CBSE Class 10 Quadratic Equations Sure Shot Questions Set B

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Solve the following quadratic equation (if they exist) by the method of completing the square:

1. 8x²-22x-21=0

2. 2x²-x+1/8=0

3. 4√3x²+5x-2√3=0

4. √2x²+7x+5√2=0

5.9x²-15x+6=0

6. 2x²-5x+3=0

7. 4x²+3x+5=0

8. 5x²-6x-2=0

9. 4x²+4bx-(a²-b²)=0

10. a²x²-3abx+2b²=0

11. x²-(√3+1)x+√3=0

12. x²-4ax+4a²-b²=0

13. x2-(√2+1)x+√2=0

14. √3x²+10x+7√3=0

15.√ 2x²-3x-2√ 2=0

16. 4x²+4 √3x+3=0

17. 2x²+x+4=0

18. 2x²+x-4=0

19. 3x²+11x+10=0

20. 2x²-7x+3=0

21. 5x²-19x+17=0

22. 2x²-6x-6=0

23. 2x²-9x+7=0

24. 6x²+7x-10=0

25.x²-4 √2x+6=0

Question. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) \( 2x^2 - 3x + 5 = 0 \) (ii) \( 3x^2 - 4\sqrt{3}x + 4 = 0 \)
(iii) \( 2x^2 - 6x + 3 = 0 \)
Answer: (i) \( 2x^2 - 3x + 5 = 0 \)
Comparing the given quadratic equation with \( ax^2 + bx + c = 0 \), we have:
\( a = 2 \)
\( b = -3 \)
\( c = 5 \)
\( \therefore \) The discriminant \( = b^2 - 4ac \)
\( = (-3)^2 - 4(2)(5) \)
\( = 9 - 40 \)
\( = -31 < 0 \)
Since \( b^2 - 4ac \) is negative.
\( \therefore \) The given quadratic equation has no real roots.
(ii) \( 3x^2 - 4\sqrt{3}x + 4 = 0 \)
Comparing the given quadratic equation with \( ax^2 + bx + c = 0 \), we get
\( a = 3 \)
\( b = -4\sqrt{3} \)
\( c = 4 \)
\( \therefore b^2 - 4ac = [-4\sqrt{3}]^2 - 4(3)(4) \)
\( = (16 \times 3) - 48 \)
\( = 48 - 48 = 0 \)
Thus, the given quadratic equation has two real roots which are equal. Here, the roots are:
\( \frac{-b}{2a} \) and \( \frac{-b}{2a} \)
i.e., \( \frac{-(-4\sqrt{3})}{2 \times 3} \) and \( \frac{-(-4\sqrt{3})}{2 \times 3} \)
\( \Rightarrow \frac{4\sqrt{3}}{2 \times 3} \) and \( \frac{4\sqrt{3}}{2 \times 3} \)
\( \Rightarrow \frac{2}{\sqrt{3}} \) and \( \frac{2}{\sqrt{3}} \) [\( \because 3 = \sqrt{3} \times \sqrt{3} \)]
Thus, \( x = \frac{2}{\sqrt{3}} \) and \( x = \frac{2}{\sqrt{3}} \)
(iii) \( 2x^2 - 6x + 3 = 0 \)
Comparing it with the general quadratic equation, we have:
\( a = 2 \)
\( b = -6 \)
\( c = 3 \)
\( \therefore b^2 - 4ac = (-6)^2 - 4(2)(3) \)
\( = 36 - 24 \)
\( = 12 > 0 \)
\( \therefore \) The given quadratic equation has two real and distinct roots, which are given by
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( \Rightarrow x = \frac{-(-6) \pm \sqrt{12}}{2 \times 2} \)
\( = \frac{6 \pm 2\sqrt{3}}{4} \)
\( = \frac{3 \pm \sqrt{3}}{2} \)
Thus, the roots are:
\( x = \frac{3 + \sqrt{3}}{2} \) and \( x = \frac{3 - \sqrt{3}}{2} \)

Question. Find the values of k for each of the following quadratic equations, so that they have two equal roots:
(i) \( 2x^2 + kx + 3 = 0 \) (ii) \( kx(x - 2) + 6 = 0 \)
Answer: (i) \( 2x^2 + kx + 3 = 0 \)
Comparing the given quadratic equation with \( ax^2 + bx + c = 0 \), we get
\( a = 2 \)
\( b = k \)
\( c = 3 \)
\( \therefore b^2 - 4ac = (k)^2 - 4(2)(3) \)
\( = k^2 - 24 \)
For a quadratic equation to have equal roots, \( b^2 - 4ac = 0 \)
\( \therefore k^2 - 24 = 0 \Rightarrow k = \pm \sqrt{24} \)
\( \Rightarrow k = \pm 2\sqrt{6} \)
Thus, the required values of k are \( 2\sqrt{6} \) and \( -2\sqrt{6} \).
(ii) \( kx(x - 2) + 6 = 0 \)
Comparing \( kx(x - 2) + 6 = 0 \) i.e., \( kx^2 - 2kx + 6 = 0 \) with \( ax^2 + bx + c = 0 \), we get
\( a = k \)
\( b = -2k \)
\( c = 6 \)
\( \therefore b^2 - 4ac = (-2k)^2 - 4(k)(6) \)
\( = 4k^2 - 24k \)
Since, the roots are real and equal,
\( \therefore b^2 - 4ac = 0 \)
\( \Rightarrow 4k^2 - 24k = 0 \)
\( \Rightarrow 4k(k - 6) = 0 \)
\( \Rightarrow 4k = 0 \) or \( k - 6 = 0 \)
\( \Rightarrow k = 0 \) or \( k = 6 \)
But k cannot be 0, otherwise, the given equation is no more quadratic. Thus, the required value of k = 6.

Question. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m\(^2\)? If so, find its length and breadth.
Answer: Let the breadth be \( x \) metres.
\( \therefore \) Length = \( 2x \) metres
Now, Area = Length \( \times \) Breadth
= \( 2x \times x \) metre\(^2\)
= \( 2x^2 \) sq. metre.
According to the given condition,
\( 2x^2 = 800 \)
\( \Rightarrow x^2 = \frac{800}{2} = 400 \)
\( \Rightarrow x = \pm \sqrt{400} = \pm 20 \)
Therefore, \( x = 20 \) and \( x = -20 \)
But \( x = -20 \) is not possible (\( \because \) breadth cannot be negative).
\( \therefore x = 20 \)
\( \Rightarrow 2x = 2 \times 20 = 40 \)
Thus, length = 40 m and breadth = 20 m

Question. Is the following situation possible? If so, determine their present ages.
The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Answer: Let the age of one friend = \( x \) years
\( \therefore \) The age of the other friend = \( (20 - x) \) years [\( \because \) Sum of their ages is 20 years]
Four years ago
Age of one friend = \( (x - 4) \) years
Age of the other friend = \( (20 - x - 4) \) years
= \( (16 - x) \) years
According to the condition,
\( (x - 4) \times (16 - x) = 48 \)
\( \Rightarrow 16x - 64 - x^2 + 4x = 48 \)
\( \Rightarrow -x^2 + 20x - 64 - 48 = 0 \)
\( \Rightarrow -x^2 + 20x - 112 = 0 \)
\( \Rightarrow x^2 - 20x + 112 = 0 \) ...(1)
Here, \( a = 1, b = -20 \) and \( c = 112 \)
\( \therefore b^2 - 4ac = (-20)^2 - 4(1)(112) \)
= 400 - 448
= -48 < 0
Since \( b^2 - 4ac \) is less than 0.
\( \therefore \) The quadratic equation (1) has no real roots.
Thus, the given equation is not possible.

Question. Is it possible to design a rectangular park of perimeter 80 m and area 400 m\(^2\)? If so, find its length and breadth.
Answer: Let the breadth of the rectangle be \( x \) metres.
Since, the perimeter of the rectangle = 80 m.
\( \therefore 2[\text{Length} + \text{Breadth}] = 80 \)
\( 2[\text{Length} + x] = 80 \)
\( \Rightarrow \text{Length} + x = \frac{80}{2} = 40 \)
\( \Rightarrow \text{Length} = (40 - x) \) metres
\( \therefore \text{Area of the rectangle} = \text{Length} \times \text{breadth} \)
= \( (40 - x) \times x \) sq. m
= \( 40x - x^2 \)
Now, according to the given condition,
Area of the rectangle = 400 m\(^2\)
\( \therefore 40x - x^2 = 400 \)
\( \Rightarrow -x^2 + 40x - 400 = 0 \)
\( \Rightarrow x^2 - 40x + 400 = 0 \) ...(1)
Comparing (1) with \( ax^2 + bx + c = 0 \), we get
\( a = 1 \)
\( b = -40 \)
\( c = 400 \)
\( \therefore b^2 - 4ac = (-40)^2 - 4(1)(400) \)
= 1600 - 1600 = 0
Thus, the equation (1) has two equal and real roots.
\( \because x = \frac{-b}{2a} \) and \( x = \frac{-b}{2a} \)
\( \therefore \text{breadth} = \frac{-(-40)}{2(1)} = \frac{40}{2} = 20 \)
\( \therefore \text{Breadth, } x = 20 \text{ m} \)
\( \therefore \text{Length} = (40 - x) = (40 - 20) \text{ m} = 20 \text{ m} \).

Question. If the roots of the quadratic equation \( ax^2 + bx + c = 0 \) are equal then show that \( b^2 = 4ac \).
Answer: For equal roots, we have
\( b^2 - 4ac = 0 \)
\( \therefore b^2 = 4ac \)

Question. Find the value of ‘k’ for which the quadratic equation \( kx^2 - 5x + k = 0 \) have real roots.
Answer: Comparing \( kx^2 - 5x + k = 0 \) with \( ax^2 + bx + c = 0 \), we have:
\( a = k \)
\( b = -5 \)
\( c = k \)
\( \therefore b^2 - 4ac = (-5)^2 - 4(k)(k) \)
= \( 25 - 4k^2 \)
For equal roots, \( b^2 - 4ac = 0 \)
\( \therefore 25 - 4k^2 = 0 \)
\( \Rightarrow 4k^2 = 25 \)
\( \Rightarrow k^2 = \frac{25}{4} \)
\( \Rightarrow k = \pm \sqrt{\frac{25}{4}} = \pm \frac{5}{2} \)

Question. If –4 is a root of the quadratic equation \( x^2 + px - 4 = 0 \) and \( x^2 + px + k = 0 \) has equal roots, find the value of k.
Answer: \( \because \) (–4) is a root of \( x^2 + px - 4 = 0 \)
\( \therefore (-4)^2 + p(-4) - 4 = 0 \)
\( \Rightarrow 16 - 4p - 4 = 0 \)
\( \Rightarrow 4p = 12 \) or \( p = 3 \)
Now, \( x^2 + px + k = 0 \)
\( \Rightarrow x^2 + 3x + k = 0 \) [\( \because p = 3 \)]
Now, \( a = 1, b = 3 \) and \( c = +k \)
\( \therefore b^2 - 4ac = (3)^2 - 4(1)(k) \)
= 9 - 4k
For equal roots, \( b^2 - 4ac = 0 \)
\( \Rightarrow 9 - 4k = 0 \Rightarrow 4k = 9 \)
\( \Rightarrow k = \frac{9}{4} \)

Question. If one root of the quadratic equation \( 2x^2 - 3x + p = 0 \) is 3, find the other root of the quadratic equation. Also find the value of p.
Answer: We have:
\( 2x^2 - 3x + p = 0 \) ...(1)
\( \therefore a = 2, b = -3 \) and \( c = p \)
Since, the sum of the roots = \( \frac{-b}{a} \)
= \( \frac{-(-3)}{2} = \frac{3}{2} \)
\( \because \) One of the roots = 3
\( \therefore \) The other root = \( \frac{3}{2} - 3 = \frac{3 - 6}{2} = \frac{-3}{2} \)
Now, substituting \( x = 3 \) in (1), we get
\( 2(3)^2 - 3(3) + p = 0 \)
\( \Rightarrow 18 - 9 + p = 0 \)
\( \Rightarrow 9 + p = 0 \Rightarrow p = -9 \)

Question. If one of the roots of \( x^2 + px - 4 = 0 \) is –4 then find the product of its roots and the value of p.
Answer: If –4 is a root of the quadratic equation,
\( x^2 + px - 4 = 0 \)
\( \therefore (-4)^2 + (-4)(p) - 4 = 0 \)
\( \Rightarrow 16 - 4p - 4 = 0 \)
\( \Rightarrow 12 - 4p = 0 \Rightarrow p = 3 \)
Now, in \( ax^2 + bx + c = 0 \), the product of the roots = \( \frac{c}{a} \)
\( \therefore \) Product of the roots in \( x^2 - px - 4 = 0 \)
= \( \frac{-4}{1} = -4 \)

Question. For what value of k, does the given equation have real and equal roots?
\( (k + 1)x^2 - 2(k - 1)x + 1 = 0 \).
Answer: Comparing the given equation with \( ax^2 + bx + c = 0 \), we have:
\( a = k + 1 \)
\( b = -2(k - 1) \)
\( c = 1 \)
For equal roots, \( b^2 - 4ac = 0 \)
\( \therefore [-2(k - 1)]^2 - 4(k + 1)(1) = 0 \)
\( \Rightarrow 4(k - 1)^2 - 4(k + 1) = 0 \)
\( \Rightarrow 4(k^2 + 1 - 2k) - 4k - 4 = 0 \)
\( \Rightarrow 4k^2 + 4 - 8k - 4k - 4 = 0 \)
\( \Rightarrow 4k^2 - 12k = 0 \)
\( \Rightarrow 4k(k - 3) = 0 \)
\( \Rightarrow k = 0 \) or \( k = 3 \)

Question. Using quadratic formula, solve the following quadratic equation for x:
\( x^2 - 2ax + (a^2 - b^2) = 0 \)
Answer: Comparing \( x^2 - 2ax + (a^2 - b^2) = 0 \), with \( ax^2 + bx + c = 0 \), we have:
\( a = 1, b = -2a, c = a^2 - b^2 \)
\( \therefore x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( \Rightarrow x = \frac{-(-2a) \pm \sqrt{(-2a)^2 - 4(1)(a^2 - b^2)}}{2(1)} \)
\( = \frac{2a \pm \sqrt{4a^2 - 4a^2 + 4b^2}}{2} \)
\( = \frac{2a \pm \sqrt{4b^2}}{2} = \frac{2a \pm 2b}{2} = a \pm b \)
\( \therefore x = (a + b) \) or \( x = (a - b) \)

Question. If one of the roots of the quadratic equation \( 2x^2 + kx - 6 = 0 \) is 2, find the value of k. Also find the other root.
Answer: Given equation:
\( 2x^2 + kx - 6 = 0 \)
one root = 2
Substituting \( x = 2 \) in \( 2x^2 + kx - 6 = 0 \)
We have:
\( 2(2)^2 + k(2) - 6 = 0 \)
\( \Rightarrow 8 + 2k - 6 = 0 \)
\( \Rightarrow 2k + 2 = 0 \Rightarrow k = -1 \)
\( \therefore 2x^2 + kx - 6 = 0 \Rightarrow 2x^2 - x - 6 = 0 \)
Sum of the roots = \( \frac{-b}{a} = \frac{1}{2} \)
\( \therefore \) other root = \( \frac{1}{2} - 2 = -\frac{3}{2} \)

Question. Determine the value of k for which the quadratic equation \( 4x^2 - 4kx + 1 = 0 \) has equal roots.
Answer: We have:
\( 4x^2 - 4kx + 1 = 0 \)
Comparing with \( ax^2 + bx + c = 0 \), we have
\( a = 4, b = -4k \) and \( c = 1 \)
\( \therefore b^2 - 4ac = (-4k)^2 - 4(4k)(1) = 16k^2 - 16 \)
For equal roots
\( b^2 - 4ac = 0 \)
\( \therefore 16k^2 - 16 = 0 \)
\( \Rightarrow 16k^2 = 16 \Rightarrow k^2 = 1 \)
\( \Rightarrow k = \pm 1 \)

Question. For what value of k, does the quadratic equation \( x^2 - kx + 4 = 0 \) have equal roots?
Answer: Comparing \( x^2 - kx + 4 = 0 \) with \( ax^2 + bx + c = 0 \), we get
\( a = 1 \)
\( b = -k \)
\( c = 4 \)
\( \therefore b^2 - 4ac = (-k)^2 - 4(1)(4) = k^2 - 16 \)
For equal roots,
\( b^2 - 4ac = 0 \)
\( \Rightarrow k^2 - 16 = 0 \)
\( \Rightarrow k^2 = 16 \)
\( \Rightarrow k = \pm 4 \)

Question. What is the nature of roots of the quadratic equation \( 4x^2 - 12x + 9 = 0 \)?
Answer: Comparing \( 4x^2 - 12x + 9 = 0 \) with \( ax^2 + bx + c = 0 \) we get
\( a = 4 \)
\( b = -12 \)
\( c = 9 \)
\( \therefore b^2 - 4ac = (-12)^2 - 4(4)(9) \)
= 144 - 144 = 0
Since \( b^2 - 4ac = 0 \)
\( \therefore \) The roots are real and equal.

Question. Write the value of k for which the quadratic equation \( x^2 - kx + 9 = 0 \) has equal roots. (AI CBSE 2009 C)
Answer: Comparing \( x^2 - kx + 9 = 0 \) with \( ax^2 + bx + c = 0 \), we get
\( a = 1 \)
\( b = -k \)
\( c = 9 \)
\( \therefore b^2 - 4ac = (-k)^2 - 4(1)(9) = k^2 - 36 \)
For equal roots,
\( b^2 - 4ac = 0 \)
\( \Rightarrow k^2 - 36 = 0 \Rightarrow k^2 = 36 \)
\( \Rightarrow k = \pm 6 \)

Question. For what value of k are the roots of the quadratic equation \( 3x^2 + 2kx + 27 = 0 \) real and equal? (CBSE 2008 C)
Answer: Comparing \( 3x^2 + 2kx + 27 = 0 \) with \( ax^2 + bx + c = 0 \), we have:
\( a = 3 \)
\( b = 2k \)
\( c = 27 \)
\( \therefore b^2 - 4ac = (2k)^2 - 4(3)(27) \)
= \( 4k^2 - (12 \times 27) \)
For the roots to be real and equal
\( b^2 - 4ac = 0 \)
\( \Rightarrow 4k^2 - (12 \times 27) = 0 \)
\( \Rightarrow 4k^2 = 12 \times 27 \)
\( \Rightarrow k^2 = \frac{12 \times 27}{4} = 81 \)
\( \Rightarrow k = \pm 9 \)

Question. For what value of k are the roots of the quadratic equation \( kx^2 + 4x + 1 = 0 \) equal and real? (CBSE 2008 C)
Answer: Comparing \( kx^2 + 4x + 1 = 0 \), with \( ax^2 + bx + c = 0 \), we get
\( a = k \)
\( b = 4 \)
\( c = 1 \)
\( \therefore b^2 - 4ac = (4)^2 - 4(k)(1) = 16 - 4k \)
For equal and real roots, we have
\( b^2 - 4ac = 0 \)
\( \Rightarrow 16 - 4k = 0 \)
\( \Rightarrow 4k = 16 \)
\( \Rightarrow k = \frac{16}{4} = 4 \)

Question. For what value of k does \( (k - 12)x^2 + 2(k - 12)x + 2 = 0 \) have equal roots? (AI CBSE 2008 C)
Answer: Comparing \( (k - 12)x^2 + 2(k - 12)x + 2 = 0 \) with \( ax^2 + bx + c = 0 \), we have:
\( a = (k - 12) \)
\( b = 2(k - 12) \)
\( c = 2 \)
\( \therefore b^2 - 4ac = [2(k - 12)]^2 - 4(k - 12)(2) \)
= \( 4(k - 12)^2 - 8(k - 12) \)
= \( 4(k - 12) [k - 12 - 2] \)
= \( 4(k - 12)(k - 14) \)
For equal roots,
\( b^2 - 4ac = 0 \)
\( \Rightarrow 4(k - 12)(k - 14) = 0 \)
\( \Rightarrow \) Either \( 4(k - 12) = 0 \Rightarrow k = 12 \)
or \( k - 14 = 0 \Rightarrow k = 14 \)
But \( k = 12 \) makes \( k - 12 = 0 \) which is not required.
\( \therefore k \neq 12 \)
\( \Rightarrow k = 14 \)

Question. For what value of k does the equation \( 9x^2 + 3kx + 4 = 0 \) has equal roots? (AI CBSE 2008 C)
Answer: Comparing \( 9x^2 + 3kx + 4 = 0 \) with \( ax^2 + bx + c = 0 \), we get
\( a = 9 \)
\( b = 3k \)
\( c = 4 \)
\( \therefore b^2 - 4ac = (3k)^2 - 4(9)(4) = 9k^2 - 144 \)
For equal roots,
\( b^2 - 4ac = 0 \)
\( \Rightarrow 9k^2 - 144 = 0 \)
\( \Rightarrow 9k^2 = 144 \)
\( \Rightarrow k^2 = \frac{144}{9} = 16 \)
\( \Rightarrow k = \pm 4 \)

II. SHORT ANSWER TYPE QUESTIONS

Question. Solve \( 2x^2 - 5x + 3 = 0 \).
Answer: We have:
\( 2x^2 - 5x + 3 = 0 \) ...(1)
Comparing (1) with \( ax^2 + bx + c = 0 \),
\( \therefore a = 2, b = -5, c = 3 \)
\( \therefore b^2 - 4ac = (-5)^2 - 4(2)(3) = 25 - 24 = 1 \)
Since, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( \therefore x = \frac{-(-5) \pm \sqrt{1}}{2(2)} = \frac{5 \pm 1}{4} \)
Taking, +ve sign,
\( x = \frac{5 + 1}{4} = \frac{6}{4} = \frac{3}{2} \)
Taking, –ve sign,
\( x = \frac{5 - 1}{4} = \frac{4}{4} = 1 \)
Thus, the required roots are \( x = \frac{3}{2} \) and \( x = 1 \)

Question. Solve the following quadratic equation:
\( 2x^2 + 4x - 8 = 0 \)
Answer: We have:
\( 2x^2 + 4x - 8 = 0 \)
Dividing by 2, we get
\( x^2 + 2x - 4 = 0 \) ...(1)
Comparing (1) with \( ax^2 + bx + c = 0 \),
\( a = 1, b = 2, c = -4 \)
\( \therefore b^2 - 4ac = (2)^2 - 4(1)(-4) = 4 + 16 = 20 \)
Since, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( \therefore x = \frac{-2 \pm \sqrt{20}}{2(1)} = \frac{-2 \pm 2\sqrt{5}}{2} = \frac{2[-1 \pm \sqrt{5}]}{2} = -1 \pm \sqrt{5} \)
Taking +ve sign, we get
\( x = (-1 + \sqrt{5}) \)
Taking –ve sign we get,
\( x = (-1 - \sqrt{5}) \)
Thus, the required roots are \( x = (-1 + \sqrt{5}) \) and \( x = (-1 - \sqrt{5}) \).

Question. Solve: \( \frac{x + 1}{x - 1} + \frac{x - 2}{x + 2} = 3 \). [CBSE 2012]
Answer: We have,
\( \frac{x + 1}{x - 1} + \frac{x - 2}{x + 2} = 3 \)
\( \therefore \frac{(x + 2)(x + 1) + (x - 2)(x - 1)}{(x - 1)(x + 2)} = 3 \)
\( \Rightarrow \frac{(x^2 + 2x + x + 2) + (x^2 - x - 2x + 2)}{x^2 - x + 2x - 2} = 3 \)
\( \Rightarrow \frac{(x^2 + 3x + 2) + (x^2 - 3x + 2)}{x^2 + x - 2} = 3 \)
\( \Rightarrow x^2 + 3x + 2 + x^2 - 3x + 2 = 3(x^2 + x - 2) \)
\( \Rightarrow 2x^2 + 4 = 3x^2 + 3x - 6 \)
\( \Rightarrow 3x^2 + 3x - 6 - 2x^2 - 4 = 0 \)
\( \Rightarrow x^2 + 3x - 10 = 0 \)
\( \Rightarrow x^2 + 5x - 2x - 10 = 0 \)
\( \Rightarrow x(x + 5) - 2(x + 5) = 0 \)
\( \Rightarrow (x + 5)(x - 2) = 0 \)
Either \( x + 5 = 0 \Rightarrow x = -5 \)
or \( x - 2 = 0 \Rightarrow x = 2 \)
Thus, the required roots are \( x = -5 \) and \( x = 2 \)

Question. Solve (using quadratic formula): \( x^2 + 5x + 5 = 0 \)
Answer: We have:
\( x^2 + 5x + 5 = 0 \)
Comparing (1) with \( ax^2 + bx + c = 0 \), we have:
\( a = 1, b = 5, c = 5 \)
\( \therefore b^2 - 4ac = (5)^2 - 4(1)(5) = 25 - 20 = 5 \)
Since, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( \therefore x = \frac{-5 \pm \sqrt{5}}{2(1)} = \frac{-5 \pm \sqrt{5}}{2} \)
Taking +ve sign, we have:
\( x = \frac{-5 + \sqrt{5}}{2} \)
Taking –ve sign, we have:
\( x = \frac{-5 - \sqrt{5}}{2} \)
Thus, the required roots are: \( x = \frac{-5 + \sqrt{5}}{2} \) and \( x = \frac{-5 - \sqrt{5}}{2} \)

Question. Solve for x: \( 36x^2 - 12ax + (a^2 - b^2) = 0 \).
Answer: We have:
\( 36x^2 - 12ax + (a^2 - b^2) = 0 \) ...(1)
Comparing (1) with \( Ax^2 + Bx + C = 0 \), we have:
\( A = 36, B = -12a, C = (a^2 - b^2) \)
\( \therefore B^2 - 4AC = [-12a]^2 - 4(36)[a^2 - b^2] \)
= \( 144a^2 - 144(a^2 - b^2) \)
= \( 144a^2 - 144a^2 + 144b^2 = 144b^2 \)
Since, \( x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \)
\( \therefore x = \frac{-(-12a) \pm \sqrt{144b^2}}{2(36)} = \frac{12a \pm 12b}{72} \)
Taking +ve sign, we have:
\( x = \frac{12a + 12b}{72} = \frac{12}{72}[a + b] = \frac{1}{6}(a + b) \)
Taking –ve sign, we get
\( x = \frac{12a - 12b}{72} = \frac{12}{72}(a - b) = \frac{1}{6}(a - b) \)
Thus, the required roots are: \( x = \frac{1}{6}(a + b) \) and \( x = \frac{1}{6}(a - b) \)

Question. Find the roots of the quadratic equation \( 2x^2 - \sqrt{5}x - 2 = 0 \), using the quadratic formula. [NCERT Exemplar]
Answer: Comparing the given equation with the general equation \( ax^2 + bx + c = 0 \), we have
\( a = 2, b = -\sqrt{5}, c = -2 \)
\( \Rightarrow b^2 - 4ac = (-\sqrt{5})^2 - 4(2)(-2) = 5 + 16 = 21 \)
Now, using the quadratic formula, we have:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( \Rightarrow x = \frac{-(-\sqrt{5}) \pm \sqrt{21}}{2(2)} = \frac{\sqrt{5} \pm \sqrt{21}}{4} \)
Taking the, positive sign, we get \( x = \frac{\sqrt{5} + \sqrt{21}}{4} \)
Taking the negative sign, we get \( x = \frac{\sqrt{5} - \sqrt{21}}{4} \)
Thus, \( x = \frac{\sqrt{5} + \sqrt{21}}{4} \) and \( \frac{\sqrt{5} - \sqrt{21}}{4} \)

Question. Solve: \( 16x^2 - 8a^2x + (a^4 - b^4) = 0 \) for x.
Answer: We have:
\( 16x^2 - 8a^2x + a^4 - b^4 = 0 \) ...(1)
Comparing (1) with \( ax^2 + bx + c = 0 \), we get
\( a = 16, b = -8a^2, c = (a^4 - b^4) \)
\( \therefore b^2 - 4ac = [-8a^2]^2 - 4(16)(a^4 - b^4) \)
= \( 64a^4 - 64(a^4 - b^4) \)
= \( 64a^4 - 64a^4 + 64b^4 = 64b^4 \)
Since, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( \therefore x = \frac{-(-8a^2) \pm \sqrt{64b^4}}{2(16)} = \frac{8a^2 \pm 8b^2}{32} = \frac{8[a^2 \pm b^2]}{32} = \frac{a^2 \pm b^2}{4} \)
Now, taking +ve sign, we get \( x = \frac{a^2 + b^2}{4} \)
Taking –ve sign, we get \( x = \frac{a^2 - b^2}{4} \)
Thus, the required roots are: \( x = \frac{a^2 + b^2}{4} \) and \( x = \frac{a^2 - b^2}{4} \)

Question. Solve for x: \( 9x^2 - 6ax + a^2 - b^2 = 0 \).
Answer: We have:
\( 9x^2 - 6ax + (a^2 - b^2) = 0 \) ...(1)
Comparing (1) with \( ax^2 + bx + c = 0 \), we get
\( a = 9, b = -6a \) and \( c = (a^2 - b^2) \)
\( \therefore b^2 - 4ac = (-6a)^2 - 4(9)(a^2 - b^2) \)
= \( 36a^2 - 36(a^2 - b^2) \)
= \( 36a^2 - 36a^2 + 36b^2 = 36b^2 = (6b)^2 \)
Since, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( \therefore x = \frac{-(-6a) \pm \sqrt{(6b)^2}}{2(9)} = \frac{6a \pm 6b}{18} = \frac{6[a \pm b]}{18} = \frac{a \pm b}{3} \)
Taking the +ve sign, we get \( x = \frac{a + b}{3} \)
Taking the –ve sign, we get \( x = \frac{a - b}{3} \)
Thus, the required roots are: \( x = \frac{a + b}{3} \) and \( x = \frac{a - b}{3} \)

Question. Evaluate \( \sqrt{20 + \sqrt{20 + \sqrt{20 + \dots}}} \)
Answer: Let \( \sqrt{20 + \sqrt{20 + \sqrt{20 + \dots}}} = x \)
The given expression can be written as
\( x = \sqrt{20 + x} \)
squaring both side, we have
\( x^2 = (\sqrt{20 + x})^2 \)
\( \Rightarrow x^2 = 20 + x \)
\( \Rightarrow x^2 - x - 20 = 0 \), where \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Here: \( a = 1, b = -1 \) and \( c = -20 \)
\( \therefore x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \times 1 \times (-20)}}{2(1)} = \frac{1 \pm \sqrt{1 + 80}}{2} = \frac{1 \pm \sqrt{81}}{2} = \frac{1 \pm 9}{2} \)
Since the given expression is positive,
\( \therefore \) Rejecting the negative sign, we have:
\( x = \frac{1 + 9}{2} = \frac{10}{2} = 5 \)
Thus, \( \sqrt{20 + \sqrt{20 + \sqrt{20 + \dots}}} = 5 \)

Question. Using quadratic formula, solve the following quadratic equation for x:
\( x^2 - 4ax + 4a^2 - b^2 = 0 \)
Answer: Comparing the given equation with \( ax^2 + bx + c = 0 \), we have:
\( a = 1 \)
\( b = -4a \)
\( c = 4a^2 - b^2 \)
\( \therefore b^2 - 4ac = [-(4a)]^2 - 4(1)[4a^2 - b^2] \)
= \( 16a^2 - 4(4a^2 - b^2) \)
= \( 16a^2 - 16a^2 + 4b^2 = 4b^2 = (2b)^2 \)

Since, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\(\therefore x = \frac{-(-4a) \pm \sqrt{(2b)^2}}{2(1)} \)
\(\Rightarrow x = \frac{4a \pm 2b}{2} \)
\(\Rightarrow x = \frac{2}{2}[2a \pm b] = 2a \pm b \)
Taking the +ve sign, \( x = 2a + b \)
Taking the -ve sign, \( x = 2a - b \)
Thus, the required roots are:
\( x = 2a + b \) and \( x = 2a - b \)

Question. Using quadratic formula, solve the following quadratic equation for x: \( x^2 - 2ax + (a^2 - b^2) = 0 \)
Answer: Comparing the given equation with \( ax^2 + bx + c = 0 \), we have:
\( a = 1, b = -2a \) and \( c = (a^2 - b^2) \)
\(\therefore b^2 - 4ac = (-2a)^2 - 4(1)(a^2 - b^2) \)
\(= 4a^2 - 4(a^2 - b^2) \)
\(= 4a^2 - 4a^2 + 4b^2 = 4b^2 \)
Since, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\(\therefore x = \frac{-(-2a) \pm \sqrt{4b^2}}{2(1)} \)
\(\Rightarrow x = \frac{2a \pm 2b}{2} \)
\(\Rightarrow x = \frac{2}{2}[a \pm b] = a \pm b \)
Taking the +ve sign, we get \( x = a + b \)
Taking the -ve sign, we get \( x = a - b \)
Thus, the required roots are:
\( x = a + b \) and \( x = a - b \)

Question. Find the roots of the equation: \( \frac{1}{x+3} + \frac{1}{2x-1} = \frac{11}{7x+9} ; x \neq -3, \frac{1}{2}, -\frac{9}{7} \) (CBSE 2009 C)
Answer: We have:
\( \frac{1}{x+3} + \frac{1}{2x-1} = \frac{11}{7x+9} \)
\(\Rightarrow \frac{2x - 1 + x + 3}{(x + 3)(2x - 1)} = \frac{11}{7x+9} \)
\(\Rightarrow \frac{3x + 2}{2x^2 - x + 6x - 3} = \frac{11}{7x+9} \)
\(\Rightarrow \frac{3x + 2}{2x^2 + 5x - 3} = \frac{11}{7x+9} \)
\(\Rightarrow (3x + 2)(7x + 9) = 11(2x^2 + 5x - 3) \)
\(\Rightarrow 21x^2 + 27x + 14x + 18 = 22x^2 + 55x - 33 \)
\(\Rightarrow 21x^2 + 41x + 18 = 22x^2 + 55x - 33 \)
\(\Rightarrow (21 - 22)x^2 + (41 - 55)x + (18 + 33) = 0 \)
\(\Rightarrow -x^2 + (-14x) + (51) = 0 \)
\(\Rightarrow x^2 + 14x - 51 = 0 \)
\(\Rightarrow x^2 + 17x - 3x - 51 = 0 \)
\(\Rightarrow x(x + 17) - 3(x + 17) = 0 \)
\(\Rightarrow (x + 17)(x - 3) = 0 \)
Either \( x + 17 = 0 \Rightarrow x = -17 \)
or \( x - 3 = 0 \Rightarrow x = 3 \)
Thus, the required roots of the given equation are: 3 and -17

Question. Find the roots of the equation: \( \frac{1}{x} - \frac{1}{x-3} = \frac{4}{3}; x \neq 0, 3 \) (CBSE 2009 C)
Answer: We have:
\( \frac{1}{x} - \frac{1}{x-3} = \frac{4}{3} \)
\(\Rightarrow \frac{(x - 3) - x}{x(x - 3)} = \frac{4}{3} \)
\(\Rightarrow \frac{-3}{x^2 - 3x} = \frac{4}{3} \)
\(\Rightarrow -3 \times 3 = 4 \times (x^2 - 3x) \)
\(\Rightarrow -9 = 4x^2 - 12x \)
\(\Rightarrow 4x^2 - 12x + 9 = 0 \) ...(1)
Comparing (1), with \( ax^2 + bx + c = 0 \), we get
\( a = 4, b = -12, c = 9 \)
\(\therefore b^2 - 4ac = (-12)^2 - 4(4)(9) \)
\(= 144 - 144 = 0 \)
Now, the roots are:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\(\Rightarrow x = \frac{-(-12) \pm 0}{2(4)} \)
\(\Rightarrow x = \frac{12}{8} = \frac{3}{2} \)
Thus, the roots are: \( \frac{3}{2} \) and \( \frac{3}{2} \)

Question. Find the roots of the equation: \( \frac{1}{x-2} + \frac{1}{x} = \frac{8}{2x+5}; x \neq 0, 2, -\frac{5}{2} \) (CBSE 2008 C)
Answer: We have:
\( \frac{1}{x-2} + \frac{1}{x} = \frac{8}{2x+5} \)
\(\Rightarrow \frac{x + x - 2}{x(x - 2)} = \frac{8}{2x+5} \)
\(\Rightarrow \frac{2x - 2}{x^2 - 2x} = \frac{8}{2x+5} \)
\(\Rightarrow (2x - 2)(2x + 5) = 8(x^2 - 2x) \)
\(\Rightarrow 4x^2 + 10x - 4x - 10 = 8x^2 - 16x \)
\(\Rightarrow -4x^2 + 22x - 10 = 0 \)
\(\Rightarrow 2x^2 - 11x + 5 = 0 \) ...(1)
Comparing (1) with \( ax^2 + bx + c = 0 \), we have:
\( a = 2, b = -11, c = 5 \)
\(\therefore b^2 - 4ac = (-11)^2 - 4(2)(5) \)
\(= 121 - 40 = 81 \)
Now, the roots are given by
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\(\Rightarrow x = \frac{-(-11) \pm \sqrt{81}}{2(2)} \)
\(= \frac{11 \pm 9}{4} \)
Taking the +ve sign,
\( x = \frac{11 + 9}{4} = \frac{20}{4} = 5 \)
Taking the -ve sign,
\( x = \frac{11 - 9}{4} = \frac{2}{4} = \frac{1}{2} \)
Thus, the required roots are: 5 and \( \frac{1}{2} \).

Question. Find the roots of the following equation: \( \frac{1}{x+4} - \frac{1}{x-7} = \frac{11}{30}; x \neq -4, 7 \) (CBSE 2012)
Answer: We have:
\( \frac{1}{x+4} - \frac{1}{x-7} = \frac{11}{30} \)
\(\Rightarrow \frac{(x - 7) - (x + 4)}{(x + 4)(x - 7)} = \frac{11}{30} \)
\(\Rightarrow \frac{x - 7 - x - 4}{x^2 - 3x - 28} = \frac{11}{30} \)
\(\Rightarrow -11 \times 30 = 11 \times (x^2 - 3x - 28) \)
\(\Rightarrow -30 = x^2 - 3x - 28 \)
\(\Rightarrow x^2 - 3x - 28 + 30 = 0 \)
\(\Rightarrow x^2 - 3x + 2 = 0 \)
\(\Rightarrow x^2 - 2x - x + 2 = 0 \)
\(\Rightarrow x(x - 2) - 1(x - 2) = 0 \)
\(\Rightarrow (x - 1)(x - 2) = 0 \)
Either \( x - 1 = 0 \Rightarrow x = 1 \)
or \( x - 2 = 0 \Rightarrow x = 2 \)
Thus, the required roots are: 1 and 2.

Question. If \(\alpha\) and \(\beta\) are roots of the equation \( x^2 - 1 = 0 \), form an equation whose roots are \( \frac{2\alpha}{\beta} \) and \( \frac{2\beta}{\alpha} \).
Answer: \(\because \alpha\) and \(\beta\) are roots of \( x^2 - 1 = 0 \) and \( x^2 - 1 = 0 \) can be written as \( x^2 + 0x - 1 = 0 \) where \( a = 1, b = 0 \) and \( c = -1 \).
\(\therefore\) Sum of roots = \( \frac{-b}{a} = \frac{0}{1} = 0 \)
\(\Rightarrow \alpha + \beta = 0 \)
Also, product of roots = \( \frac{c}{a} = \frac{-1}{1} = -1 \)
\(\Rightarrow \alpha\beta = -1 \)
Now, the roots of the new equation are \( \frac{2\alpha}{\beta} \) and \( \frac{2\beta}{\alpha} \)
\(\therefore\) Sum of the roots of the new equation = \( \frac{2\alpha}{\beta} + \frac{2\beta}{\alpha} \)
\(= \frac{2\alpha^2 + 2\beta^2}{\alpha\beta} = \frac{2(\alpha^2 + \beta^2)}{\alpha\beta} \)
\(= \frac{2[(\alpha + \beta)^2 - 2\alpha\beta]}{\alpha\beta} \)
\(= 2 \left[ \frac{(0)^2 - 2(-1)}{(-1)} \right] \text{ [}\because \alpha + \beta = 0 \text{ and } \alpha\beta = (-1)\text{]} \)
\(= 2 \left[ \frac{0 + 2}{-1} \right] \)
\(= 2 \times (-2) = -4 \)
Product of the roots of the new equation = \( \frac{2\alpha}{\beta} \times \frac{2\beta}{\alpha} = 4 \)
Since, a quadratic equation is given by
\( x^2 - (\text{Sum of the roots})x + (\text{Product of the roots}) = 0 \)
\(\therefore\) The required quadratic equation is
\( x^2 - (-4)x + 4 = 0 \)
or \( x^2 + 4x + 4 = 0 \)
Remember
\(\alpha^2 + \beta^2 + 2\alpha\beta = (\alpha + \beta)^2 \)
\(\Rightarrow \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \)

III. LONG ANSWER TYPE QUESTIONS

Question. A two-digit number is 5 times the sum of its digits and is also equal to 5 more than twice the product of its digits. Find the number.
Answer: Let the tens digit = \( x \)
And the ones digit = \( y \)
\(\therefore\) The number = \( 10x + y \)
According to the conditions,
\( 10x + y = 5(x + y) \) ...(1)
\( 10x + y = 2xy + 5 \) ...(2)
From (1), we have
\( 10x + y = 5x + 5y \)
\(\Rightarrow 10x + y - 5x - 5y = 0 \)
\(\Rightarrow 5x - 4y = 0 \)
\(\Rightarrow 5x = 4y \) or \( y = \frac{5}{4}x \)
Substituting \( y = \frac{5}{4}x \) in (2), we get
\( 10x + \frac{5}{4}x = 2x(\frac{5}{4}x) + 5 \)
\(\Rightarrow 40x + 5x = 10x^2 + 20 \) [Multiplying both sides by 4]
\(\Rightarrow 45x = 10x^2 + 20 \)
\(\Rightarrow 10x^2 - 45x + 20 = 0 \)
\(\Rightarrow 2x^2 - 9x - 4 = 0 \)
\(\Rightarrow 2x^2 - 8x - x + 4 = 0 \)
\(\Rightarrow 2x(x - 4) - 1(x - 4) = 0 \)
\(\Rightarrow (x - 4)(2x - 1) = 0 \)
Either \( x - 4 = 0 \Rightarrow x = 4 \)
or \( 2x - 1 = 0 \Rightarrow x = \frac{1}{2} \)
But a digit cannot be a fraction,
\(\therefore x = 4 \Rightarrow\) The tens digit = 4
Now, the ones digit \( y = \frac{5 \times 4}{4} = 5 \)
\(\therefore x = 4 \) and \( y = 5 \)
\(\therefore\) The required number = \( 10 \times 4 + 5 \)
\(= 40 + 5 = 45 \)

Question. The denominator is one more than twice the numerator. If the sum of the fraction and its reciprocal is \( 2 \frac{16}{21} \), find the fraction.
Answer: Let the numerator be \( x \)
\(\therefore\) Denominator = \( (2x + 1) \)
\(\therefore\) Fraction = \( \frac{x}{2x + 1} \)
Reciprocal of the fraction = \( \frac{2x + 1}{x} \)
According to the condition,
\( \frac{x}{2x + 1} + \frac{2x + 1}{x} = 2 \frac{16}{21} \)
\(\Rightarrow \frac{x \times x + (2x + 1)(2x + 1)}{x(2x + 1)} = \frac{58}{21} \)
\(\Rightarrow \frac{x^2 + (2x + 1)^2}{2x^2 + x} = \frac{58}{21} \)
\(\Rightarrow \frac{x^2 + 4x^2 + 4x + 1}{2x^2 + x} = \frac{58}{21} \)
\(\Rightarrow \frac{5x^2 + 4x + 1}{2x^2 + x} = \frac{58}{21} \)
\(\Rightarrow 21(5x^2 + 4x + 1) = 58(2x^2 + x) \)
\(\Rightarrow 105x^2 + 84x + 21 = 116x^2 + 58x \)
\(\Rightarrow 105x^2 - 116x^2 + 84x - 58x + 21 = 0 \)
\(\Rightarrow -11x^2 + 26x + 21 = 0 \)
\(\Rightarrow 11x^2 - 26x - 21 = 0 \) ...(1)
Comparing (1) with \( ax^2 + bx + c = 0 \), we have:
\( a = 11, b = -26, c = -21 \)
\(\therefore b^2 - 4ac = (-26)^2 - 4 \times 11(-21) \)
\(= 676 + 924 = 1600 \)
Since, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\(\therefore x = \frac{-(-26) \pm \sqrt{1600}}{2(11)} \)
\(\Rightarrow x = \frac{26 \pm 40}{22} \)
Taking the +ve sign,
\( x = \frac{26 + 40}{22} = \frac{66}{22} = 3 \)
Taking the -ve sign,
\( x = \frac{26 - 40}{22} = \frac{-14}{22} = \frac{-7}{11} \)
But the numerator cannot be \( -\frac{7}{11} \)
\(\therefore x = 3 \Rightarrow\) The numerator = 3
\(\therefore\) Denominator = \( 2(3) + 1 = 7 \)
Thus, the required fraction = \( \frac{3}{7} \)

Question. The sum of a number and its reciprocal is \( \frac{10}{3} \). Find the number.
Answer: Let the required number = \( x \)
\(\therefore\) Its reciprocal = \( \frac{1}{x} \)
According to the condition, we have:
[The number] + [Reciprocal of the number] = \( \frac{10}{3} \)
\(\Rightarrow x + \frac{1}{x} = \frac{10}{3} \)
\(\Rightarrow \frac{x^2 + 1}{x} = \frac{10}{3} \)
\(\Rightarrow 3(x^2 + 1) = 10x \)
\(\Rightarrow 3x^2 + 3 - 10x = 0 \)
\(\Rightarrow 3x^2 - 10x + 3 = 0 \)
\(\Rightarrow 3x^2 - 9x - x + 3 = 0 \)
\(\Rightarrow 3x(x - 3) - 1(x - 3) = 0 \)
\(\Rightarrow (x - 3)(3x - 1) = 0 \)
Either \( x - 3 = 0 \Rightarrow x = 3 \)
or \( 3x - 1 = 0 \Rightarrow x = \frac{1}{3} \)
Thus, the required number is 3 or \( \frac{1}{3} \).

Question. The hypotenuse of a right triangle is \( 3\sqrt{10} \) cm. If the smaller side is tripled and the longer side doubled, new hypotenuse will be \( 9\sqrt{5} \) cm. How long are the sides of the triangle?
Answer: Let the smaller side = \( x \)
\(\therefore\) Longer side = \( \sqrt{(\text{Hypotenuse})^2 - (\text{smaller side})^2} \)
\(= \sqrt{(3\sqrt{10})^2 - x^2} \)
\(= \sqrt{9 \times 10 - x^2} \)
\(= \sqrt{90 - x^2} \)
According to the condition, we have
\([3(\text{Smaller side})]^2 + [2(\text{Longer side})]^2 = [\text{New Hypotenuse}]^2 \)
\(\Rightarrow [3(x)]^2 + [2(\sqrt{90 - x^2})]^2 = [9\sqrt{5}]^2 \)
\(\Rightarrow 9x^2 + 4(90 - x^2) = 81 \times 5 \)
\(\Rightarrow 9x^2 + 360 - 4x^2 = 405 \)
\(\Rightarrow 5x^2 = 405 - 360 \)
\(\Rightarrow 5x^2 = 45 \)
\(\Rightarrow x^2 = \frac{45}{5} = 9 \)
\(\Rightarrow x = \pm \sqrt{9} = \pm 3 \)
But \( x = -3 \) is not required, because the side of a triangle cannot be negative.
\(\therefore x = 3 \Rightarrow\) Smaller side = 3 cm
\(\therefore\) Longer side = \( \sqrt{90 - 3^2} \)
\(= \sqrt{90 - 9} \)
\(= \sqrt{81} = 9 \) cm
Thus, the required sides of the triangle are 3 cm and 9 cm.

Please click the link below to download CBSE Class 10 Quadratic Equations Sure Shot Questions Set B.