CBSE Class 10 Mathematics Quadratic Equations MCQs Set 10

Question. The roots of the quadratic equation \(2x^2 - x - 6 = 0\) are
(a) –2, 3/2
(b) 2, – 3/2
(c) –2, –3/2
(d) 2, 3/2
Answer: B

Question. Which of the following equations has the sum of its roots as 3?
(a) \(2x^2 - 3x + 6 = 0\)
(b) \(-x^2 + 3x - 3 = 0\)
(c) \(\sqrt{2}x^2 - \frac{3}{\sqrt{2}}x + 1 = 0\)
(d) \(3x^2 - 3x + 3 = 0\)
Answer: B

Question. If \((x + 4) (x - 4) = 9\), then the values of \(x\) are
(a) \(\pm 5\)
(b) \(\pm \frac{1}{5}\)
(c) \(-\frac{1}{3}, \frac{1}{5}\)
(d) \(\pm 4\)
Answer: A

Question. The sum of the squares of two consecutive natural numbers is 41. Represent this situation in the form of a quadratic equation.
(a) \(x^2 + x - 20 = 0\)
(b) \(x^2 - x - 20 = 0\)
(c) \(x^2 + x + 20 = 0\)
(d) \(x^2 - x + 20 = 0\)
Answer: A

Question. The roots of the equation \(x^2 - 2x - (r^2 - 1) = 0\) are
(a) \(1 - r, -r - 1\)
(b) \(1 - r, r + 1\)
(c) \(1, r\)
(d) \(1 - r, r\)
Answer: B

Question. If \(\frac{1}{3}\) is a root of the equation \(x^2 + kx - \frac{5}{9} = 0\), then find the value of \(k\).
(a) \(\frac{3}{4}\)
(b) \(\frac{4}{3}\)
(c) \(\frac{2}{3}\)
(d) \(\frac{3}{2}\)
Answer: B

Question. If –2 is a root of the quadratic equation \(3x^2 + px - 8 = 0\) and the quadratic equation \(4x^2 - 2px + k = 0\) has equal roots, then find the value of \(k\).
(a) –1
(b) 2
(c) –2
(d) 1
Answer: D

Question. The roots of the quadratic equation \(2x^2 - 3x - 5 = 0\) are
(a) both equal
(b) opposite integers
(c) rational and unequal
(d) not real
Answer: C

Question. Which of the following is a root of the quadratic equation \(\sqrt{3}x^2 + 10x + 7\sqrt{3} = 0\)?
(a) \(-\sqrt{3}\)
(b) \(\sqrt{3}\)
(c) \(7\sqrt{3}\)
(d) \(-7\sqrt{3}\)
Answer: A

Question. The discriminant of the equation \(x^2 + 9x - 13 = 0\) is
(a) 157
(b) 141
(c) 133
(d) 129
Answer: C

Question. The number of real roots of the equation \((x - 1)^2 + (x - 2)^2 + (x - 3)^2 = 0\) is
(a) 2
(b) 1
(c) 0
(d) 3
Answer: C

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

Question. Solve the following quadratic equation for \(x\) : \(4\sqrt{3}x^2 + 5x - 2\sqrt{3} = 0\)
(a) \(\frac{\sqrt{3}}{4}, -\frac{2}{\sqrt{3}}\)
(b) \(-\frac{\sqrt{3}}{4}, \frac{2}{\sqrt{3}}\)
(c) \(\frac{\sqrt{3}}{4}, \frac{2}{\sqrt{3}}\)
(d) \(-\frac{\sqrt{3}}{4}, -\frac{2}{\sqrt{3}}\)
Answer: A

Question. If \(x = \sqrt{2 + \sqrt{2 + \sqrt{2 + ...\infty}}}\) and \(x\) is a natural number, then
(a) \(x^2 + x - 2 = 0\)
(b) \(x^2 + 2x + 2 = 0\)
(c) \(x^2 - x - 2 = 0\)
(d) \(x^2 - x + 2 = 0\)
Answer: C

Question. The roots of the quadratic equation \(5(x - 3)^2 = 20\) are
(a) 1, – 5
(b) 1, 5
(c) –1, –5
(d) –1, 5
Answer: B

Question. For what value of \(t, x = \frac{2}{3}\) is a root of \(7x^2 + tx - 3 = 0\)?
(a) \(\frac{1}{6}\)
(b) \(-\frac{1}{6}\)
(c) \(\frac{1}{5}\)
(d) \(\frac{1}{8}\)
Answer: B

Question. The roots of the quadratic equation \(\frac{x^2 - 8}{x^2 + 20} = \frac{1}{2}\) are
(a) \(\pm 3\)
(b) \(\pm 2\)
(c) \(\pm 6\)
(d) \(\pm 4\)
Answer: C

Question. Find the roots of the following quadratic equation. \(2\sqrt{3}x^2 - 5x + \sqrt{3} = 0\).
(a) \(-\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{3}}\)
(b) \(\frac{\sqrt{3}}{2}, -\frac{1}{\sqrt{3}}\)
(c) \(\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{3}}\)
(d) \(-\frac{\sqrt{3}}{2}, -\frac{1}{\sqrt{3}}\)
Answer: C

Question. The roots of the equation \(\sqrt{x^2 + 15} = 8\) are
(a) \(x = 7\)
(b) \(x = \pm 7\)
(c) \(x = – 7\)
(d) \(x = 0\)
Answer: B

Case Based MCQs

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