow 4m^2c^2 - 4(c^2 - a^2 + m^2c^2 - m^2a^2) = 0 \)
\( \Rightarrow m^2c^2 - c^2 + a^2 - m^2c^2 + m^2a^2 = 0 \)
\( \Rightarrow -c^2 + a^2(1 + m^2) = 0 \Rightarrow c^2 = a^2(1 + m^2) \). Hence Proved.