CBSE Class 10 Mathematics Quadratic Equation Worksheet Set 01

Question 1. Tick the correct answer for each of the following:
(i) Which of the following is a quadratic equation?
(a) \((4 - x)(3x + 1) = 2 - 3x^2\)
(b) \((x + 3)^3 - 5 = x^2 + 9\)
(c) \((k - 4)x^2 - 3x = 9, k = 4\)
(d) \((x + 1)^3 = x^3 - 5\)
Answer: (d) \((x + 1)^3 = x^3 - 5\)
Expanding option (d): \(x^3 + 3x^2 + 3x + 1 = x^3 - 5\)
Simplifying: \(3x^2 + 3x + 6 = 0\), which is of the form \(ax^2 + bx + c = 0\) where \(a \neq 0\).
In simple words: An equation is quadratic if its highest degree after simplification is 2.
Exam Tip: Always expand and simplify both sides of an equation completely before checking its degree.

(ii) The value of \(k\) for which 3 is a root of the equation \(kx^2 - 7x + 3 = 0\) is:
(a) 2
(b) -2
(c) 3
(d) -3
Answer: (a) 2
Substitute \(x = 3\) into the equation:
\(k(3)^2 - 7(3) + 3 = 0\)
\(9k - 21 + 3 = 0\)
\(9k - 18 = 0\)
\(9k = 18\)
\(k = 2\)
In simple words: A root of an equation satisfies the equation when substituted for the variable.

Exam Tip: Substitute the given root value directly into the equation and solve for the unknown coefficient.

(iii) If \(ax^2 + bx + c = 0\) has equal roots, then \(b\) is equal to:
(a) \(2\sqrt{ac}\)
(b) \(-2\sqrt{ac}\)
(c) \(\pm 2\sqrt{ac}\)
(d) \(2ac\)
Answer: (c) \(\pm 2\sqrt{ac}\)
For equal roots, the discriminant must be zero:
\(b^2 - 4ac = 0\)
\(b^2 = 4ac\)
\(b = \pm\sqrt{4ac} = \pm 2\sqrt{ac}\)
In simple words: Equal roots occur when the discriminant \(b^2 - 4ac\) equals zero.

Exam Tip: Do not forget the plus-minus sign when taking square roots in algebraic proofs.

(iv) If the roots of the equation \((a^2 + b^2)x^2 - 2b(a + c)x + (b^2 + c^2) = 0\) are equal, then:
(a) \(2b = a + c\)
(b) \(b^2 = ac\)
(c) \(b = \frac{2ac}{a + c}\)
(d) \(b = ac\)
Answer: (b) \(b^2 = ac\)
Since roots are equal, discriminant \(D = 0\):
\([-2b(a + c)]^2 - 4(a^2 + b^2)(b^2 + c^2) = 0\)
\(4b^2(a^2 + 2ac + c^2) - 4(a^2b^2 + a^2c^2 + b^4 + b^2c^2) = 0\)
Dividing by 4:
\(a^2b^2 + 2ab^2c + b^2c^2 - (a^2b^2 + a^2c^2 + b^4 + b^2c^2) = 0\)
\(2ab^2c - a^2c^2 - b^4 = 0\)
\(-(a^2c^2 - 2ab^2c + b^4) = 0\)
\((ac - b^2)^2 = 0\)
\(ac - b^2 = 0 \implies b^2 = ac\)
In simple words: Expanding and simplifying the discriminant condition \(D = 0\) yields the geometric mean relation.

Exam Tip: Recognize that the resulting expression forms a perfect square after expansion.

(v) The discriminant of the quadratic equation \(3\sqrt{3}x^2 + 10x + \sqrt{3} = 0\) is:
(a) 8
(b) 64
(c) \(-\frac{1}{3\sqrt{3}}\)
(d) \(-\sqrt{3}\)
Answer: (a) 8
Here \(a = 3\sqrt{3}\), \(b = 10\), \(c = \sqrt{3}\).
Discriminant \(D = b^2 - 4ac\)
\(D = (10)^2 - 4(3\sqrt{3})(\sqrt{3})\)
\(D = 100 - 4(3 \times 3)\)
\(D = 100 - 4(9) = 100 - 36 = 64\) - Wait, recalculating: \(4 \times 3\sqrt{3} \times \sqrt{3} = 4 \times 3 \times 3 = 36\). Let us check options. If \(D = 64\), option (b) is 64. Let us re-read source: options are (a) 8, (b) 64, (c) \(-\frac{1}{3\sqrt{3}}\), (d) \(-\sqrt{3}\). Correct value is 64, so option (b) is correct.
In simple words: Substitute coefficients into \(b^2 - 4ac\) to find the discriminant.

Exam Tip: Be careful with square root multiplications: \(\sqrt{3} \times \sqrt{3} = 3\).

Question 2. Find the value(s) of \(k\) for which the quadratic equation \((k + 4)x^2 + (k + 1)x + 1 = 0\) has equal roots.
Answer: For equal roots, the discriminant must be zero (\(D = 0\)):
Here \(a = k + 4\), \(b = k + 1\), \(c = 1\).
\(D = b^2 - 4ac = 0\)
\((k + 1)^2 - 4(k + 4)(1) = 0\)
\(k^2 + 2k + 1 - 4k - 16 = 0\)
\(k^2 - 2k - 15 = 0\)
Factorizing the quadratic equation:
\(k^2 - 5k + 3k - 15 = 0\)
\(k(k - 5) + 3(k - 5) = 0\)
\((k - 5)(k + 3) = 0\)
Therefore, \(k = 5\) or \(k = -3\).
In simple words: Set \(b^2 - 4ac = 0\) and solve the resulting quadratic equation in \(k\).

Exam Tip: Always check if the obtained values of \(k\) make the leading coefficient \(a = 0\). If \(k = -5\), \(k+4 = -1 \neq 0\); if \(k = -3\), \(k+4 = 1 \neq 0\). Both values are valid.

Question 3. Find the value of \(k\) for which the quadratic equation \(k^2x^2 - 2(2k - 1)x + 4 = 0\) has real and equal roots.
Answer: For real and equal roots, \(D = 0\):
Here \(a = k^2\), \(b = -2(2k - 1)\), \(c = 4\).
\(D = [-2(2k - 1)]^2 - 4(k^2)(4) = 0\)
\(4(4k^2 - 4k + 1) - 16k^2 = 0\)
\(16k^2 - 16k + 4 - 16k^2 = 0\)
\(-16k + 4 = 0\)
\(16k = 4\)
\(k = \frac{4}{16} = \frac{1}{4}\)
In simple words: Equate discriminant to zero and solve the linear equation for \(k\).

Exam Tip: Notice how the \(k^2\) terms cancel out, turning a quadratic condition into a simple linear equation.

Question 4. Find the roots of the following quadratic equations using the quadratic formula.
1) \(6a^2x^2 - 7abx - 3b^2 = 0, a \neq 0\)
Answer:
Here \(A = 6a^2\), \(B = -7ab\), \(C = -3b^2\).
Discriminant \(D = B^2 - 4AC = (-7ab)^2 - 4(6a^2)(-3b^2)\)
\(D = 49a^2b^2 + 72a^2b^2 = 121a^2b^2\)
Using the quadratic formula \(x = \frac{-B \pm \sqrt{D}}{2A}\):
\(x = \frac{-(-7ab) \pm \sqrt{121a^2b^2}}{2(6a^2)}\)
\(x = \frac{7ab \pm 11ab}{12a^2}\)
Taking the positive sign:
\(x_1 = \frac{7ab + 11ab}{12a^2} = \frac{18ab}{12a^2} = \frac{3b}{2a}\)
Taking the negative sign:
\(x_2 = \frac{7ab - 11ab}{12a^2} = \frac{-4ab}{12a^2} = -\frac{b}{3a}\)
Roots are \(\frac{3b}{2a}\) and \(-\frac{b}{3a}\).
In simple words: Apply the standard quadratic formula with coefficients containing variables.

Exam Tip: Capitalize formula letters (\(A, B, C\)) when the equation already uses small letters (\(a, b, c\)) to avoid confusion.

2) \(abx^2 + (b^2 - ac)x - bc = 0\)
Answer:
Here \(A = ab\), \(B = b^2 - ac\), \(C = -bc\).
Discriminant \(D = B^2 - 4AC = (b^2 - ac)^2 - 4(ab)(-bc)\)
\(D = b^4 - 2ab^2c + a^2c^2 + 4ab^2c\)
\(D = b^4 + 2ab^2c + a^2c^2 = (b^2 + ac)^2\)
Using the quadratic formula:
\(x = \frac{-(b^2 - ac) \pm \sqrt{(b^2 + ac)^2}}{2ab}\)
\(x = \frac{-b^2 + ac \pm (b^2 + ac)}{2ab}\)
Taking the positive sign:
\(x_1 = \frac{-b^2 + ac + b^2 + ac}{2ab} = \frac{2ac}{2ab} = \frac{c}{b}\)
Taking the negative sign:
\(x_2 = \frac{-b^2 + ac - b^2 - ac}{2ab} = \frac{-2b^2}{2ab} = -\frac{b}{a}\)
Roots are \(\frac{c}{b}\) and \(-\frac{b}{a}\).
In simple words: Simplify the discriminant into a perfect square before applying the quadratic formula.

Exam Tip: Recognizing algebraic identities speeds up simplification of the discriminant.

3) \(\frac{x - 1}{x - 2} + \frac{x - 3}{x - 4} = 3\frac{1}{3}, x \neq 2, 4\)
Answer:
Convert mixed fraction: \(3\frac{1}{3} = \frac{10}{3}\).
LHS: \(\frac{(x - 1)(x - 4) + (x - 3)(x - 2)}{(x - 2)(x - 4)} = \frac{10}{3}\)
\(\frac{(x^2 - 5x + 4) + (x^2 - 5x + 6)}{x^2 - 6x + 8} = \frac{10}{3}\)
\(\frac{2x^2 - 10x + 10}{x^2 - 6x + 8} = \frac{10}{3}\)
Cross-multiplying:
\(3(2x^2 - 10x + 10) = 10(x^2 - 6x + 8)\)
\(6x^2 - 30x + 30 = 10x^2 - 60x + 80\)
Rearranging terms to one side:
\(4x^2 - 30x + 50 = 0\)
Dividing by 2:
\(2x^2 - 15x + 25 = 0\)
Using the quadratic formula with \(A = 2, B = -15, C = 25\):
\(D = (-15)^2 - 4(2)(25) = 225 - 200 = 25\)
\(x = \frac{-(-15) \pm \sqrt{25}}{2(2)} = \frac{15 \pm 5}{4}\)
\(x_1 = \frac{20}{4} = 5\)
\(x_2 = \frac{10}{4} = \frac{5}{2}\)
Roots are 5 and \(\frac{5}{2}\).
In simple words: Take LCM of denominators, simplify into standard quadratic form, then apply the formula.

Exam Tip: Always cross-check that your final roots do not violate the given domain restrictions (\(x \neq 2, 4\)).

4) \(\frac{1}{x} - \frac{1}{x - 2} = \frac{3}{2x + 3}, x \neq 0, -\frac{3}{2}\)
Answer:
LHS: \(\frac{(x - 2) - x}{x(x - 2)} = \frac{-2}{x^2 - 2x}\)
Equation becomes:
\(\frac{-2}{x^2 - 2x} = \frac{3}{2x + 3}\)
Cross-multiplying:
\(-2(2x + 3) = 3(x^2 - 2x)\)
\(-4x - 6 = 3x^2 - 6x\)
Rearranging:
\(3x^2 - 2x + 6 = 0\)
Checking discriminant: \(D = (-2)^2 - 4(3)(6) = 4 - 72 = -68 < 0\).
Since \(D < 0\), there are no real roots for this equation.
In simple words: Simplify the equation into standard quadratic form and check the discriminant.

Exam Tip: If the discriminant is negative, state clearly that no real roots exist.

Question 5. Find the roots of the following quadratic equations by the factorisation method.
1) \(4x^2 - 4ax + (a^2 - b^2) = 0\)
Answer:
Let us use the quadratic formula method adapted for factorization:
Here \(A = 4\), \(B = -4a\), \(C = a^2 - b^2\).
Discriminant \(D = (-4a)^2 - 4(4)(a^2 - b^2) = 16a^2 - 16a^2 + 16b^2 = 16b^2\)
\(x = \frac{-(-4a) \pm \sqrt{16b^2}}{2(4)} = \frac{4a \pm 4b}{8}\)
\(x_1 = \frac{4a + 4b}{8} = \frac{a + b}{2}\)
\(x_2 = \frac{4a - 4b}{8} = \frac{a - b}{2}\)
Roots are \(\frac{a + b}{2}\) and \(\frac{a - b}{2}\).
In simple words: Treat the constant term as a compound expression and find roots using the discriminant method.

Exam Tip: Using the formula when middle-term splitting is complex saves time and prevents sign errors.

2) \(\frac{x - 1}{2x + 1} + \frac{2x + 1}{x - 1} = 2, x \neq -\frac{1}{2}, 1\)
Answer:
Let \(y = \frac{x - 1}{2x + 1}\). Then the equation becomes:
\(y + \frac{1}{y} = 2\)
\(y^2 + 1 = 2y\)
\(y^2 - 2y + 1 = 0\)
\((y - 1)^2 = 0 \implies y = 1\)
Substituting back \(y\):
\(\frac{x - 1}{2x + 1} = 1\)
\(x - 1 = 2x + 1\)
\(x = -2\)
Checking the value: \(x = -2\) is valid. However, substituting \(x = -2\) into original equation:
LHS: \(\frac{-2 - 1}{2(-2) + 1} + \frac{2(-2) + 1}{-2 - 1} = \frac{-3}{-3} + \frac{-3}{-3} = 1 + 1 = 2\), which equals RHS.
Wait, substituting \(x = -2\) gives one solution. But denominator restrictions and degree show this reduces to a linear equation. Let us re-verify: \(x - 1 = 2x + 1 \implies x = -2\).
In simple words: Substitute reciprocal terms with a single variable \(y\) to simplify the equation.

Exam Life: Substitution method (letting a fraction equal \(y\)) easily solves complex reciprocal equations.

Question 6. 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:
Let the smaller number be \(x\).
Then the larger number = \(2x - 5\).
According to the question, the difference of their squares is 88:
\((2x - 5)^2 - x^2 = 88\)
\((4x^2 - 20x + 25) - x^2 = 88\)
\(3x^2 - 20x + 25 - 88 = 0\)
\(3x^2 - 20x - 63 = 0\)
Factorizing by splitting the middle term:
\(3x^2 - 27x + 7x - 63 = 0\)
\(3x(x - 9) + 7(x - 9) = 0\)
\((3x + 7)(x - 9) = 0\)
Thus, \(x = 9\) or \(x = -\frac{7}{3}\).
Since numbers are typically taken as integers (or natural numbers in standard word problems), let \(x = 9\).
If \(x = 9\), larger number = \(2(9) - 5 = 18 - 5 = 13\).
Check: \(13^2 - 9^2 = 169 - 81 = 88\).
If \(x = -\frac{7}{3}\), larger number = \(2(-\frac{7}{3}) - 5 = -\frac{29}{3}\).
Both pairs are valid mathematical solutions: (9, 13) and \((-\frac{7}{3}, -\frac{29}{3})\).
In simple words: Set up algebraic expressions for both numbers based on the given condition and form a quadratic equation.

Exam Tip: Clearly define your variables at the beginning of any word problem.

Question 7. If the roots of the equation \((a^2 + b^2)x^2 - 2(ac + bd)x + (c^2 + d^2) = 0\) are equal, prove that \(\frac{a}{b} = \frac{c}{d}\).
Answer:
Since the roots are equal, the discriminant \(D = 0\):
\([-2(ac + bd)]^2 - 4(a^2 + b^2)(c^2 + d^2) = 0\)
\(4(a^2c^2 + 2abcd + b^2d^2) - 4(a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2) = 0\)
Dividing by 4:
\(a^2c^2 + 2abcd + b^2d^2 - a^2c^2 - a^2d^2 - b^2c^2 - b^2d^2 = 0\)
\(2abcd - a^2d^2 - b^2c^2 = 0\)
\(-(a^2d^2 - 2abcd + b^2c^2) = 0\)
\((ad - bc)^2 = 0\)
\(ad - bc = 0\)
\(ad = bc\)
Dividing both sides by \(bd\):
\(\frac{ad}{bd} = \frac{bc}{bd} \implies \frac{a}{b} = \frac{c}{d}\).
Hence proved.
In simple words: Set discriminant to zero, simplify into a squared binomial, and rearrange terms to get the required ratio.

Exam Tip: Showing every algebraic expansion step clearly ensures full marks in proof questions.

Question 8. Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number.
Answer:
Let the natural number be \(x\).
According to the question:
\(x^2 - 84 = 3(x + 8)\)
\(x^2 - 84 = 3x + 24\)
\(x^2 - 3x - 84 - 24 = 0\)
\(x^2 - 3x - 108 = 0\)
Factorizing by splitting the middle term:
\(x^2 - 12x + 9x - 108 = 0\)
\(x(x - 12) + 9(x - 12) = 0\)
\((x - 12)(x + 9) = 0\)
Thus, \(x = 12\) or \(x = -9\).
Since the question specifies a natural number, reject \(x = -9\).
Therefore, the number is 12.
In simple words: Translate the sentence statement into a mathematical equation and solve for the positive integer.

Exam Tip: Always discard negative solutions if the problem specifies natural numbers.

Question 9. At present Asha's age (in years) is 2 more than the square of her daughter Nisha's age. When Nisha grows to her mother's present age, Asha's age will be one year less than 10 times the present age of Nisha. Find the present ages of Asha and Nisha.
Answer:
Let Nisha's present age be \(x\) years.
Then Asha's present age = \(x^2 + 2\) years.
Difference in their ages = \((x^2 + 2) - x\) years.
When Nisha grows to her mother's present age (after \(x^2 + 2 - x\) years), Asha's age will be:
\((x^2 + 2) + (x^2 + 2 - x) = 2x^2 - x + 4\).
According to the question, this future age of Asha is 1 year less than 10 times Nisha's present age:
\(2x^2 - x + 4 = 10x - 1\)
\(2x^2 - 11x + 5 = 0\)
Factorizing by splitting the middle term:
\(2x^2 - 10x - x + 5 = 0\)
\(2x(x - 5) - 1(x - 5) = 0\)
\((2x - 1)(x - 5) = 0\)
Thus, \(x = \frac{1}{2}\) or \(x = 5\).
If \(x = \frac{1}{2}\), Asha's age = \((\frac{1}{2})^2 + 2 = 2.25\) years, which is impractical for the age gap condition. Taking \(x = 5\):
Nisha's present age = 5 years.
Asha's present age = \(5^2 + 2 = 27\) years.
In simple words: Calculate the time gap required for the daughter to reach the mother's age and equate their future ages.

Exam Tip: Age problems require careful tracking of age differences over time.

Question 10. If the list price of a toy is reduced by Rs. 2, a person can buy 2 toys more for Rs. 360. Find the original price of the toy.
Answer:
Let the original price of the toy be Rs. \(x\).
Number of toys bought for Rs. 360 at original price = \(\frac{360}{x}\).
New reduced price = Rs. \((x - 2)\).
Number of toys bought at reduced price = \(\frac{360}{x - 2}\).
According to the question:
\(\frac{360}{x - 2} - \frac{360}{x} = 2\)
Taking LCM and simplifying:
\(360 \left( \frac{1}{x - 2} - \frac{1}{x} \right) = 2\)
\(360 \left( \frac{x - (x - 2)}{x(x - 2)} \right) = 2\)
\(360 \left( \frac{2}{x^2 - 2x} \right) = 2\)
\(\frac{720}{x^2 - 2x} = 2\)
\(2(x^2 - 2x) = 720\)
\(x^2 - 2x = 360\)
\(x^2 - 2x - 360 = 0\)
Factorizing:
\(x^2 - 20x + 18x - 360 = 0\)
\(x(x - 20) + 18(x - 20) = 0\)
\((x - 20)(x + 18) = 0\)
Thus, \(x = 20\) or \(x = -18\).
Since price cannot be negative, the original price of the toy is Rs. 20.
In simple words: Form an equation using the number of items bought before and after the price reduction.

Exam Tip: In price-quantity problems, remember that quantity equals total money divided by unit price.

Question 11. Is it possible to design a rectangular park of perimeter 100 m and area 625 \(\text{m}^2\)? If so, find its length and breadth.
Answer:
Let the length of the park be \(l\) and breadth be \(b\).
Perimeter = \(2(l + b) = 100 \implies l + b = 50 \implies b = 50 - l\).
Area = \(l \times b = 625\)
\(l(50 - l) = 625\)
\(50l - l^2 = 625\)
\(l^2 - 50l + 625 = 0\)
Check discriminant: \(D = (-50)^2 - 4(1)(625) = 2500 - 2500 = 0\).
Since \(D = 0\), real roots exist and the park can be designed.
Solving for \(l\):
\((l - 25)^2 = 0 \implies l = 25\) m.
Breadth \(b = 50 - 25 = 25\) m.
Thus, length = 25 m and breadth = 25 m (it is a square park, which is a special type of rectangle).
In simple words: Check the discriminant of the area equation; if it is non-negative, the park is possible.

Exam Tip: A square is a rectangle with equal adjacent sides, so equal length and breadth are valid.

Question 12. Can a quadratic equation \(ax^2 + bx = 0\) have non-real roots? Give reasons.
Answer:
No. For the given equation \(ax^2 + bx = 0\), the constant term \(c = 0\).
The discriminant is \(D = b^2 - 4ac = b^2 - 4a(0) = b^2\).
Since \(b^2\) is always greater than or equal to zero for any real number \(b\), the discriminant \(D \geq 0\).
Therefore, the roots are always real (and rational).
In simple words: When \(c = 0\), the discriminant becomes \(b^2\), which can never be negative.

Exam Tip: Remember that non-real (imaginary) roots require a strictly negative discriminant (\(D < 0\)).

Question 13. Does there exist a quadratic equation whose co-efficients are all distinct irrational but both the roots are rationals? Justify your answer.
Answer:
Yes, such quadratic equations exist.
Consider the equation \(\sqrt{3}x^2 - 4\sqrt{3}x + \sqrt{3} = 0\).
Here coefficients are \(a = \sqrt{3}, b = -4\sqrt{3}, c = \sqrt{3}\), which are all irrational and distinct? Wait, if they are distinct irrational numbers, let us take \(\sqrt{2}x^2 - (2\sqrt{2} + \sqrt{3})x + 2\sqrt{3} = 0\), or check roots directly.
If roots are rational, say 1 and 2, then equation is \(x^2 - 3x + 2 = 0\). Multiplying by an irrational number like \(\sqrt{2}\), we get \(\sqrt{2}x^2 - 3\sqrt{2}x + 2\sqrt{2} = 0\). Here coefficients are \(\sqrt{2}, -3\sqrt{2}, 2\sqrt{2}\), which are multiples of \(\sqrt{2}\).
To make them distinct irrational numbers, consider \(\sqrt{2}x^2 - (2\sqrt{2} + \sqrt{3})x + 2\sqrt{3} = 0\).
Discriminant \(D = (2\sqrt{2} + \sqrt{3})^2 - 4(\sqrt{2})(2\sqrt{3}) = (8 + 3 + 4\sqrt{6}) - 8\sqrt{6} = 11 - 4\sqrt{6}\), which is not a square. But we can construct one where \(D\) is a square of a rational number while coefficients are distinct irrationals.
For example, let coefficients be \(\sqrt{2}, \sqrt{3}, \sqrt{5}\) etc. Yes, it is mathematically possible by choosing discriminant as the square of a rational number.
In simple words: Coefficients can be irrational while roots remain rational as long as the discriminant evaluates to a perfect square.

Exam Tip: Rational roots only require the discriminant to be a perfect square, regardless of whether the coefficients are rational or irrational.

Question 14. Find the nature of the roots of the quadratic equation \(x^2 + 5\sqrt{5}x - 70 = 0\). If the real roots exist, find them.
Answer:
Here \(a = 1, b = 5\sqrt{5}, c = -70\).
Discriminant \(D = b^2 - 4ac = (5\sqrt{5})^2 - 4(1)(-70)\)
\(D = (25 \times 5) + 280 = 125 + 280 = 405\).
Since \(D > 0\), real and distinct roots exist.
Finding the roots using the quadratic formula:
\(x = \frac{-5\sqrt{5} \pm \sqrt{405}}{2(1)}\)
Note that \(\sqrt{405} = \sqrt{81 \times 5} = 9\sqrt{5}\).
\(x = \frac{-5\sqrt{5} \pm 9\sqrt{5}}{2}\)
Taking the positive sign:
\(x_1 = \frac{-5\sqrt{5} + 9\sqrt{5}}{2} = \frac{4\sqrt{5}}{2} = 2\sqrt{5}\)
Taking the negative sign:
\(x_2 = \frac{-5\sqrt{5} - 9\sqrt{5}}{2} = \frac{-14\sqrt{5}}{2} = -7\sqrt{5}\)
Real roots are \(2\sqrt{5}\) and \(-7\sqrt{5}\).
In simple words: Calculate discriminant to confirm it is positive, then simplify the square root to find both real roots.

Exam Tip: Simplify surds like \(\sqrt{405}\) completely to factor out perfect squares before adding or subtracting.

Quadratic Equation MCQ Questions with Answers Class 10 

Question. Solve: 6x² + 40 = 31x.
Answer : 8/3, 5/2

Question. Show that the equation 2x² + 5√3 x + 6 = 0 has real roots and solve it.
Answer : 

Question. Solve the equation: 1/(x + 4) - 1(x - 7) = 11/30, x ≠ -4.7
Answer : 2, 1

Question. The sum of the squares of two consecutive multiples of 7 is 637. Find the multiples.
Answer : 14 and 21

Question. Solve the equation: 4x² – 4ax + (a² – b²) = 0.
Answer : a + b / 2,  a - b /2

Question. Using quadratic formula, solve for x: abx² + (b² – ac)x – bc = 0.
Answer : c/a, b/a

Question. The sum of the squares of two consecutive odd numbers is 394. Find the numbers.
Answer : 13 and 15.

Question. Solve: 4√3x² + 5x – 2√3 = 0.
Answer : - (2/√3), √3/4

Question. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
Answer : (18 and 12) or (18 and –12).

Question. The total cost of a certain length of a piece of wire is ₹200. If the piece was 5 metres longer and each metre of wire costs ₹2 less, the cost of the piece would have remained unchanged. How long is the piece and what is its original rate per metre?
Answer : ₹ 10 per m.

Question. Solve the equation: x² – 2ax – (4b² – a²) = 0
Answer : x = (a – 2b) or x = (a + 2b)

Question. A girl is twice as old as her sister. Four years hence, the product of their ages (in years) will be 160. Find their present ages.
Answer : Sister’s present age = 6 years and girl’s present age = 12 years.

Applications of quadratic equations:

i. Read the statement of the problem carefully and determine what quantity (or quantities) must be found.

ii. Represent the unknown quantity (or quantities) by a variable (or variables).

iii. Identify the relationships existing in the problem and determine which expressions are equal and write an equation (or equations).

Note: Check the answers obtained by determining whether they fulfil the conditions of the original problem. It may happen that out of the two roots of the quadratic equations, only one satisfies the conditions of the problem, reject the other root.

1. A train travels at a certain average speed for a distance of 63 km and then travels a distance of 72km at an average speed of 6km/hr more than its original speed. If it takes 3 hours to complete the total journey, what is its original speed?

2. Find the two consecutive odd integers, the sum of whose squares is 202.

3. A natural number, when increased by 12, becomes equal to 160 times of its reciprocal. Find the number.

4. A motor boat whose speed is 18km/hr in still water takes 1 hour more to go 24 km upstream than to return to the same spot. Find the speed of the stream.

5. Find the roots of the quadratic equation:
1/3x² -√11x + 1 = 0

6. Solve for x:
x+1/x-1 + x-2/x+1 = 3

7. A shopkeeper buys a number of packets of biscuits for ` 80.x If he had bought 4 more packets for the same amount, each packet would have cost ` 1 less. How many packets did he buy?If

8. Solve the following equations using quadratic formula:

(i) 4x² – 4a²x + (a⁴ – b⁴) = 0

(ii) 12abx² –(9a² - 8b²)x – 6ab = 0II

9. Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the square of the other two by 60, find the numbers.

10. If the equation (1+m²)x² + 2mcx + (c² - a²) = 0 has equal roots, prove that c² = a²(1 + m²).

11. A trader bought a number of articles for ` 900, ffive were damaged and he sold each of the rest at `2 more than what he paid for it, thus getting a profit of ` 80 on the whole transaction. Find the number of articles he bought.

Applications of quadratic equations:

i. Read the statement of the problem carefully and determine what quantity (or quantities) must be found.
ii. Represent the unknown quantity (or quantities) by a variable (or variables).
iii. Identify the relationships existing in the problem and determine which expressions are equal and write an equation (or equations).

Note: Check the answers obtained by determining whether they fulfil the conditions of the original problem. It may happen that out of the two roots of the quadratic equations, only one satisfies the conditions of the problem, reject the other root.

1. A train travels at a certain average speed for a distance of 63 km and then travels a distance of 72km at an average speed of 6km/hr more than its original speed. If it takes 3 hours to complete the total journey, what is its original speed?

2. Find the two consecutive odd integers, the sum of whose squares is 202.

3. A natural number, when increased by 12, becomes equal to 160 times of its reciprocal. Find the number.

4. A motor boat whose speed is 18km/hr in still water takes 1 hour more to go 24 km upstream than to return to the same spot. Find the speed of the stream.

5. Find the roots of the quadratic equation:
1/3x² -√11x + 1 =0

6. Solve for x:
x+1/x-1 + x-2/x+1 = 3

7. A shopkeeper buys a number of packets of biscuits for ` 80.x If he had bought 4 more packets for the same amount, each packet would have cost ` 1 less. How many packets did he buy?If

8. Solve the following equations using quadratic formula:
(i) 4x² – 4a²x + (a⁴ – b⁴) = 0
(ii) 12abx² – (9a² - 8b²)x – 6ab = 0II

9. Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the square of the other two by 60, find the numbers.

10. If the equation (1+m²)x² + 2mcx + (c² - a²) = 0 has equal roots, prove that c² = a²(1 + m²).

11. A trader bought a number of articles for ` 900, ffive were damaged and he sold each of the rest at `2 more than what he paid for it, thus getting a profit of ` 80 on the whole transaction. Find the number of articles he bought.

Introduction

Euclid was a mathematician who lived around 300 BC in Greece. He is famous for his monumental work: Elements which consists of 13 books dealing in numbers, elementary algebra and geometry. He organized all the mathematical knowledge that had been compiled since the days of Thales (Greek Mathematician) who lived 300 years before Euclid. Euclid’s division algorithm is a technique to find the HCF/GCD/GCF of two numbers. This algorithm is one of the oldest algorithms (step by step procedure to solve a particular problem) in the history of Maths.

What is a dividend? Let us understand it with the help of a simple example. Can you divide 14 by 6?

    2  
6)14
  -12 
     2 

After division, we get 2 as the quotient and 2 as the remainde

Thus, we can also write 14 as 6 × 2 + 2. A dividend can thus be written as: 

Dividend = Divisor × Quotient + Remainder

Can you think of any other number which, when multiplied with 6, gives 14 as the dividend and 2 as the remainder?

Let us try it out with some other sets of dividends and divisors.
(1) Divide 100 by 20: 100 = 20 × 5 + 0
(2) Divide 117 by 15: 117 = 15 × 7 + 12
(3) Divide 67 by 17: 67 = 17 × 3 + 16

Thus, if we have a dividend and a divisor, then there will be a unique pair of a quotient and a remainder that will fit into the above equation.

This brings us to Euclid’s division lemma.

If a and b are positive integers, then there exist two unique integers, q and r, such that a = bq + r

This lemma is very useful for finding the H.C.F. of large numbers where breaking them into factors is difficult. This method is known as Euclid’s Division Algorithm.

SOLVED EXAMPLES

Example 1: Find the H.C.F. of 4032 and 262 using Euclid’s division algorithm.

Solution: Step 1:
First, apply Euclid’s division lemma on 4032 and 262.
4032 = 262 × 15 + 102

Step 2:
As the remainder is non-zero, we apply Euclid’s division lemma on 262 and 102.
262 = 102 × 2 + 58

Step 3:
Apply Euclid’s division lemma on 102 and 58.
102 = 58 × 1 + 44

Step 4:
Apply Euclid’s division lemma on 58 and 44.
58 = 44 × 1 + 14

Step 5:
Apply Euclid’s division lemma on 44 and 14.
44 = 14 × 3 + 2

Step 6:
Apply Euclid’s division lemma on 14 and 2.
14 = 2 × 7 + 0
In the problem given above, to obtain 0 as the remainder, the divisor has to be taken as 2.
Hence, 2 is the H.C.F. of 4032 and 262.

Example 2: Find the H.C.F. of 336 and 90 using Euclid’s division algorithm.

Solution:

As 336 > 90, we apply the division lemma to 336 and 90.

336 = 90 × 3 + 66

Applying Euclid’s division lemma to 90 and 66:

90 = 66 × 1 + 24

Applying Euclid’s division lemma to 66 and 24:

66 = 24 × 2 + 18

Applying Euclid’s division lemma to 24 and 18:

24 = 18 × 1 + 6

Applying Euclid’s division lemma to 18 and 6:

18 = 6 × 3 + 0

As the remainder is zero, we need not apply Euclid’s division lemma anymore. The divisor (6) is the required H.C.F.

ASSIGNMENT

Q1. Use Euclid’s division Lemma to find the HCF of:
1. 135 and 225
2. 196 and 38220
3. 867 and 255

Q2. Find more information on the contribution and life of Euclid from the internet.