CBSE Class 10 Mathematics Quadratic Equations Assignment Set 06

For the things of this world cannot be made known without a knowledge of mathematics.

Question 1. Solve by factorization
a. \(4x^2 - 4a^2x + (a^4 - b^4) = 0\)
Answer: Given equation is \(4x^2 - 4a^2x + (a^4 - b^4) = 0\)
Rewrite the constant term as a product of factors: \(a^4 - b^4 = (a^2 - b^2)(a^2 + b^2)\)
Split the middle term using \(2(a^2 + b^2) + 2(a^2 - b^2) = 4a^2\):
\(4x^2 - [2(a^2 + b^2) + 2(a^2 - b^2)]x + (a^2 - b^2)(a^2 + b^2) = 0\)
Group the terms:
\(2x[2x - (a^2 + b^2)] - (a^2 - b^2)[2x - (a^2 + b^2)] = 0\)
\([2x - (a^2 + b^2)][2x - (a^2 - b^2)] = 0\)
Setting each factor to zero:
\(2x - (a^2 + b^2) = 0 \implies x = \frac{a^2 + b^2}{2}\)
\(2x - (a^2 - b^2) = 0 \implies x = \frac{a^2 - b^2}{2}\)
In simple words: We split the middle term of the quadratic equation by breaking down the coefficients and factoring by grouping.

Exam Tip: Recognize \(a^4 - b^4\) as a difference of squares to simplify factoring.

Question 1. b. \(x^2 + \left(\frac{a}{a+b} + \frac{a+b}{a}\right)x + 1 = 0\)
Answer: Given equation is \(x^2 + \left(\frac{a}{a+b} + \frac{a+b}{a}\right)x + 1 = 0\)
Expand the middle term:
\(x^2 + \left(\frac{a}{a+b}\right)x + \left(\frac{a+b}{a}\right)x + 1 = 0\)
Factor by grouping:
\(x\left(x + \frac{a}{a+b}\right) + \frac{a+b}{a}\left(x + \frac{a}{a+b}\right) = 0\)
\(\left(x + \frac{a}{a+b}\right)\left(x + \frac{a+b}{a}\right) = 0\)
Setting each factor to zero:
\(x = -\frac{a}{a+b}\) or \(x = -\frac{a+b}{a}\), where \(a \neq 0\) and \(a+b \neq 0\).
In simple words: Expand the bracket containing the sum of fractions, then factor by grouping the terms in pairs.

Exam Tip: Notice that the product of the two fractions inside the coefficient is 1, which matches the constant term.

Question 1. c. \(\frac{1}{a+b+x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x}\), where \(a+b \neq 0\)
Answer: Given equation is \(\frac{1}{a+b+x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x}\)
Transpose \(\frac{1}{x}\) to the left side:
\(\frac{1}{a+b+x} - \frac{1}{x} = \frac{1}{a} + \frac{1}{b}\)
Take LCM on both sides:
\(\frac{x - (a+b+x)}{x(a+b+x)} = \frac{b+a}{ab}\)
\(\frac{-(a+b)}{x(a+b+x)} = \frac{a+b}{ab}\)
Cancel \((a+b)\) from both sides (since \(a+b \neq 0\)):
\(\frac{-1}{x(a+b+x)} = \frac{1}{ab}\)
Cross-multiply:
\(x(a+b+x) = -ab\)
\(x^2 + ax + bx + ab = 0\)
Factor by grouping:
\(x(x+a) + b(x+a) = 0\)
\((x+a)(x+b) = 0\)
\(x = -a\) or \(x = -b\).
In simple words: Move the term with \(x\) to the same side of the equation to simplify subtraction before cross-multiplying.

Exam Tip: Always group terms with identical denominators together to simplify rational equations quickly.

Question 1. d. \((x - 3)(x - 4) = \frac{34}{33^2}\)
Answer: Given equation is \((x - 3)(x - 4) = \frac{34}{33^2}\)
Expand the left side:
\(x^2 - 7x + 12 = \frac{34}{1089}\)
\(x^2 - 7x + 12 - \frac{34}{1089} = 0\)
\(x^2 - 7x + \frac{13043}{1089} = 0\)
Split the middle term \(7x\) into \(\frac{98}{33}x\) and \(\frac{133}{33}x\):
\(x^2 - \frac{98}{33}x - \frac{133}{33}x + \frac{13043}{1089} = 0\)
Factor by grouping:
\(x\left(x - \frac{98}{33}\right) - \frac{133}{33}\left(x - \frac{98}{33}\right) = 0\)
\(\left(x - \frac{98}{33}\right)\left(x - \frac{133}{33}\right) = 0\)
\(x = \frac{98}{33}\) or \(x = \frac{133}{33}\).
In simple words: Expand the binomials first, then use fractions to find the factors that satisfy the quadratic formula or splitting method.

Exam Tip: Be careful with large denominators; splitting fractions method works best when matching numerators to factors.

Question 1. e. \(x = \frac{1}{2 - \frac{1}{2 - \frac{1}{2 - x}}}\), where \(x \neq 2\)
Answer: Given continued fraction equation is \(x = \frac{1}{2 - \frac{1}{2 - \frac{1}{2 - x}}}\)
Simplify the innermost denominator fraction:
\(2 - \frac{1}{2 - x} = \frac{2(2-x) - 1}{2-x} = \frac{4 - 2x - 1}{2-x} = \frac{3 - 2x}{2-x}\)
Substitute this back into the expression:
\(x = \frac{1}{2 - \frac{2-x}{3-2x}}\)
Simplify the denominator:
\(2 - \frac{2-x}{3-2x} = \frac{2(3-2x) - (2-x)}{3-2x} = \frac{6 - 4x - 2 + x}{3-2x} = \frac{4 - 3x}{3-2x}\)
Substitute this back:
\(x = \frac{3-2x}{4-3x}\)
Cross-multiply:
\(x(4 - 3x) = 3 - 2x\)
\(4x - 3x^2 = 3 - 2x\)
Rearrange into standard quadratic form:
\(3x^2 - 6x + 3 = 0\)
Divide by 3:
\(x^2 - 2x + 1 = 0\)
\((x - 1)^2 = 0\)
\(x = 1, 1\).
In simple words: Simplify the continued fraction level by level from the bottom up, then solve the resulting quadratic equation.

Exam Tip: Always work from the innermost denominator outward to avoid algebraic errors in complex fractions.

Question 2. By the method of completion of squares show that the equation \(4x^2 + 3x + 5 = 0\) has no real roots.
Answer: Given equation is \(4x^2 + 3x + 5 = 0\)
Divide the entire equation by 4 to make the coefficient of \(x^2\) equal to 1:
\(x^2 + \frac{3}{4}x + \frac{5}{4} = 0\)
Shift the constant term to the right side:
\(x^2 + \frac{3}{4}x = -\frac{5}{4}\)
Add the square of half of the coefficient of \(x\) (which is \(\left(\frac{1}{2} \times \frac{3}{4}\right)^2 = \left(\frac{3}{8}\right)^2 = \frac{9}{64}\)) to both sides:
\(x^2 + \frac{3}{4}x + \frac{9}{64} = -\frac{5}{4} + \frac{9}{64}\)
Rewrite the left side as a complete square and simplify the right side:
\(\left(x + \frac{3}{8}\right)^2 = \frac{-80 + 9}{64}\)
\(\left(x + \frac{3}{8}\right)^2 = -\frac{71}{64}\)
Taking the square root on both sides:
\(x + \frac{3}{8} = \pm \sqrt{-\frac{71}{64}}\)
Since the square root of a negative number is not a real number, the equation has no real roots.
In simple words: Completing the square results in a negative value on the right side, which means no real solutions exist.

Exam Tip: A negative number on the right-hand side of a squared binomial means the discriminant is negative and roots are non-real.

Question 3. The sum of areas of two squares is \(468\text{ m}^2\). If the difference of their perimeters is \(24\text{ cm}\), find the sides of the two squares.
Answer: Let the side of the larger square be \(x \text{ m}\) and the side of the smaller square be \(y \text{ m}\).
From the first condition (sum of areas):
\(x^2 + y^2 = 468 \quad \text{--- (Equation 1)}\)
From the second condition (difference of perimeters):
\(4x - 4y = 24\)
Divide by 4:
\(x - y = 6 \implies x = y + 6 \quad \text{--- (Equation 2)}\)
Substitute Equation 2 into Equation 1:
\((y + 6)^2 + y^2 = 468\)
\(y^2 + 12y + 36 + y^2 = 468\)
\(2y^2 + 12y + 36 - 468 = 0\)
\(2y^2 + 12y - 432 = 0\)
Divide by 2:
\(y^2 + 6y - 216 = 0\)
Factor the quadratic equation:
\((y + 18)(y - 12) = 0\)
\(y = 12\) (since side length cannot be negative, \(y = -18\) is rejected).
Substitute \(y = 12\) into Equation 2:
\(x = 12 + 6 = 18 \text{ m}\)
Therefore, the sides of the two squares are \(18\text{ m}\) and \(12\text{ m}\).
In simple words: Set up two equations based on area and perimeter formulas, substitute one into the other, and solve the quadratic equation.

Exam Tip: Always check that side lengths are positive; reject any negative values obtained from the quadratic roots.

Question 4. A dealer sells a toy for Rs.24 and gains as much percent as the cost price of the toy. Find the cost price of the toy.
Answer: Let the cost price (C.P.) of the toy be Rs. \(x\).
Given that the gain percentage equals the cost price, so Gain = \(x\%\).
Amount of Gain = \(\frac{x}{100} \times x = \frac{x^2}{100}\)
We know that:
\(\text{Selling Price (S.P.)} = \text{C.P.} + \text{Gain}\)
\(24 = x + \frac{x^2}{100}\)
Multiply the entire equation by 100:
\(2400 = 100x + x^2\)
Rearrange into standard quadratic form:
\(x^2 + 100x - 2400 = 0\)
Factor the quadratic equation:
\((x + 120)(x - 20) = 0\)
\(x = 20\) or \(x = -120\)
Since cost price cannot be negative, \(x = -120\) is rejected.
Therefore, the cost price of the toy is Rs. 20.
In simple words: Express the selling price in terms of cost price and percentage gain, then solve the resulting quadratic equation.

Exam Tip: Remember that percentage gain is calculated on the cost price, so Gain = \((\text{C.P.} \times \text{Percentage}) / 100\).

Question 5. A fox and an eagle lived at the top of a cliff of height \(6\text{ m}\), whose base was at a distance of \(10\text{ m}\) from a point A on the ground. The fox descends the cliff and went straight to the point A. The eagle flew vertically up to a height \(x\) metres and then flew in a straight line to a point A, the distance traveled by each being the same. Find the value of \(x\).
Answer: Let the height of the cliff be \(6\text{ m}\) and the horizontal distance to point A be \(10\text{ m}\).
The distance traveled by the fox from the top of the cliff to point A (using Pythagoras theorem in the base triangle):
\(\text{Distance of fox} = 6 + 10 = 16\text{ m}\) (down the cliff then along the ground, or directly along the hypotenuse if interpreted as descending the cliff face of height 6m and walking 10m).
According to the problem text and standard geometric setup:
Height of cliff = \(6\text{ m}\), base = \(10\text{ m}\). Hypotenuse of cliff base triangle = \(\sqrt{6^2 + 10^2} = \sqrt{36 + 100} = \sqrt{136}\).
Wait, reading the exact equation from the source solution:
Distance traveled by the fox = \(6 + 10 = 16\text{ m}\) (descends cliff vertically by 6m, then walks 10m horizontally to A).
The eagle flies vertically up by height \(x\), so its height from the ground is \(6 + x\). Then it flies to point A.
Distance traveled by the eagle = (vertical distance up \(x\)) + (distance from new height to point A) = \(x + \sqrt{(6+x)^2 + 10^2}\)... Wait, looking at the source OCR solution:
\((6+x)^2 + (10)^2 = (16 - x)^2\)
Solving this equation:
\(36 + 12x + x^2 + 100 = 256 - 32x + x^2\)
\(136 + 12x = 256 - 32x\)
\(12x + 32x = 256 - 136\)
\(44x = 120\)
\(x = \frac{120}{44} = \frac{30}{11} \approx 2.72\text{ m}\).
Therefore, the value of \(x\) is \(2.72\text{ m}\).
In simple words: Equate the total travel distance of the fox and the eagle, then solve the linear equation for \(x\).

Exam Tip: Carefully set up total path lengths by breaking them into vertical and horizontal components.

Question 6. A lotus is \(2\text{ m}\) above the water in a pond. Due to wind the lotus slides on the side and only the stem completely submerges in the water at a distance of \(10\text{ m}\) from the original position. Find the depth of water in the pond.
Answer: Let the depth of the water be \(x\text{ m}\).
The height of the lotus above water is \(2\text{ m}\), so the total length of the stem is \((x + 2)\text{ m}\).
When the lotus is pushed by the wind, it forms a right-angled triangle where:
- Depth of water = \(x\) (perpendicular)
- Distance from original position = \(10\text{ m}\) (base)
- Length of stem = \(x + 2\) (hypotenuse)
Using Pythagoras theorem:
\((x + 2)^2 = x^2 + 10^2\)
\(x^2 + 4x + 4 = x^2 + 100\)
Subtract \(x^2\) from both sides:
\(4x + 4 = 100\)
\(4x = 96\)
\(x = 24\text{ m}\)
Therefore, the depth of the water in the pond is \(24\text{ m}\).
In simple words: Apply the Pythagorean theorem to the right-angled triangle formed by the water depth, displacement distance, and the stem length.

Exam Tip: Remember that the hypotenuse is the total length of the stem, which is depth plus the height above water \((x+2)\).

Question 7. Solve \(x = \sqrt{6 + \sqrt{6 + \sqrt{6 + \dots }}}\)
Answer: Given equation is \(x = \sqrt{6 + \sqrt{6 + \sqrt{6 + \dots }}}\)
Since the expression inside the outer square root repeats infinitely, we can write:
\(x = \sqrt{6 + x}\)
Square both sides:
\(x^2 = 6 + x\)
Rearrange into standard quadratic form:
\(x^2 - x - 6 = 0\)
Factor the quadratic equation:
\((x - 3)(x + 2) = 0\)
\(x = 3\) or \(x = -2\)
Since square roots yield non-negative values, \(x = -2\) is rejected.
Therefore, \(x = 3\).
In simple words: Substitute the repeating infinite radical part with \(x\), square both sides, and solve the resulting quadratic equation.

Exam Tip: For nested radical problems of the form \(\sqrt{N + \sqrt{N + \dots}}\), find two consecutive integers whose product is \(N\); the larger integer is always the positive root.

Question 8. The hypotenuse of a right triangle is \(20\text{ m}\). If the difference between the length of the other sides is \(4\text{ m}\), find the sides.
Answer: Let the two legs of the right-angled triangle be \(x\) and \(y\), with \(x > y\).
Given hypotenuse = \(20\text{ m}\).
Using Pythagoras theorem:
\(x^2 + y^2 = 20^2\)
\(x^2 + y^2 = 400 \quad \text{--- (Equation 1)}\)
Given the difference between the sides is \(4\text{ m}\):
\(x - y = 4 \implies x = y + 4 \quad \text{--- (Equation 2)}\)
Substitute Equation 2 into Equation 1:
\((y + 4)^2 + y^2 = 400\)
\(y^2 + 8y + 16 + y^2 = 400\)
\(2y^2 + 8y + 16 - 400 = 0\)
\(2y^2 + 8y - 384 = 0\)
Divide by 2:
\(y^2 + 4y - 192 = 0\)
Factor the quadratic equation:
\((y + 16)(y - 12) = 0\)
\(y = 12\) (since side length cannot be negative, \(y = -16\) is rejected).
Substitute \(y = 12\) into Equation 2:
\(x = 12 + 4 = 16 \text{ m}\)
Therefore, the lengths of the other sides are \(12\text{ m}\) and \(16\text{ m}\).
In simple words: Combine the Pythagorean theorem with the difference between the two sides to form a quadratic equation.

Exam Tip: Always verify your side lengths using \(a^2 + b^2 = c^2\) at the end of the problem.

Question 9. Find the positive value of \(k\) for which \(x^2 + kx + 64 = 0\) and \(x^2 - 8x + k = 0\) will have real roots.
Answer: For a quadratic equation \(ax^2 + bx + c = 0\) to have real roots, its discriminant must be greater than or equal to zero (\(b^2 - 4ac \geq 0\)).
For the first equation \(x^2 + kx + 64 = 0\):
\(k^2 - 4(1)(64) \geq 0\)
\(k^2 - 256 \geq 0\)
\(k^2 \geq 256\)
\(k \geq 16\) or \(k \leq -16 \quad \text{--- (Condition 1)}\)
For the second equation \(x^2 - 8x + k = 0\):
\((-8)^2 - 4(1)(k) \geq 0\)
\(64 - 4k \geq 0\)
\(4k \leq 64\)
\(k \leq 16 \quad \text{--- (Condition 2)}\)
Combining Condition 1 and Condition 2, the only common value is \(k = 16\).
In simple words: Set the discriminant \(b^2 - 4ac \geq 0\) for both equations and find the common value of \(k\).

Exam Tip: Real roots require the discriminant to be non-negative (\(\Delta \geq 0\)).

Question 10. A teacher on attempting to arrange the students for mass drill in the form of a solid square found that 24 students were left over. When he increased the size of the square by one student he found he was short of 25 students. Find the number of students.
Answer: Let the side of the original square arrangement be \(x\).
Number of students = \(x^2 + 24\)
When the side of the square is increased by 1, the new side is \(x + 1\).
Number of students required for the new square = \((x + 1)^2 - 25\)
Since the total number of students remains constant:
\(x^2 + 24 = (x + 1)^2 - 25\)
\(x^2 + 24 = x^2 + 2x + 1 - 25\)
\(x^2 + 24 = x^2 + 2x - 24\)
Subtract \(x^2\) from both sides:
\(24 = 2x - 24\)
\(2x = 48\)
\(x = 24\)
Side of the square = 24.
Total number of students = \(x^2 + 24 = 24^2 + 24 = 576 + 24 = 600\).
Therefore, the total number of students is 600.
In simple words: Equate the two expressions for the total number of students to find the side length of the square.

Exam Tip: "Left over" means add to the square area, while "short of" means subtract from the larger square area.

Question 11. A pole has to be erected at a point on the boundary of a circular park of diameter \(13\text{ m}\) in such a way that the differences of its distances from two diametrically opposite fixed gates A & B on the boundary is \(7\text{ m}\). Is it possible to do so? If answer is yes at what distances from the two gates should the pole be erected.
Answer: Let the diameter of the circular park be \(AB = 13\text{ m}\). Let P be the position of the pole on the boundary.
Since AB is the diameter, angle APB is a right angle (\(\angle APB = 90^\circ\)) by Thales's theorem.
Let the distance from gate B to the pole be \(BP = x\).
Given the difference between distances from the two gates is \(7\text{ m}\):
\(AP - BP = 7 \implies AP = x + 7\)
In the right-angled triangle APB, using Pythagoras theorem:
\(AP^2 + BP^2 = AB^2\)
\((x + 7)^2 + x^2 = 13^2\)
\(x^2 + 14x + 49 + x^2 = 169\)
\(2x^2 + 14x + 49 - 169 = 0\)
\(2x^2 + 14x - 120 = 0\)
Divide by 2:
\(x^2 + 7x - 60 = 0\)
Factor the quadratic equation:
\((x + 12)(x - 5) = 0\)
\(x = 5\) (since distance cannot be negative, \(x = -12\) is rejected).
So, \(BP = 5\text{ m}\) and \(AP = 5 + 7 = 12\text{ m}\).
Yes, it is possible, and the pole should be erected at a distance of \(5\text{ m}\) from gate B and \(12\text{ m}\) from gate A.
In simple words: Use the property that an angle inscribed in a semicircle is a right angle, then apply the Pythagorean theorem.

Exam Tip: The diameter subtends a right angle at any point on the circumference of a circle.

Question 12. If the roots of the equation \((a-b)x^2 + (b-c)x + (c-a) = 0\) are equal. Prove that \(2a = b + c\).
Answer: Given quadratic equation is \((a-b)x^2 + (b-c)x + (c-a) = 0\)
Since the roots are equal, the discriminant must be zero (\(B^2 - 4AC = 0\)):
Here, \(A = a-b\), \(B = b-c\), and \(C = c-a\).
\((b-c)^2 - 4(a-b)(c-a) = 0\)
Expand both terms:
\(b^2 - 2bc + c^2 - 4(ac - a^2 - bc + ab) = 0\)
\(b^2 - 2bc + c^2 - 4ac + 4a^2 + 4bc - 4ab = 0\)
Combine like terms:
\(4a^2 + b^2 + c^2 - 4ab + 2bc - 4ac = 0\)
Notice that this expression can be written as a complete square using the identity \((2a - b - c)^2 = 0\):
\((2a - (b + c))^2 = 0\)
Taking the square root on both sides:
\(2a - b - c = 0\)
\(2a = b + c\).
Hence proved.
In simple words: Set the discriminant to zero for equal roots and simplify the algebraic identity to prove the relation.

Exam Tip: Equal roots mean \(b^2 - 4ac = 0\); look for algebraic expansion identities to factor the resulting expression.

Question 13. X and Y are centers of circles of radius \(9\text{ cm}\) and \(2\text{ cm}\) and \(XY = 17\text{ cm}\). Z is the centre of a circle of radius \(4\text{ cm}\), which touches the above circles externally. Given that \(\angle XZY = 90^\circ\), write an equation in \(r\) and solve it for \(r\).
Answer: Let the radius of the third circle with center Z be \(r\) (or given as \(4\text{ cm}\) in the text, but let's follow the OCR equation setup where radii are related).
Given radii of circles at X and Y are \(9\text{ cm}\) and \(2\text{ cm}\), and distance \(XY = 17\text{ cm}\).
Let \(r\) be the radius of circle Z. The distances from Z to X and Y for externally touching circles are:
\(XZ = 9 + r\)
\(YZ = 2 + r\) (or matching the OCR setup where \(Z\) has radius 4, but here variable \(r\) is solved).
Using Pythagoras theorem in right-angled triangle \(XZY\) (\(\angle XZY = 90^\circ\)):
\(XZ^2 + YZ^2 = XY^2\)
\((r + 9)^2 + (r + 2)^2 = 17^2\)
\(r^2 + 18r + 81 + r^2 + 4r + 4 = 289\)
\(2r^2 + 22r + 85 - 289 = 0\)
\(2r^2 + 22r - 204 = 0\)
Divide by 2:
\(r^2 + 11r - 102 = 0\)... Wait, checking the OCR factors: \((r + 17)(r - 6) = 0\), which gives \(r^2 + 11r - 102 = 0\).
Factor the quadratic equation:
\((r + 17)(r - 6) = 0\)
\(r = 6\) (since radius cannot be negative, \(r = -17\) is rejected).
Therefore, the radius \(r = 6\text{ cm}\).
In simple words: Apply the Pythagorean theorem to the triangle formed by the centers of the three mutually touching circles.

Exam Tip: When circles touch externally, the distance between their centers is the sum of their radii.