CBSE Class 10 Mathematics Quadratic Equations Assignment Set 05

Level - 1 (01 Marks Question)

Q.1 Check whether the following are quadratic equation or not
i) \((x - 3)(2x + 1) = x(x + 5)\)
Answer: Expanding both sides:
\(2x^2 + x - 6x - 3 = x^2 + 5x\)
\(2x^2 - 5x - 3 = x^2 + 5x\)
\(x^2 - 10x - 3 = 0\)
Since it is of the form \(ax^2 + bx + c = 0\) where \(a \neq 0\), it is a quadratic equation.
In simple words: Simplify both sides and check if the highest power of the variable is 2.

Exam Tip: Always expand completely and bring all terms to one side before deciding.

ii) \((x + 2)^3 = 2x(x^2 - 1)\)
Answer: Expanding both sides using \((a+b)^3 = a^3 + b^3 + 3a^2b + 3ab^2\):
\(x^3 + 8 + 6x^2 + 12x = 2x^3 - 2x\)
Bringing all terms to the right side:
\(x^3 - 6x^2 - 14x - 8 = 0\)
Since the highest power is 3, it is a cubic equation, not a quadratic equation.
In simple words: The highest exponent is 3, so it is not a quadratic equation.

Exam Tip: Do not judge by initial appearance; always expand parentheses fully.

Q.2 Solve by factorisation method
\(x^2 - 7x + 12 = 0\)
Answer: Split the middle term \(-7x\) into \(-4x\) and \(-3x\):
\(x^2 - 4x - 3x + 12 = 0\)
\(x(x - 4) - 3(x - 4) = 0\)
\((x - 4)(x - 3) = 0\)
Setting each factor to zero:
\(x - 4 = 0 \implies x = 4\)
\(x - 3 = 0 \implies x = 3\)
In simple words: Find two numbers that multiply to 12 and add up to -7. These are -4 and -3.

Exam Tip: Always verify your roots by substituting them back into the original equation.

Q.3 Find the discriminant
\(x^2 - 3x - 10 = 0\)
Answer: Comparing with \(ax^2 + bx + c = 0\), we get \(a = 1\), \(b = -3\), \(c = -10\).
Discriminant \(D = b^2 - 4ac\)
\(D = (-3)^2 - 4(1)(-10)\)
\(D = 9 + 40 = 49\)
In simple words: Substitute coefficients into the formula \(b^2 - 4ac\).

Exam Tip: Be careful with negative signs when calculating \(4ac\).

Q.4 Find the nature of root
\(2x^2 + 3x - 4 = 0\)
Answer: Here \(a = 2\), \(b = 3\), \(c = -4\).
Discriminant \(D = b^2 - 4ac\)
\(D = 3^2 - 4(2)(-4)\)
\(D = 9 + 32 = 41\)
Since \(D > 0\), the roots are real and distinct.
In simple words: A positive discriminant means two different real roots.

Exam Tip: Remember: \(D > 0\) (real and distinct), \(D = 0\) (real and equal), \(D < 0\) (no real roots).

Q.5 Find the value k so that quadratic equation \(3x^2 - kx + 27 = 0\) has equal root
Answer: For equal roots, the discriminant must be zero: \(D = 0\).
Here \(a = 3\), \(b = -k\), \(c = 27\).
\(D = b^2 - 4ac = 0\)
\((-k)^2 - 4(3)(27) = 0\)
\(k^2 - 324 = 0\)
\(k^2 = 324\)
\(k = \pm 18\)
In simple words: Set the discriminant to zero and solve for k.

Exam Tip: Do not forget both positive and negative values when taking square roots (\(\pm\)).

Q.6 Determine whether given value of x is a solution or not
(1) \(x^2 - 3x - 1 = 0\) ; \(x = 1\)
Answer: Substitute \(x = 1\) into the LHS of the equation:
LHS = \((1)^2 - 3(1) - 1\)
LHS = \(1 - 3 - 1 = -3\)
Since LHS \(\neq\) RHS (0), \(x = 1\) is not a solution.
In simple words: Plug the value of x into the equation. If it equals zero, it is a solution.

Exam Tip: Clearly show the substitution step and state whether LHS equals RHS.

Level 2 (02 Marks)

Example 1 Solve the quadratic equation \(9x^2 - 3x - 20 = 0\) by factorisation method and find the roots of quadratic equation.
Answer: Given equation: \(9x^2 - 3x - 20 = 0\)
Multiply \(a\) and \(c\): \(9 \times (-20) = -180\). We need two numbers that multiply to \(-180\) and add to \(-3\). These are \(-15\) and \(12\).
\(9x^2 - 15x + 12x - 20 = 0\)
\(3x(3x - 5) + 4(3x - 5) = 0\)
\((3x - 5)(3x + 4) = 0\)
Either \(3x - 5 = 0\) or \(3x + 4 = 0\)
\(3x = 5 \implies x = \frac{5}{3}\)
\(3x = -4 \implies x = \frac{-4}{3}\)
In simple words: Split the middle term into two parts whose product is \(9 \times (-20)\) and sum is \(-3\).

Exam Tip: For equations where \(a \neq 1\), always find factors of the product \(ac\).

Q.1 Solve by quadratic equation \(16x^2 - 24x - 1 = 0\) by using quadratic formula.
Answer: Here \(a = 16\), \(b = -24\), \(c = -1\).
Discriminant \(D = b^2 - 4ac\)
\(D = (-24)^2 - 4(16)(-1)\)
\(D = 576 + 64 = 640\)
Quadratic formula: \(x = \frac{-b \pm \sqrt{D}}{2a}\)
\(x = \frac{-(-24) \pm \sqrt{640}}{2(16)}\)
\(x = \frac{24 \pm 8\sqrt{10}}{32}\)
\(x = \frac{3 \pm \sqrt{10}}{4}\)
In simple words: Apply the quadratic formula directly after finding the discriminant.

Q.2 Determine the values of k for which the quadratic equation \(2x^2 + 3x + K = 0\) have both roots real.
Answer: For roots to be real, the discriminant must be greater than or equal to zero: \(D \geq 0\).
Here \(a = 2\), \(b = 3\), \(c = K\).
\(D = b^2 - 4ac \geq 0\)
\(3^2 - 4(2)(K) \geq 0\)
\(9 - 8K \geq 0\)
\(9 \geq 8K\)
\(K \leq \frac{9}{8}\)
In simple words: Real roots condition is \(b^2 - 4ac \geq 0\).

Exam Tip: Pay close attention to inequality direction when dividing by a negative number.

Q.3 Find the roots of equation \(2x^2 + x - 6 = 0\).
Answer: Given equation: \(2x^2 + x - 6 = 0\)
Product \(ac = 2 \times (-6) = -12\), sum = \(1\). Factors are \(4\) and \(-3\).
\(2x^2 + 4x - 3x - 6 = 0\)
\(2x(x + 2) - 3(x + 2) = 0\)
\((2x - 3)(x + 2) = 0\)
\(2x - 3 = 0 \implies x = \frac{3}{2}\)
\(x + 2 = 0 \implies x = -2\)
In simple words: Split the middle term \(x\) into \(4x - 3x\).

Exam Tip: Check your factorization by expanding the terms mentally.

Q.4 Find the roots of equation \(x - \frac{1}{x} = 3, \; x \neq 0\)
Answer: Multiply the entire equation by \(x\):
\(x^2 - 1 = 3x\)
\(x^2 - 3x - 1 = 0\)
Using the quadratic formula where \(a = 1\), \(b = -3\), \(c = -1\):
\(D = (-3)^2 - 4(1)(-1) = 9 + 4 = 13\)
\(x = \frac{-(-3) \pm \sqrt{13}}{2(1)}\)
\(x = \frac{3 \pm \sqrt{13}}{2}\)
In simple words: Clear the denominator by multiplying by x, then apply the quadratic formula.

Exam Tip: Always state the restriction (\(x \neq 0\)) as given in the problem.

Level 3 (03 Marks)

Q.1 The sum of the squares of two consecutive positive integers is 265. Find the integers.
Answer: Let the two consecutive positive integers be \(x\) and \(x + 1\).
According to the question:
\(x^2 + (x + 1)^2 = 265\)
\(x^2 + x^2 + 2x + 1 = 265\)
\(2x^2 + 2x + 1 - 265 = 0\)
\(2x^2 + 2x - 264 = 0\)
Divide by 2:
\(x^2 + x - 132 = 0\)
\((x + 12)(x - 11) = 0\)
Since integers are positive, \(x = 11\).
The integers are \(11\) and \(12\).
In simple words: Assume the numbers are x and x+1, set up the equation, and solve.

Exam Tip: Reject negative values since the problem specifies positive integers.

Q.2 Divide 39 into two parts such that their product is 324.
Answer: Let one part be \(x\). Then the other part is \(39 - x\).
According to the question:
\(x(39 - x) = 324\)
\(39x - x^2 = 324\)
\(x^2 - 39x + 324 = 0\)
Using factorization: \(-27\) and \(-12\) multiply to 324 and add to -39.
\((x - 27)(x - 12) = 0\)
\(x = 27\) or \(x = 12\)
The two parts are 12 and 27.

In simple words: If one part is x, the other is 39 - x. Multiply them and solve the quadratic equation.

Exam Tip: Both values represent the same two parts (12 and 27), so either value for x is correct.

Q.3 The sum of number and its reciprocal is \(\frac{17}{4}\). Find the number.
Answer: Let the number be \(x\).
According to the question:
\(x + \frac{1}{x} = \frac{17}{4}\)
\(\frac{x^2 + 1}{x} = \frac{17}{4}\)
\(4(x^2 + 1) = 17x\)
\(4x^2 - 17x + 4 = 0\)
Using factorization: \(4x^2 - 16x - x + 4 = 0\)
\(4x(x - 4) - 1(x - 4) = 0\)
\((4x - 1)(x - 4) = 0\)
\(x = \frac{1}{4}\) or \(x = 4\)
In simple words: Set up the equation \(x + 1/x = 17/4\) and cross-multiply to form a quadratic equation.

Exam Tip: Notice that if the number is 4, its reciprocal is \(\frac{1}{4}\), and vice versa.

Q.4 The length of rectangle is 5cm more than its breadth if its area is 150 Sq cm. Find the length and breadth.
Answer: Let the breadth be \(x\) cm. Then length = \((x + 5)\) cm.
Area = \(\text{Length} \times \text{Breadth}\)
\(x(x + 5) = 150\)
\(x^2 + 5x - 150 = 0\)
\((x + 15)(x - 10) = 0\)
Since breadth cannot be negative, \(x = 10\).
Breadth = \(10\) cm, Length = \(10 + 5 = 15\) cm.
In simple words: Breadth = x, Length = x + 5. Area is length times breadth.

Exam Tip: Always include proper units (cm) in the final answer for geometric word problems.

Q.5 The altitude of a right triangle is 7cm less than its base. If the hypotenuse is 13 cm. Find the other two sides.
Answer: Let the base of the right triangle be \(x\) cm.
Altitude = \((x - 7)\) cm.
Hypotenuse = \(13\) cm.
Using Pythagoras theorem: \(\text{Base}^2 + \text{Altitude}^2 = \text{Hypotenuse}^2\)
\(x^2 + (x - 7)^2 = 13^2\)
\(x^2 + x^2 - 14x + 49 = 169\)
\(2x^2 - 14x + 49 - 169 = 0\)
\(2x^2 - 14x - 120 = 0\)
Divide by 2:
\(x^2 - 7x - 60 = 0\)
\((x - 12)(x + 5) = 0\)
Since length cannot be negative, \(x = 12\).
Base = \(12\) cm, Altitude = \(12 - 7 = 5\) cm.
In simple words: Apply Pythagoras theorem with base x and altitude x - 7.

Exam Tip: Always state Pythagoras theorem explicitly before substituting values.