CBSE Class 10 Quadratic Equations Sure Shot Questions Set 07

Case Based MCQs

Case I : Read the following passage and answer the questions from 39 to 43.
Formation of Quadratic Equation
Quadratic equations started around 3000 B.C. with the Babylonians. They were one of the world’s first civilisation, and came up with some great ideas like agriculture, irrigation and writing. There were many reasons why Babylonians needed to solve quadratic equations. For example to know what amount of crop you can grow on the square field.
Now represent the following situations in the form of quadratic equation.

Question. The sum of squares of two consecutive integers is 650.
(a) \( x^2 + 2x - 650 = 0 \)
(b) \( 2x^2 + 2x - 649 = 0 \)
(c) \( x^2 - 2x - 650 = 0 \)
(d) \( 2x^2 + 6x - 550 = 0 \)
Answer: (b)

Question. The sum of two numbers is 15 and the sum of their reciprocals is 3/10.
(a) \( x^2 + 10x - 150 = 0 \)
(b) \( 15x^2 - x + 150 = 0 \)
(c) \( x^2 - 15x + 50 = 0 \)
(d) \( 3x^2 - 10x + 15 = 0 \)
Answer: (c)

Question. Two numbers differ by 3 and their product is 504.
(a) \( 3x^2 - 504 = 0 \)
(b) \( x^2 - 504x + 3 = 0 \)
(c) \( 504x^2 + 3 = x \)
(d) \( x^2 + 3x - 504 = 0 \)
Answer: (d)

Question. A natural number whose square diminished by 84 is thrice of 8 more of given number.
(a) \( x^2 + 8x - 84 = 0 \)
(b) \( 3x^2 - 84x + 3 = 0 \)
(c) \( x^2 - 3x - 108 = 0 \)
(d) \( x^2 - 11x + 60 = 0 \)
Answer: (c)

Question. A natural number when increased by 12, equals 160 times its reciprocal.
(a) \( x^2 - 12x + 160 = 0 \)
(b) \( x^2 - 160x + 12 = 0 \)
(c) \( 12x^2 - x - 160 = 0 \)
(d) \( x^2 + 12x - 160 = 0 \)
Answer: (d)

Case II : Read the following passage and answer the questions from 44 to 48.
Factorization Method
Amit is preparing for his upcoming semester exam. For this, he has to practice the chapter of Quadratic Equations. So he started with factorization method. Let two linear factors of \( ax^2 + bx + c \) be \( (px + q) \) and \( (rx + s) \).
\( \therefore ax^2 + bx + c = (px + q)(rx + s) \)
\( = prx^2 + (ps + qr)x + qs \).
Now, help Amit in factorizing the following quadratic equations and find the roots.

Question. \( 6x^2 + x - 2 = 0 \)
(a) 1, 6
(b) \( \frac{1}{2}, \frac{-2}{3} \)
(c) \( \frac{1}{3}, \frac{-1}{2} \)
(d) \( \frac{3}{2}, -2 \)
Answer: (b)

Question. \( 2x^2 + x - 300 = 0 \)
(a) \( 30, \frac{2}{15} \)
(b) \( 60, \frac{-2}{5} \)
(c) \( 12, \frac{-25}{2} \)
(d) None of these
Answer: (c)

Question. \( x^2 - 8x + 16 = 0 \)
(a) 3, 3
(b) 3, –3
(c) 4, –4
(d) 4, 4
Answer: (d)

Question. \( 6x^2 - 13x + 5 = 0 \)
(a) \( 2, \frac{3}{5} \)
(b) \( -2, \frac{-5}{3} \)
(c) \( \frac{1}{2}, \frac{-3}{5} \)
(d) \( \frac{1}{2}, \frac{5}{3} \)
Answer: (d)

Question. \( 100x^2 - 20x + 1 = 0 \)
(a) \( \frac{1}{10}, \frac{1}{10} \)
(b) –10, –10
(c) \( -10, \frac{1}{10} \)
(d) \( \frac{-1}{10}, \frac{-1}{10} \)
Answer: (a)

Case-III : Read the following passage and answer the questions from 49 to 53.
Nature of Roots
A quadratic equation can be defined as an equation of degree 2. This means that the highest exponent of the polynomial in it is 2. The standard form of a quadratic equation is \( ax^2 + bx + c = 0 \), where \( a, b \), and \( c \) are real numbers and \( a \neq 0 \). Every quadratic equation has two roots depending on the nature of its discriminant, \( D = b^2 - 4ac \).

Question. Which of the following quadratic equation have no real roots?
(a) \( -4x^2 + 7x - 4 = 0 \)
(b) \( -4x^2 + 7x - 2 = 0 \)
(c) \( -2x^2 + 5x - 2 = 0 \)
(d) \( 3x^2 + 6x + 2 = 0 \)
Answer: (a)

Question. Which of the following quadratic equation have rational roots?
(a) \( x^2 + x - 1 = 0 \)
(b) \( x^2 - 5x + 6 = 0 \)
(c) \( 4x^2 - 3x - 2 = 0 \)
(d) \( 6x^2 - x + 11 = 0 \)
Answer: (b)

Question. Which of the following quadratic equation have irrational roots?
(a) \( 3x^2 + 2x + 2 = 0 \)
(b) \( 4x^2 - 7x + 3 = 0 \)
(c) \( 6x^2 - 3x - 5 = 0 \)
(d) \( 2x^2 + 3x - 2 = 0 \)
Answer: (c)

Question. Which of the following quadratic equations have equal roots?
(a) \( x^2 - 3x + 4 = 0 \)
(b) \( 2x^2 - 2x + 1 = 0 \)
(c) \( 5x^2 - 10x + 1 = 0 \)
(d) \( 9x^2 + 6x + 1 = 0 \)
Answer: (d)

Question. Which of the following quadratic equations has two distinct real roots?
(a) \( x^2 + 3x + 1 = 0 \)
(b) \( -x^2 + 3x - 3 = 0 \)
(c) \( 4x^2 + 8x + 4 = 0 \)
(d) \( 3x^2 + 6x + 4 = 0 \)
Answer: (a)

Assertion & Reasoning Based MCQs

Directions : In these questions, a statement of Assertion is followed by a statement of Reason is given. Choose the correct answer out of the following choices :
(a) Assertion and Reason both are correct statements and Reason is the correct explanation of Assertion.
(b) Assertion and Reason both are correct statements but Reason is not the correct explanation of Assertion.
(c) Assertion is correct statement but Reason is wrong statement.
(d) Assertion is wrong statement but Reason is correct statement.

Question. Assertion : \( 2x^2 - 4x + 3 = 0 \) is a quadratic equation.
Reason : All polynomials of degree \( n \), when \( n \) is a whole number can be treated as quadratic equation.
(a) A
(b) B
(c) C
(d) D
Answer: (c)

Question. Assertion : \( 3y^2 + 17y - 30 = 0 \) have distinct roots.
Reason : The quadratic equation \( ax^2 + bx + c = 0 \) have distinct roots (real roots) if \( D > 0 \).
(a) A
(b) B
(c) C
(d) D
Answer: (a)

Question. Assertion : \( 9x^2 - 3x - 20 = 0 \Rightarrow (3x - 5)(3x + 4) = 0 \). If the roots are calculated by splitting the middle term.
Reason : To factorise \( ax^2 + bx + c = 0 \), we write it in the form \( ax^2 + b_1x + b_2x + c = 0 \) such that \( b_1 + b_2 = b \) and \( b_1b_2 = ac \).
(a) A
(b) B
(c) C
(d) D
Answer: (a)

Question. Assertion : The value of \( k \) for which the equation \( kx^2 - 12x + 4 = 0 \) has equal roots, is 9.
Reason : The equation \( ax^2 + bx + c = 0, (a \neq 0) \) has equal roots, if \( (b^2 - 4ac) > 0 \).
(a) A
(b) B
(c) C
(d) D
Answer: (c)

Question. Assertion : Both the roots of the equation \( x^2 - x + 1 = 0 \) are real.
Reason : The roots of the equation \( ax^2 + bx + c = 0 \) are real if and only if \( b^2 - 4ac \geq 0 \).
(a) A
(b) B
(c) C
(d) D
Answer: (d)

Question. Assertion : \( 2\sqrt{2} \) is a root of the quadratic equation \( x^2 - 4\sqrt{2}x + 8 = 0 \).
Reason : The root of a quadratic equation satisfies it.
(a) A
(b) B
(c) C
(d) D
Answer: (a)

Question. Assertion : \( \frac{1}{(x-1)(x-2)} + \frac{1}{(x-2)(x-3)} = \frac{2}{3} \) \( (x \neq 1,2,3) \) is a quadratic equation.
Reason : An equation of the form \( ax^2 + bx + c = 0 \), where \( a, b, c \in R \) is a quadratic equation.
(a) A
(b) B
(c) C
(d) D
Answer: (b)

Very Short Answer Type Questions 

Question. If 1 is a root of the equations \(ay^2 + ay + 3 = 0\) and \(y^2 + y + b = 0\), then find the value of \(ab\).
Answer: Given, 1 is the root of \(ay^2 + ay + 3 = 0\) and \(y^2 + y + b = 0\)
\(\therefore y = 1\) will satisfy these equations.
i.e., \(a(1)^2 + a \times 1 + 3 = 0 \Rightarrow 2a + 3 = 0 \Rightarrow a = \frac{-3}{2}\)
\((1)^2 + 1 + b = 0 \Rightarrow 2 + b = 0 \Rightarrow b = -2\)
\(\therefore ab = \frac{-3}{2} \times (-2) = 3\)

Question. If the quadratic equation \(mx^2 + 2x + m = 0\) has two equal roots, then find the values of \(m\).
Answer: For roots of \(mx^2 + 2x + m = 0\) to be equal, Discriminant, \(D = 0\)
\(\therefore (2)^2 - 4(m)(m) = 0\)
\(\Rightarrow 4 - 4m^2 = 0 \Rightarrow m^2 = 1 \Rightarrow m = \pm 1\).

Question. Form the quadratic equation from \(x = \sqrt{6 + \sqrt{6 + \sqrt{6 + ...}}}\), where \(x\) is a natural number.
Answer: We have, \(x = \sqrt{6 + \sqrt{6 + \sqrt{6 + ...}}}\)
Squaring both sides, we get \(x^2 = 6 + \sqrt{6 + \sqrt{6 + ...}}\)
\(\Rightarrow x^2 = 6 + x\)
\(\Rightarrow x^2 - x - 6 = 0\), which is the required quadratic equation.

Question. Find the signs of the roots of the quadratic equation \(x^2 + kx + k = 0\), where \(k > 0\).
Answer: Let \(\alpha\) and \(\beta\) be the roots of \(x^2 + kx + k = 0\).
Then, \(\alpha + \beta = -k\) and \(\alpha \beta = k\).
This is possible only when \(\alpha\) and \(\beta\) are both negative.

Question. Check whether the equation \((x^2 + 1)(x + 2) = (x + 3)^2\) is quadratic or not?
Answer: We have, \((x^2 + 1)(x + 2) = (x + 3)^2\)
\(\Rightarrow x^3 + 2x^2 + x + 2 = x^2 + 6x + 9\)
\(\Rightarrow x^3 + x^2 - 5x - 7 = 0\), which is not of the form \(ax^2 + bx + c = 0, a \neq 0\).
Hence, given equation is not a quadratic equation.

Question. Find the roots of the quadratic equation \(2x^2 - 7x - 85 = 0\) by factorisation.
Answer: We have, \(2x^2 - 7x - 85 = 0\)
\(\Rightarrow 2x^2 + 10x - 17x - 85 = 0\)
\(\Rightarrow 2x(x + 5) - 17(x + 5) = 0\)
\(\Rightarrow (x + 5)(2x - 17) = 0 \Rightarrow x = -5\) or \(x = \frac{17}{2}\).

Question. If a number is added to twice its square, then the resultant is 21. Find the number.
Answer: Let the number be \(x\).
According to question, \(x + 2x^2 = 21\)
\(\Rightarrow 2x^2 + x - 21 = 0 \Rightarrow 2x^2 - 6x + 7x - 21 = 0\)
\(\Rightarrow 2x(x - 3) + 7(x - 3) = 0\)
\(\Rightarrow (x - 3)(2x + 7) = 0 \Rightarrow x = 3\) or \(x = \frac{-7}{2}\).

Question. Find the discriminant of the quadratic equation \(x^2 + px + 2q = 0\).
Answer: The given equation is, \(x^2 + px + 2q = 0\)
Here, \(a = 1, b = p\) and \(c = 2q\)
\(\therefore D = b^2 - 4ac = p^2 - 4 \times 1 \times 2q = p^2 - 8q\)

Question. Check whether \(x = 0\) and \(x = 1\) are the solutions of the equation \(x^2 + x + 1 = 0\) or not?
Answer: Given equation is, \(x^2 + x + 1 = 0\)
When \(x = 0\), L.H.S. \(= 0^2 + 0 + 1 = 1 \neq 0 = \text{R.H.S}\)
\(\therefore x = 0\) is not the solution of given equation.
When \(x = 1\), L.H.S. \(= 1^2 + 1 + 1 = 3 \neq 0 = \text{R.H.S}\)
\(\therefore x = 1\) is not the solution of given equation.

Question. Find the value(s) of \(k\) so that the quadratic equation \(x^2 - 4kx + k = 0\) has equal roots.
Answer: Given, \(x^2 - 4kx + k = 0\)
Since, given equation has equal roots. \(\therefore D = 0\)
\(\Rightarrow (-4k)^2 - 4(1)(k) = 0 \Rightarrow 16k^2 - 4k = 0\)
\(\Rightarrow 4k(4k - 1) = 0 \Rightarrow k = 0\) or \(k = \frac{1}{4}\).