CBSE Class 10 Maths HOTs Quadratic Equations Set M

Very Short Answer Type Questions

Question. Find the values of ‘k’ for which \( x = 2 \) is a solution of the equation \( kx^2 + 2x – 3 = 0 \).
Answer: Given equation is, \( kx^2 + 2x – 3 = 0 \)
if \( x = 2 \), then
\( \Rightarrow k (2)^2 + 2 (2) – 3 = 0 \)
\( \Rightarrow 4k = – 1 \)
\( \Rightarrow k = -\frac{1}{4} \)
Hence, the value of k is \( -\frac{1}{4} \)

Question. Find the value/s of k for which the quadratic equation \( 3x^2 + kx + 3 = 0 \) has real and equal roots.
Answer: Given, quadratic equation is: \( 3x^2 + kx + 3 = 0 \)
For real and equal roots \( b^2 – 4ac = 0 \)
Here, \( a = 3, b = k \) and \( c = 3 \)
\( \therefore b^2 – 4ac = (k)^2 – 4 \times 3 \times 3 = 0 \)
\( \Rightarrow k^2 = 36 \)
\( \Rightarrow k = \pm 6 \)
Hence, the value of k is –6.

Question. For what values of k does the quadratic equation \( 4x^2 – 12x – k = 0 \) have no real roots ?
Answer: Given equation is: \( 4x^2 – 12x – k = 0 \)
For equation to have no real roots, \( D < 0 \)
or \( b^2 – 4ac < 0 \)
Here, \( a = 4, b = – 12, c = – k \)
\( (– 12)^2 – 4 \times 4 \times (– k) < 0 \)
\( 144 + 16k < 0 \)
\( 16k < –144 \)
\( k < –9 \)
Hence, the value of k should be less than –9.

Question. Find the nature of the roots of the quadratic equation \( 2x^2 – 4x + 3 = 0 \).
Answer: Given: quadratic equation :
\( 2x^2 – 4x + 3 = 0 \)
Here, \( a = 2, b = – 4, c = 3 \)
Discriminant, \( D = b^2 – 4ac \)
\( = (– 4)^2 – 4 \times 2 \times 3 \)
\( = 16 – 24 = – 8 < 0 \)
\( \therefore D < 0 \)
Hence, the given equation does not have real roots.

Question. Is 0.2 a root of the equation \( x^2 – 0.4 = 0 \)? Justify.
Answer: No, because 0.2 does not satisfy the quadratic equation i.e.
\( x^2 – 0.4 = (0.2)^2 – 0.4 \)
\( = 0.04 – 0.4 \)
\( = -0.36 \neq 0 \)

Question. For what values of k, the roots of the equation \( x^2 + 4x + k = 0 \) are real?
Answer: Since, the roots of the equation \( x^2 + 4x + k = 0 \) are real,
i.e. \( D \ge 0 \)
\( b^2 – 4ac \ge 0 \)
Here, \( a = 1, b = 4, c = k \)
\( \Rightarrow (4)^2 – 4 \times 1 \times k \ge 0 \)
\( \Rightarrow 16 – 4k \ge 0 \)
\( \Rightarrow k \le 4 \)
Hence, the value of ‘k’ is less than or equal to 4.

Question. If \( x = 2 \) and \( m = 3 \), the equation is \( 3x^2 – 2kx + 2m = 0 \), find k.
Answer: \( 3x^2 – 2kx + 2m = 0 \)
\( x = 2 \) and \( m = 3 \) [Given]
So, \( 3(2)^2 – 2k(2) + 2(3) = 0 \)
\( \Rightarrow 12 – 4k + 6 = 0 \)
\( \Rightarrow – 4k + 18 = 0 \)
\( \Rightarrow k = \frac{9}{2} \cdot \)

Question. If one root of the quadratic equation \( 6x^2 – x – k = 0 \) is \( \frac{2}{3} \), then find the value of k’.
Answer: Given: quadratic equation: \( 6x^2 – x – k = 0 \).
Its one of its root: \( \frac{2}{3} \)
If \( x = \frac{2}{3} \) is root of the given equation, then it will satisfy the given equation:
Then, \( 6\left(\frac{2}{3}\right)^2 - \frac{2}{3} - k = 0 \)
\( \Rightarrow 6 \times \frac{4}{9} - \frac{2}{3} - k = 0 \)
\( \Rightarrow \frac{8}{3} - \frac{2}{3} - k = 0 \)
\( \Rightarrow k = 2 \)
Hence, the value of k is 2.

Question. For what values of ‘a’, does the quadratic equation \( x^2 – ax + 1 = 0 \) not have real roots?
Answer: Given quadratic equation is \( x^2 – ax + 1 = 0 \)
On comparing the given equation with \( Ax^2 + Bx + C = 0 \), we get:
\( A = 1, B = – a, C = 1 \)
For real roots, \( D > 0 \)
\( B^2 – 4AC > 0 \)
i.e. \( (– a)^2 – 4 \times 1 \times 1 = 0 \)
\( a^2 > 4 \)
or \( a > \sqrt{4} \)
or \( – 2 > a > 2 \)
Hence, the value of ‘a’ lies between – 2 and 2.

Question. Find the value of k for which the roots of the quadratic equation \( 2x^2 + kx + 8 = 0 \) will have equal value.
Answer: Given: quadratic equation is \( 2x^2 + kx + 8 = 0 \)
For roots of the equation to be equal.
\( D = 0 \)
i.e., \( b^2 – 4ac = 0 \)
Here, \( a = 2, b = k \) and \( c = 8 \)
\( \Rightarrow k^2 – 4 \times 2 \times 8 = 0 \)
\( \Rightarrow k^2 = 64 \)
\( \Rightarrow k = \pm 8 \)
Hence, the value of k is 8 or – 8.

Question. Solve for x : \( 6x^2 + 11x + 3 = 0 \)
Answer: \( 6x^2 + 11x + 3= 0 \)
\( \Rightarrow 6x^2 + 9x + 2x + 3 = 0 \)
\( \Rightarrow 3x (2x + 3) + 1 (2x + 3) = 0 \)
\( \Rightarrow (2x + 3) (3x + 1) = 0 \)
\( \Rightarrow 2x + 3 = 0 \) or \( 3x + 1 = 0 \)
i.e., \( x = -\frac{3}{2} \) or \( x = -\frac{1}{3} \)

Question. Solve for x : \( 8x^2 – 2x – 3 = 0 \)
Answer: \( 8x^2 – 2x – 3 = 0 \)
\( \Rightarrow 8x^2 – 6x + 4x – 3 = 0 \)
\( \Rightarrow 2x(4x – 3) + 1(4x – 3) = 0 \)
\( \Rightarrow (4x – 3) (2x + 1) = 0 \)
\( \Rightarrow 4x – 3 = 0 \) or \( 2x + 1 = 0 \)
i.e., \( x = \frac{3}{4} \) or \( -\frac{1}{2} \)

Question. Solve the following quadratic equation for x: \( 4x^2 + 4bx – (a^2 – b^2) = 0 \)
Answer: \( 4x^2 + 4bx + b^2 – a^2= 0 \)
\( \Rightarrow (2x + b)^2 – (a)^2= 0 \)
\( \Rightarrow (2x + b + a)(2x + b – a) = 0 \)
\( \Rightarrow x = -\frac{a+b}{2}, x = \frac{a-b}{2} \)

Short Answer Type Questions

Question. For what positive values of k, does the quadratic equation \( 3x^2 – kx + 3 = 0 \) not have real roots ?
Answer: Given: quadratic equation \( 3x^2 – kx + 3 = 0 \), has no real roots.
On comparing the given equation with \( ax^2 + bx + c = 0 \), we have:
\( a = 3, b = – k, c = 3 \)
Then, discriminant,
\( D = b^2 – 4ac \)
\( = (– k)^2 – 4 \times 3 \times 3 \)
\( = k^2 – 36 \)
But for no real roots, \( D < O \)
Then \( k^2 – 36 < 0 \)
\( k^2 < 36 \)
\( k < \pm 6 \)
\( k > – 6 \) or \( k < 6 \)
Hence, the value of \( k < 6 \) (positive value) for no real roots.

Question. Solve for x : \( \frac{1}{a + b + x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x} \); \( a \neq b \neq 0, x \neq 0, x \neq -(a + b) \)
Answer: \( \frac{1}{a + b + x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x} \)
\( \Rightarrow \frac{1}{a + b + x} - \frac{1}{x} = \frac{1}{a} + \frac{1}{b} \)
\( \Rightarrow \frac{x - (a + b + x)}{x(a + b + x)} = \frac{a + b}{ab} \)
\( \Rightarrow \frac{-(a + b)}{x(a + b + x)} = \frac{a + b}{ab} \)
\( \Rightarrow \frac{-1}{ax + bx + x^2} = \frac{1}{ab} \)
\( \Rightarrow - ab = ax + bx + x^2 \)
\( \Rightarrow x^2 + ax + bx + ab = 0 \)
\( \Rightarrow x(x + a) + b(x + a) = 0 \)
\( \Rightarrow (x + b) (x + a) = 0 \)
\( \Rightarrow x = -a, -b \)
Hence, the values of x are -a and -b.

Question. Solve the following quadratic equation : \( 6a^2x^2 – 7abx – 3b^2 = 0 \)
Answer: Given: quadratic equation is :
\( 6a^2x^2 – 7abx – 3b^2 = 0 \)
On comparing the given equation with \( AX^2 + BX + C = 0 \), we get:
\( A = 6a^2, B = –7ab, C = –3b^2 \)
Then, discrimnants.
\( D = \sqrt{B^2 - 4AC} \)
\( = \sqrt{(-7ab)^2 - 4 \times 6a^2 \times (-3b^2)} \)
\( = \sqrt{49a^2b^2 + 72a^2b^2} \)
\( = \sqrt{121a^2b^2} \)
\( = 11 ab \)
Then, roots of the equation, \( x = \frac{-B \pm D}{2A} \)
\( = \frac{-(-7ab) \pm 11ab}{2 \times 6a^2} \)
\( = \frac{18ab}{12a^2} \) and \( \frac{-4ab}{12a^2} \)
\( = \frac{3b}{2a} \) and \( \frac{-b}{3a} \)
Hence, the roots of the given equation are: \( \frac{3b}{2a} \) and \( \frac{-b}{3a} \).

Question. Solve for x : \( \sqrt{3}x^2 + 10x - 8\sqrt{3} = 0 \)
Answer: Given: \( \sqrt{3}x^2 + 10x - 8\sqrt{3} = 0 \)
On comparing the above equation with \( ax^2 + bx + c = 0 \),
we get : \( a = \sqrt{3} \) , \( b = 10 \) and \( c = – 8\sqrt{3} \)
Then, discriminant, \( D = b^2 – 4ac \)
\( = (10)^2 – 4 \times \sqrt{3} \times ( – 8 \sqrt{3} ) \)
\( = 100 + 96 \)
\( = 196 \)
Roots of equation, \( x = \frac{- b \pm \sqrt{D}}{2a} \)
\( = \frac{- 10 \pm \sqrt{196}}{2 \times \sqrt{3}} \)
\( = \frac{- 10 \pm 14}{2 \sqrt{3}} \)
\( = \frac{4}{2 \sqrt{3}} \) or \( -\frac{24}{2 \sqrt{3}} \)
\( = \frac{2}{\sqrt{3}} \) or \( -\frac{12}{\sqrt{3}} \)
\( = \frac{2 \sqrt{3}}{3} \) or \( - 4 \sqrt{3} \)
Hence, the roots of the given equation are \( \frac{2 \sqrt{3}}{3} \) and \( - 4 \sqrt{3} \).

Question. A quadratic equation with integral coefficient has integral roots. Justify your answer.
Answer: No, the given statement is not always true.
Consider the quadratic equation
\( 8x^2 – 2x – 1 = 0 \)
By splitting the middle term,
\( 8x^2 – 4x + 2x – 1 = 0 \)
\( 4x(2x – 1) + 1(2x – 1) = 0 \)
\( (4x + 1)(2x – 1) = 0 \)
If \( 4x + 1 = 0 \Rightarrow x = – \frac{1}{4} \)
\( 2x – 1 = 0 \Rightarrow x = \frac{1}{2} \)
So, the given equation has integral coefficients but no integral roots.
Hence, the given statement is false.

Question. Does there exist a quadratic equation whose coefficients are rational but both of its roots are irrational? Justify your answer.
Answer: Yes, there exists a quadratic equation whose coefficients are rational but both of its roots are irrational .
Consider the quadratic equation
\( x^2 – 6x + 7 = 0 \)
Here, \( D = b^2 – 4ac \)
\( = (–6)^2 – 4(1)(7) \)
\( \Rightarrow D = 36 – 28 = 8 \)
Since, discriminant is not a perfect square, therefore it will have irrational roots.
The roots will be
\( \frac{-b \pm \sqrt{D}}{2a} = \frac{6 \pm \sqrt{8}}{2} = \frac{6 \pm 2\sqrt{2}}{2} \)
The roots will be \( 3 \pm \sqrt{2} \)
i.e. \( 3 + \sqrt{2} \) and \( 3 – \sqrt{2} \), are both irrational.

Question. Solve for x : \( \frac{x + 3}{x + 2} = \frac{3x - 7}{2x - 3} \), \( x \neq -2, \frac{3}{2} \)
Answer: \( \Rightarrow (x + 3) (2x – 3) = (x + 2) (3x – 7) \)
\( \Rightarrow 2x^2 + 6x – 3x – 9 = 3x^2 + 6x – 7x – 14 \)
\( \Rightarrow 2x^2 – 3x^2 + 3x + x – 9 + 14 = 0 \)
\( \Rightarrow – x^2 + 4x + 5 = 0 \)
\( \Rightarrow x^2 – 4x – 5 = 0 \)
\( \Rightarrow x^2 – 5x + x – 5 = 0 \)
(on splitting the middle term)
\( \Rightarrow x(x – 5) + 1(x – 5) = 0 \)
\( \Rightarrow (x + 1) (x – 5) = 0 \)
\( \Rightarrow x = – 1, 5 \)
Hence, the values of x are – 1 and 5.

Question. Find the roots of the quadratic equation \( \sqrt{2}x^2 + 7x + 5\sqrt{2} = 0 \).
Answer: Given, quadratic equation is
\( \sqrt{2}x^2 + 7x + 5\sqrt{2} = 0 \)
On comparing the above equation with \( ax^2 + bx + x = 0 \)
we get \( a = \sqrt{2} , b = 7, c = 5\sqrt{2} \)
Then, discriminant,
\( D = b^2 – 4ac \)
\( = (7)^2 – 4 \times \sqrt{2} \times 5\sqrt{2} \)
\( = 49 – 40 = 9 \)
Now, \( x = \frac{-b \pm \sqrt{D}}{2a} \)
\( = \frac{-7 \pm \sqrt{9}}{2\sqrt{2}} = \frac{-7 \pm 3}{2\sqrt{2}} \)
\( \therefore x = \frac{-4}{2\sqrt{2}} \) and \( \frac{-10}{2\sqrt{2}} \)
\( = \frac{-2}{\sqrt{2}} \) and \( \frac{-5}{\sqrt{2}} \)
Hence, the roots of the given equation is \( -\sqrt{2} \) and \( \frac{-5\sqrt{2}}{2} \).

Question. If b = 0 and c < 0, is it true that the roots of \( x^2 + bx + c = 0 \) are numerically equal and opposite in sign? Justify.
Answer: It is given that b = 0 and c < 0.
The given quadratic equation is:
\( x^2 + bx + c = 0 \)
On putting b = 0 in this equation, we get
\( x^2 + 0.x + c = 0 \)
\( x^2 + c = 0 \)
\( \Rightarrow x^2 = –c \)
Here, \( c < 0 \Rightarrow –c > 0 \)
\( \Rightarrow x = \pm \sqrt{-c} \)
Hence, the roots of \( x^2 + bx + c = 0 \) are numerically equal and opposite in sign.

Question. Find the value of k for which the equation \( x^2 + k(2x + k – 1) + 2 = 0 \) has real and equal roots.
Answer: Given, quadratic equation is:
\( x^2 + 2xk + (k^2 – k + 2) = 0 \)
On comparing the quadratic equation, with \( ax^2 + bx + c = 0 \), we get:
\( a = 1, b = 2k, c = k^2 – k + 2 \)
Since, the roots of the above equation are real and equal.
\( \therefore \) Discriminant, \( D = 0 \)
i.e., \( b^2 – 4ac = 0 \)
\( (2k)^2 – 4 \times 1 \times (k^2 – k + 2) = 0 \)
\( 4k^2 – 4k^2 + 4k – 8 = 0 \)
\( 4k – 8 = 0 \)
\( k = 2 \)
Hence, the value of k is 2.

Question. If \( x = \frac{2}{3} \) and \( x = – 3 \) are roots of the quadratic equation \( ax^2 + 7x + b = 0 \), find the values of a and b.
Answer: Since \( x = \frac{2}{3} \) and \( x = – 3 \) are the roots of the quadratic equation
\( ax^2 + 7x + b = 0 \)
Now, sum of roots:
\( \frac{2}{3} + (-3) = -\frac{7}{a} \)
\( \Rightarrow -\frac{7}{3} = -\frac{7}{a} \Rightarrow a = 3 \).
Product of roots:
\( \Rightarrow \frac{2}{3} \times (- 3 ) = \frac{b}{a} \)
\( \Rightarrow – 2 = \frac{b}{3} \) [\( \because a = 3 \)]
\( \Rightarrow b = – 6 \)
Hence, the values of a and b are 3 and –6 respectively.

Question. If a and b are the roots of the equation \( x^2 + ax – b = 0 \), then find a and b.
Answer: \( x^2 + ax – b = 0 \)
Sum of the roots = \( a + b \)
\( = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} = -a \)
Product of roots = \( ab \)
\( = \frac{\text{constant term}}{\text{Coefficient of } x^2} = b \) [sic]
So, \( a + b = –a \)
\( b = –2a \)
and, \( ab = –b \)
\( a = –1 \)
Putting the value of a, we get
\( b = –2 \times (–1) = 2 \)
Hence, \( a = –1 \) and \( b = 2 \).

Question. Solve for x : \( \frac{1}{x + 4} - \frac{1}{x - 7} = \frac{11}{30} \), \( x \neq -4, 7 \).
Answer: Solve for x :
\( \frac{1}{x + 4} - \frac{1}{x - 7} = \frac{11}{30} \)
Now \( \frac{(x - 7) - (x + 4)}{(x + 4)(x - 7)} = \frac{11}{30} \)
\( \frac{x - 7 - x - 4}{(x + 4)(x - 7)} = \frac{11}{30} \)
\( \frac{-11}{x^2 - 3x - 28} = \frac{11}{30} \)
\( -30 = x^2 - 3x - 28 \)
\( x^2 - 3x + 2 = 0 \)
\( (x - 2) (x - 1) = 0 \)
i.e. \( x - 1 = 0 \) or \( x - 2 = 0 \)
\( x = 1 \) or \( 2 \)

Question. Determine the condition for one root of the quadratic equation \( ax^2 + bx + c = 0 \) to be thrice the other.
Answer: Let the roots of the equation \( ax^2 + bx + c \) be \( \alpha \) and \( 3\alpha \).
Then, sum of the roots = \( \alpha + 3\alpha = 4\alpha = -\frac{b}{a} \)
\( \Rightarrow \alpha = -\frac{b}{4a} \)
Product of the roots = \( \alpha \times 3\alpha = 3\alpha^2 = \frac{c}{a} \)
Substitute value of \( \alpha \):
\( 3 \left( -\frac{b}{4a} \right)^2 = \frac{c}{a} \)
\( 3 \frac{b^2}{16a^2} = \frac{c}{a} \)
\( \frac{3b^2}{16a} = c \)
\( 3b^2 = 16ac \), which is the required condition.

Question. Write all the values of p for which the quadratic equation \( x^2 + px + 16 = 0 \) has equal roots. Find the roots of the equation so obtained.
Answer: Given equation: \( x^2 + px + 16 = 0 \)
Here, \( a = 1, b = p, c = 16 \)
Discriminant, \( D = b^2 - 4ac \)
\( = p^2 - 4 \times 1 \times 16 \)
\( = p^2 - 64 \)
If roots are equal, then:
\( D = 0 \)
i.e. \( p^2 - 64 = 0 \)
\( \Rightarrow p^2 = 64 \)
\( \Rightarrow p = \pm 8 \)
\( \therefore \) Equation is \( x^2 \pm 8x + 16 = 0 \)
\( \Rightarrow (x \pm 4)^2 = 0 \) [\( \because (a + b)^2 = a^2 \pm 2ab + b^2 \)]
\( \Rightarrow x \pm 4 = 0 \)
\( \Rightarrow x = -4, 4 \)
Hence, roots are \( x = -4 \) and \( x = 4 \) and the values of p are -8 and 8.

Question. Solve for x : \( \frac{x + 3}{x + 2} = \frac{3x - 7}{2x - 3} ; x \neq -2, \frac{3}{2} \)
Answer: \( \frac{x + 3}{x + 2} = \frac{3x - 7}{2x - 3} \)
On cross-multiplying
\( (x + 3) (2x - 3) = (3x - 7) (x + 2) \)
\( 2x^2 + 6x - 3x - 9 = 3x^2 - 7x + 6x - 14 \)
\( 2x^2 + 3x - 9 = 3x^2 - x - 14 \)
\( -x^2 + 4x + 5 = 0 \)
\( x^2 - 4x - 5 = 0 \)
\( x^2 - 5x + x - 5 = 0 \)
\( x(x - 5) + 1(x - 5) = 0 \)
\( (x + 1) (x - 5) = 0 \)
\( x = -1, 5 \)
Hence, the value of x are -1 and 5.

Question. The product of two successive integral multiples of 5 is 1050. Determine the multiples.
Answer: Let two successive integral multiples of 5 be x and (x + 5)
According to question,
\( x(x + 5) = 1050 \)
\( x^2 + 5x - 1050 = 0 \)
\( (x - 30) (x + 35) = 0 \)
\( x = 30 \) or -35
When \( x = 30 \),
Multiples are 30 and \( 30 + 5 = 35 \)
When \( x = -35 \),
Multiples are -35 and \( -35 + 5 = -30 \)

Long Short Answer Type Questions

Question. A train travels 360 km at a uniform speed. In the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Answer: Let, the actual speed of the train be 'x' km/hr.
Time taken by the train at actual speed, \( t_1 = \frac{360}{x} \) hr
Increased speed of the train = (x + 5) km / hr
Time taken by the train at the increased speed \( t_2 = \frac{360}{x + 5} \) hr
According to the given condition:
\( t_1 - t_2 = 1 \)
\( \Rightarrow \frac{360}{x} - \frac{360}{x + 5} = 1 \)
\( \Rightarrow \frac{360(x + 5 - x)}{x(x + 5)} = 1 \)
\( \Rightarrow 360 \times 5 = x^2 + 5x \)
\( \Rightarrow x^2 + 5x - 1800 = 0 \)
\( \Rightarrow x^2 + 45x - 40x - 1800 = 0 \)
\( \Rightarrow x(x + 45) - 40(x + 45) = 0 \)
\( \Rightarrow (x - 40) (x + 45) = 0 \)
\( \Rightarrow x = 40 \) [\( \because x = -45 \), is not possible as speed cannot be negative]
Hence, the actual speed of the train is 40 km/hr.

Question. A line segment AB of length 2m is divided at a point C into two parts such that \( AC^2 = AB \times CB \). Find the length of CB.
Answer: Let the length of AC be x
The, \( BC = 2 - x \)
\( AC^2 = AB \times CB \) (given)
\( \therefore x^2 = 2 \times (2 - x) \) [\( \because AB = 2m \) (given)]
\( x^2 = 4 - 2x \)
\( x^2 + 2x - 4 = 0 \)
Now, if we compare the above equation with \( ax^2 + bx + c = 0 \).
Then, \( a = 1, b = 2, c = -4 \)
Root of the equation are,
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( = \frac{-2 \pm \sqrt{(2)^2 - 4 \times 1 \times (-4)}}{2} \)
\( = \frac{-2 \pm \sqrt{20}}{2} \)
\( = -1 \pm \sqrt{5} \)
\( = -1 + \sqrt{5} \) or \( -1 - \sqrt{5} \)
Then, \( BC = 2 - x \)
when \( x = -1 + \sqrt{5} \)
\( BC = 2 - (-1 + \sqrt{5}) = 3 - \sqrt{5} \)
\( = 3 - 2.24 = 0.76 \)
When \( x = -1 - \sqrt{5} \)
\( BC = 2 - (-1 - \sqrt{5}) = 3 + \sqrt{5} \)
\( = 3 + 2.24 = 5.24 \) (which is not possible)
Hence, the length of BC is \( 3 - \sqrt{5} \) or 0.76 cm.

Question. Show that if the roots of the following quadratic equation are equal, then ad = bc
\( x^2(a^2 + b^2) + 2(ac + bd)x + (c^2 + d^2) = 0 \)
Answer: Given, quadratic equation is :
\( x^2(a^2 + b^2) + 2(ac + bd)x + (c^2 + d^2) = 0 \)
whose roots are equal.
To Prove : \( ad = bc \)
Proof : In the given equation,
\( A = a^2 + b^2 \)
\( B = 2(ac + bd) \)
\( C = c^2 + d^2 \)
Since, roots of the given equation are equal.
\( \therefore \) Discriminant \( B^2 - 4AC = 0 \)
\( [2(ac + bd)]^2 - 4 \times (a^2 + b^2) (c^2 + d^2) = 0 \)
\( \Rightarrow 4(ac + bd)^2 - 4(a^2c^2 + b^2c^2 + a^2d^2 + b^2d^2) = 0 \)
\( \Rightarrow 4(a^2c^2 + b^2d^2 + 2abcd) - 4a^2c^2 - 4b^2c^2 - 4a^2d^2 - 4b^2d^2 = 0 \)
\( \Rightarrow 8abcd - 4a^2d^2 - 4b^2c^2 = 0 \)
\( \Rightarrow a^2d^2 + b^2c^2 - 2abcd = 0 \)
\( \Rightarrow (ad - bc)^2 = 0 \)
On taking square-root on both sides
\( \Rightarrow ad - bc = 0 \)
\( \Rightarrow ad = bc \)
Hence, proved.

Question. Solve the given quadratic equation for x: \( 9x^2 - 9(a + b)x + (2a^2 + 5ab + 2b^2) = 0 \)
Answer: \( 9x^2 - 9(a + b)x + (2a^2 + 5ab + 2b^2) = 0 \)
On comparing the given equation with \( ax^2 + bx + c = 0 \), we get :
\( a = 9, b = - 9(a + b), c = 2a^2 + 5ab + 2b^2 \)
Discriminant, \( D = b^2 - 4ac \)
\( = [- 9(a + b)]^2 - 4 \times 9(2a^2 + 5ab + 2b^2) \)
\( = 81(a + b)^2 - 36(2a^2 + 5ab + 2b^2) \)
\( = 81(a^2 + b^2 + 2ab) - 72a^2 - 180ab - 72b^2 \)
\( = 81a^2 + 81b^2 + 162ab - 72a^2 - 180ab - 72b^2 \)
\( = 9a^2 + 9b^2 - 18ab \)
\( = 9(a^2 + b^2 - 2ab) \)
\( = 9(a - b)^2 = [3(a - b)]^2 \)
Then, roots : \( x = \frac{-b \pm \sqrt{D}}{2a} \)
\( = \frac{9(a + b) \pm \sqrt{3^2(a - b)^2}}{2 \times 9} \)
\( = \frac{9(a + b) \pm 3(a - b)}{18} \)
\( = \frac{3(a + b) \pm (a - b)}{6} \)
\( = \frac{3a + 3b + a - b}{6} \) or \( \frac{3a + 3b - a + b}{6} \)
\( = \frac{4a + 2b}{6} \) or \( \frac{2a + 4b}{6} \)
\( = \frac{2a + b}{3} \) or \( \frac{a + 2b}{3} \)
Hence, the roots of the given equation are \( \frac{2a + b}{3} \) and \( \frac{a + 2b}{3} \).

Question. Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number.
Answer: Let n be the required natural number.
According to the question:
Square of natural number diminished by 84 gives \( n^2 - 84 \).
Thrice of 8 more than given number = 3(8 + n).
According to the question,
\( n^2 - 84 = 3(8 + n) \)
\( \Rightarrow n^2 - 84 = 24 + 3n \)
\( \Rightarrow n^2 - 3n - 108 = 0 \)
Splitting the middle term, we have
\( \Rightarrow n^2 - 12n + 9n - 108 = 0 \)
\( \Rightarrow n(n - 12) + 9(n - 12) = 0 \)
\( \Rightarrow (n - 12)(n + 9) = 0 \)
\( n = 12 \) or \( n = -9 \)
But \( n \neq -9 \) as n is a natural number.
Hence, the required natural number is 12.

Question. A natural number, when increased by 12, equals 160 times its reciprocal. Find the number.
Answer: Let n be the required natural number.
According to the question, number when increased by 12 is \( n + 12 \).
160 times number's reciprocal = \( 160 (\frac{1}{n}) = \frac{160}{n} \)
Now, by the given condition
\( n + 12 = \frac{160}{n} \)
\( \Rightarrow n(n + 12) = 160 \)
\( \Rightarrow n^2 + 12n - 160 = 0 \)
Splitting the middle term, we have
\( n^2 + 20n - 8n - 160 = 0 \)
\( n(n + 20) - 8(n + 20) = 0 \)
\( (n + 20)(n - 8) = 0 \)
\( n = -20 \) or 8
But \( n \neq -20 \) as n is a natural number.
Hence, the required number is 8.

Question. A plane left 30 minutes later than the scheduled time and in order to reach its destination 1500 km away on time it has to increase its speed by 250 km/hr from its usual speed. Find the usual speed of the plane.
Answer: Let 'x' km / hr be the speed of the plane.
Increased speed = (x + 250) km / hr
Time taken at the usual speed = \( \frac{1500}{x} \) hr
Time taken at increased speed = \( \frac{1500}{x + 250} \) hr
Difference between the two times taken = 30 minutes = \( \frac{1}{2} \) hr
\( \therefore \frac{1500}{x} - \frac{1500}{x + 250} = \frac{1}{2} \)
\( 1500 \left[ \frac{1}{x} - \frac{1}{x + 250} \right] = \frac{1}{2} \)
\( \frac{3000(x + 250 - x)}{x(x + 250)} = 1 \)
\( x^2 + 250x = 250 \times 3000 \)
\( x^2 + 250x - 750000 = 0 \)
\( x^2 + 1000x - 750x - 750000 = 0 \)
\( x(x + 1000) - 750(x + 1000) = 0 \)
\( (x - 750) (x + 1000) = 0 \)
\( x = 750 \) or -1000
\( x = 750 \) [\( \because \) speed cannot be negative]
Hence, the usual speed of the plane is 750 km/hr.

Question. Find the dimensions of a rectangular park whose perimeter is 60 m and area 200 m\(^2\).
Answer: Let 'l' be the length and 'b' be the breadth of the rectangular park.
Perimeter of the park, \( p = 2(l + b) \)
Area of the park, \( A = l \times b \)
According to the given conditions:
\( 2(l + b) = 60 \)
\( l + b = 30 \)
\( l = 30 - b \) ....(i)
and \( lb = 200 \)
\( (30 - b)b = 200 \) [from (i)]
\( 30b - b^2 = 200 \)
\( b^2 - 30b + 200 = 0 \)
on splitting the middle term, we get:
\( b^2 - 20b - 10b + 200 = 0 \)
\( (b - 20) (b - 10) = 0 \)
\( b = 20 \) or 10
when \( b = 20m \), \( l = 10 m \)
when \( b = 10m \), \( l = 20 m \)
Hence, the length and breadth of the rectangle are 10 m and 20 m or 20 m and 10 m respectively.

Question. If Zeba were younger by 5 years than what she really is, then the square of her age (in years) would have been 11 more than five times her actual age. What is her age now?
Answer: Let actual age of Zeba be x years.
Her age when she was 5 years younger = (x - 5) years.
According to the condition given in question:
Square of her age = 11 more than 5 times her actual age
\( (x - 5)^2 = 11 + 5(x) \)
\( \Rightarrow x^2 + 25 - 10x = 11 + 5x \)
[\( \because (a - b)^2 = a^2 + b^2 - 2ab \)]
\( \Rightarrow x^2 - 10x - 5x + 25 - 11 = 0 \)
\( \Rightarrow x^2 - 15x + 14 = 0 \)
Splitting the middle term, we have
\( \Rightarrow x^2 - 14x - x + 14 = 0 \)
\( \Rightarrow x(x - 14) - 1(x - 14) = 0 \)
\( \Rightarrow (x - 14)(x - 1) = 0 \)
\( \Rightarrow x = 14 \) or \( x = 1 \)
But \( x \neq 1 \) as in that case (x - 5) will not be possible
\( \Rightarrow x = 14 \)
Hence, Zeba's age now is 14 years.

Question. 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 would be one year less than 10 times the present age of Nisha. Find the present ages of both Asha and Nisha.
Answer: Let Nisha's present age be x years.
Then, Asha's present age = \( (2 + x^2) \)
[By the given condition]
Now, when Nisha grows to her mother's present age i.e. after \( \{(x^2 + 2) - x\} \) years.
Then, Asha's age will become \( \{(x^2 + 2) - x\} \) years.
Now by the given condition,
Asha's age = 1 year less than 10 times present age of Nisha.
\( (2 + x^2) + \{(x^2 + 2) - x\} = 10x - 1 \)
\( \Rightarrow 2 + x^2 + x^2 + 2 - x = 10x - 1 \)
\( \Rightarrow 2x^2 - 11x + 5 = 0 \)
Splitting the middle term, we have
\( \Rightarrow 2x^2 - 10x - x + 5 = 0 \)
\( \Rightarrow 2x(x - 5) - 1(x - 5) = 0 \)
\( \Rightarrow (x - 5)(2x - 1) = 0 \)
\( \Rightarrow x = 5 \) or \( x = \frac{1}{2} \)
But \( x \neq \frac{1}{2} \) as then Nisha's age = \( \frac{1}{2} \). This means that her mother Asha's age = \( (x^2 + 2) = (\frac{1}{4} + 2) = 2\frac{1}{4} \) years which is not possible.
Hence, the present age of Nisha = 5 years and the present age of Asha = \( x^2 + 2 = 5^2 + 2 = 25 + 2 = 27 \) years.

Question. In a class test, the sum of Arun's marks in Hindi and English is 30. Had he got 2 marks more in Hindi and 3 marks less in English, the product of the marks would have been 210. Find his marks in the two subjects.
Answer: Let, the Arun's marks in Hindi be x.
Then, marks in English = \( 30 - x \)
According to the given condition,
\( \Rightarrow (x + 2) (30 - x - 3) = 210 \)
\( \Rightarrow (x + 2) (27 - x) = 210 \)
\( \Rightarrow 27x + 54 - x^2 - 2x = 210 \)
\( \Rightarrow x^2 - 25x + 156 = 0 \)
\( x^2 - 13x - 12x + 156 = 0 \)
\( x(x - 13) - 12(x - 13) = 0 \)
\( \Rightarrow (x - 13) (x - 12) = 0 \)
\( \Rightarrow x = 13 \) or 12
When : \( x = 13 \)
Marks in Hindi = 13
Marks in English = \( 30 - 13 = 17 \)
When : \( x = 12 \)
Marks in Hindi = 12
Marks in English = \( 30 - 12 = 18 \)
Hence, the marks obtained in the two subjects are (13, 17) or (12, 18).

Question. A motor boat whose speed is 18 km/hr in still water takes 1 hr more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.
Answer: Let, the speed of the stream be x km/hr.
Speed of the boat in still water = 18 km/hr
\( \therefore \) The speed of the boat in upstream = (18 - x) km/hr
The speed of the boat in downstream = (18 + x) km/hr.
Total distance to be covered = 24 km
\( \therefore \) Time taken in upstream, \( t_1 = \frac{24}{18 - x} \) hr
Time taken in downstream, \( t_2 = \frac{24}{18 + x} \) hr
According to the question,
\( \frac{24}{18 - x} - \frac{24}{18 + x} = 1 \)
\( 24 \left[ \frac{18 + x - (18 - x)}{(18 - x)(18 + x)} \right] = 1 \)
\( \Rightarrow 24 \times 2x = 324 - x^2 \)
\( \Rightarrow x^2 + 48x - 324 = 0 \)
\( \Rightarrow x^2 + 54x - 6x - 324 = 0 \)
\( \Rightarrow x(x + 54) - 6(x + 54) = 0 \)
\( \Rightarrow (x + 54) (x - 6) = 0 \)
\( x \neq -54 \) ( \( \because \) speed can't be negative)
\( \therefore x = 6 \)
Hence, the speed of the stream is 6 km/hr.

Question. A train travels at a certain average speed for a distance of 63 km and then travels at a distance of 72 km at an average speed of 6 km/hr more than its original speed. If it takes 3 hours to complete total journey, what is the original average speed ?
Answer: Let the original speed of the train be 'x' km/hr.
Increased speed = (x + 6) km/hr
Now, time taken to cover 60 km at original speed, \( t_1 = \frac{63}{x} \) hr
Time taken to cover 72 km at increased speed: \( t_2 = \frac{72}{x + 6} \) hr
\( \frac{63}{x} + \frac{72}{x + 6} = 3 \)
\( \Rightarrow \frac{63x + 378 + 72x}{x(x + 6)} = 3 \)
\( \Rightarrow 135x + 378 = 3(x^2 + 6x) \)
\( \Rightarrow 3x^2 + 18x - 135x - 378 = 0 \)
\( \Rightarrow 3x^2 - 117x - 378 = 0 \)
\( \Rightarrow x^2 - 42x + 3x - 126 = 0 \)
\( \Rightarrow x(x - 42) + 3(x - 42) = 0 \)
\( \Rightarrow (x - 42) (x + 3) = 0 \)
\( x = -3 \) (\( \because \) speed can't be negative)
\( \therefore x = 42 \)
Hence, the original average speed of train is 42 km/hr.

Question. The altitude of a right-angled triangle is 7 cm less than its base. If the hypotenuse is 13 cm, then find the other two sides.
Answer: Let, the base of the right angled triangle be 'x' cm.
Then, the altitude of a right-angled triangle is (x - 7) cm.
And, the hypotenuse of right angled triangle = 13 cm
Then, by the pythagoras theorem
\( H^2 = P^2 + B^2 \)
\( 13^2 = (x - 7)^2 + x^2 \)
\( 169 = x^2 + 49 - 14x + x^2 \)
\( 2x^2 - 14x - 120 = 0 \)
\( x^2 - 7x - 60 = 0 \)
\( x^2 - 12x + 5x - 60 = 0 \)
\( x(x - 12) + 5(x - 12) = 0 \)
\( (x + 5) (x - 12) = 0 \)
\( x = 12 \) (\( \because x = -5 \) is not possible)
The base of the right angled triangle = 12 cm
and altitude = 12 - 7 = 5 cm.
Hence, the other two sides of triangle are 5 cm and 12 cm.

Question. Solve for x : \( \frac{x + 3}{x - 2} - \frac{1 - x}{x} = \frac{17}{4} ; x \neq 0, 2 \)
Answer: \( \frac{x + 3}{x - 2} - \frac{1 - x}{x} = \frac{17}{4} \)
\( \Rightarrow \frac{x(x + 3) - (x - 2)(1 - x)}{x(x - 2)} = \frac{17}{4} \)
\( \Rightarrow \frac{(x^2 + 3x) - (x - 2 - x^2 + 2x)}{x^2 - 2x} = \frac{17}{4} \)
\( \Rightarrow (x^2 + 3x - 3x + 2 + x^2) \times 4 = 17(x^2 - 2x) \)
\( \Rightarrow 4(2x^2 + 2) = 17x^2 - 34x \)
\( \Rightarrow 8x^2 + 8 = 17x^2 - 34x \)
\( \Rightarrow 17x^2 - 8x^2 - 34x - 8 = 0 \)
\( \Rightarrow 9x^2 - 34x - 8 = 0 \)
\( \Rightarrow 9x^2 - 36x + 2x - 8 = 0 \)
\( \Rightarrow 9x(x - 4) + 2(x - 4) = 0 \)
\( \Rightarrow (9x + 2) (x - 4) = 0 \)
\( \Rightarrow x = -\frac{2}{9}, 4 \)
Hence, the values of x are \( -\frac{2}{9}, 4 \)

Question. Find two consecutive odd natural numbers, the sum of whose squares is 394.
Answer: Let, the first number be x
and the second number be (x + 2).
According to the given condition,
\( x^2 + (x + 2)^2 = 394 \)
\( \Rightarrow x^2 + x^2 + 4 + 4x = 394 \)
\( \Rightarrow 2x^2 + 4x - 390 = 0 \)
\( \Rightarrow x^2 + 2x - 195 = 0 \)
\( \Rightarrow x^2 + 15x - 13x - 195 = 0 \)
\( \Rightarrow x(x + 15) - 13(x + 15) = 0 \)
\( \Rightarrow (x - 13) (x + 15) = 0 \)
\( \Rightarrow x = 13, -15 \)
\( x \neq -15 \) [\( \because \) natural numbers are positive]
Hence, the two consecutive odd natural numbers are 13 and 15.

Question. A and B working together can do a work in 6 days. If A takes 5 days less than B to finish the work, in how many days can B can do the work alone?
Answer: Let B take 'x' days to complete the work.
Then, A takes (x - 5) days to complete the work done.
According to the given condition:
\( \frac{1}{x} + \frac{1}{x - 5} = \frac{1}{6} \)
\( \Rightarrow \frac{x - 5 + x}{x(x - 5)} = \frac{1}{6} \)
\( \Rightarrow 6(2x - 5) = x^2 - 5x \)
\( \Rightarrow x^2 - 5x - 12x + 30 = 0 \)
\( \Rightarrow x^2 - 17x + 30 = 0 \)
\( \Rightarrow x^2 - 15x - 2x + 30 = 0 \)
(on splitting the middle term)
\( x(x - 15) - 2(x - 15) = 0 \)
\( (x - 15)(x - 2) = 0 \)
\( x = 2 \) or 15
But \( x = 2 \) is not possible as \( x < 5 \) (A takes x-5 which would be negative).
\( \therefore x = 15 \)
Hence, B takes 15 days to complete the work alone.

Question. Find x in terms of a, b and c : \( \frac{a}{x - a} + \frac{b}{x - b} = \frac{2c}{x - c} \), \( x \neq a, b, c \)
Answer: \( \frac{a}{x - a} + \frac{b}{x - b} = \frac{2c}{x - c} \)
\( \Rightarrow \frac{a(x - b) + b(x - a)}{(x - a)(x - b)} = \frac{2c}{x - c} \)
\( \Rightarrow \frac{ax - ab + bx - ab}{x^2 - ax - bx + ab} = \frac{2c}{x - c} \)
\( \Rightarrow (ax - 2ab + bx)(x - c) = (x^2 - ax - bx + ab)(2c) \)
\( \Rightarrow ax^2 - acx - 2abx + 2abc + bx^2 - bcx = 2cx^2 - 2acx - 2bcx + 2abc \)
\( \Rightarrow (a + b)x^2 - x(ac + 2ab + bc) = 2cx^2 - x(2ac + 2bc) \)
\( \Rightarrow x^2(a + b - 2c) - x(ac + 2ab + bc - 2ac - 2bc) = 0 \)
\( \Rightarrow x^2(a + b - 2c) - x(2ab - ac - bc) = 0 \)
\( \Rightarrow x [x(a + b - 2c) - (2ab - ac - bc)] = 0 \)
\( \Rightarrow x = 0 \) or \( x(a + b - 2c) = 2ab - ac - bc \)
\( \Rightarrow x = \frac{2ab - ac - bc}{a + b - 2c} \)
Hence, the 'x' in terms of a, b, c is \( \frac{2ab - ac - bc}{a + b - 2c} \).

Question. At t minutes past 2 pm, the time needed by the minutes hand of a clock to show 3 pm was found to be 3 minutes less than \( \frac{t^2}{4} \) minutes. Find t.
Answer: It is given that at t minutes past 2 pm, the time needed by the minute hand to show 3 pm was found to be 3 minutes less than \( \frac{t^2}{4} \) min.
\( \Rightarrow t + (\frac{t^2}{4} - 3) = 60 \)
[ \( \because \) time between 2 pm and 3 pm = 1 hour = 60 min.]
\( \Rightarrow 4t + t^2 - 12 = 240 \)
\( \Rightarrow t^2 + 4t - 12 - 240 = 0 \)
\( \Rightarrow t^2 + 4t - 252 = 0 \)
Splitting the middle term, we have
\( t^2 + 18t - 14t - 252 = 0 \)
\( \Rightarrow t(t + 18) - 14(t + 18) = 0 \)
\( \Rightarrow (t + 18)(t - 14) = 0 \)
\( t = -18 \) or \( t = 14 \).
But \( t \neq -18 \) as time cannot be negative
\( \Rightarrow t = 14 \)
Hence, the required value of t is 14 minutes.

Question. Solve for x : \( \frac{2x}{x - 3} + \frac{1}{2x + 3} + \frac{3x + 9}{(x - 3)(2x + 3)} = 0 \), \( x \neq 3, -3/2 \)
Answer: \( \frac{2x}{x - 3} + \frac{1}{2x + 3} + \frac{3x + 9}{(x - 3)(2x + 3)} = 0 \)
\( \Rightarrow \frac{2x(2x + 3) + (x - 3) + 3x + 9}{(x - 3)(2x + 3)} = 0 \)
\( \Rightarrow 4x^2 + 6x + x - 3 + 3x + 9 = 0 \)
\( \Rightarrow 4x^2 + 10x + 6 = 0 \)
\( \Rightarrow 2x^2 + 5x + 3 = 0 \)
\( \Rightarrow 2x^2 + 2x + 3x + 3 = 0 \)
(on splitting the middle term)
\( \Rightarrow 2x(x + 1) + 3(x + 1) = 0 \)
\( \Rightarrow (2x + 3) (x + 1) = 0 \)
\( \Rightarrow x = -\frac{3}{2} \) or -1
Since \( x \neq -\frac{3}{2} \), x = -1.
Hence, the value of x is -1.

Question. Find the roots of the quadratic equations by using the quadratic formula in each of the following:
(A) \( 2x^2 – 3x – 5 = 0 \)
(B) \( 5x^2 + 13x + 8 = 0 \)
(C) \( -3x^2 + 5x + 12 = 0 \)
(D) \( -x^2 + 7x - 10 = 0 \)
(E) \( x^2 + 2\sqrt{2}x - 6 = 0 \)
(F) \( x^2 - 3\sqrt{5}x + 10 = 0 \)
(G) \( \frac{1}{2}x^2 - \sqrt{11}x + 1 = 0 \)
Answer:
(A) The given equation is: \( 2x^2 – 3x – 5 = 0 \)
On comparing with \( ax^2 + bx + c = 0 \), we get
\( a = 2, b = –3 \) and \( c = –5 \)
By quadratic formula:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-5)}}{2(2)} \)
\( = \frac{3 \pm \sqrt{9 + 40}}{4} \)
\( = \frac{3 \pm \sqrt{49}}{4} = \frac{3 \pm 7}{4} \)
\( \Rightarrow x = \frac{3+7}{4} \) or \( x = \frac{3-7}{4} \)
\( \Rightarrow x = \frac{10}{4} = \frac{5}{2} \) or \( x = -\frac{4}{4} = -1 \)
Hence, roots of the given equations are \( \frac{5}{2} \) and –1.
(B) The given equation is: \( 5x^2 + 13x + 8 = 0 \)
On comparing with \( ax^2 + bx + c = 0 \), we get
\( a = 5, b =13, c = 8 \)
By quadratic formula:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( = \frac{-(13) \pm \sqrt{(13)^2 - 4(5)(8)}}{2(5)} \)
\( = \frac{-13 \pm \sqrt{169 - 160}}{10} \)
\( = \frac{-13 \pm 3}{10} \)
\( \Rightarrow x = \frac{-13 + 3}{10} \) or \( x = \frac{-13 - 3}{10} \)
\( \Rightarrow x = \frac{-10}{10} = -1 \) or \( x = -\frac{16}{10} = -\frac{8}{5} \)
Hence, roots of the given equation are –1 and \( -\frac{8}{5} \).
(C) The given equation is: \( –3x^2 + 5x + 12 = 0 \)
On comparing with \( ax^2 + bx + c = 0 \), we get
\( a = –3, b = 5 \) and \( c = 12 \)
By quadratic formula,
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( = \frac{-5 \pm \sqrt{(5)^2 - 4(-3)(12)}}{2(-3)} \)
\( = \frac{-5 \pm \sqrt{25 + 144}}{-6} \)
\( = \frac{-5 \pm \sqrt{169}}{-6} = \frac{-5 \pm 13}{-6} \)
\( \Rightarrow x = \frac{-5 + 13}{-6} \) or \( x = \frac{-5 - 13}{-6} \)
\( \Rightarrow x = \frac{8}{-6} = -\frac{4}{3} \) or \( x = \frac{-18}{-6} = 3 \)
Hence, the roots of the given equation are \( -\frac{4}{3} \) and 3.
(D) The given equation is: \( –x^2 + 7x – 10 = 0 \)
On comparing with \( ax^2 + bx + c = 0 \), we get
\( a = –1, b = 7, c = –10 \)
By quadratic formula,
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( = \frac{-7 \pm \sqrt{(7)^2 - 4(-1)(-10)}}{2(-1)} \)
\( = \frac{-7 \pm \sqrt{49 - 40}}{-2} \)
\( = \frac{-7 \pm \sqrt{9}}{-2} = \frac{-7 \pm 3}{-2} \)
\( \Rightarrow x = \frac{-7 + 3}{-2} \) or \( x = \frac{-7 - 3}{-2} \)
\( \Rightarrow x = \frac{-4}{-2} = +2 \) or \( x = \frac{-10}{-2} = +5 \)
Hence, roots of the given equation are 2 and 5.
(E) The given equation is: \( x^2 + 2\sqrt{2}x – 6 = 0 \)
On comparing with \( ax^2 + bx + c = 0 \), we get
\( a = 1, b = 2\sqrt{2}, c = –6 \)
By quadratic formula,
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( = \frac{-2\sqrt{2} \pm \sqrt{(2\sqrt{2})^2 - 4(1)(-6)}}{2(1)} \)
\( = \frac{-2\sqrt{2} \pm \sqrt{8 + 24}}{2} \)
\( = \frac{-2\sqrt{2} \pm \sqrt{32}}{2} = -\sqrt{2} \pm 2\sqrt{2} \)
\( \Rightarrow x = -\sqrt{2} + 2\sqrt{2} \) or \( x = -\sqrt{2} - 2\sqrt{2} \)
\( \Rightarrow x = \sqrt{2} \) or \( x = -3\sqrt{2} \)
Hence, the roots of the given equation are \( \sqrt{2} \) and \( -3\sqrt{2} \).
(F) The given equation is: \( x^2 – 3\sqrt{5}x + 10 = 0 \)
On comparing with \( ax^2 + bx + c = 0 \), we get
\( a = 1, b = -3\sqrt{5} , c = 10 \)
By quadratic formula,
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( = \frac{-(-3\sqrt{5}) \pm \sqrt{(-3\sqrt{5})^2 - 4(1)(10)}}{2(1)} \)
\( = \frac{3\sqrt{5} \pm \sqrt{45 - 40}}{2} \)
\( = \frac{3\sqrt{5} \pm \sqrt{5}}{2} \)
\( \Rightarrow x = \frac{3\sqrt{5} + \sqrt{5}}{2} \) or \( x = \frac{3\sqrt{5} - \sqrt{5}}{2} \)
\( \Rightarrow x = \frac{4\sqrt{5}}{2} = 2\sqrt{5} \) or \( x = \frac{2\sqrt{5}}{2} = \sqrt{5} \)
Hence, the roots of the given equation are \( 2\sqrt{5} \) and \( \sqrt{5} \).
(G) The given equation is: \( \frac{1}{2}x^2 – \sqrt{11}x + 1 = 0 \)
On comparing with \( ax^2 + bx + c = 0 \), we get
\( a = \frac{1}{2}, b = -\sqrt{11}, c = 1 \)
By quadratic formula,
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( = \frac{-(-\sqrt{11}) \pm \sqrt{(-\sqrt{11})^2 - 4(\frac{1}{2})(1)}}{2(\frac{1}{2})} \)
\( = \frac{\sqrt{11} \pm \sqrt{11 - 2}}{1} \)
\( = \sqrt{11} \pm \sqrt{9} = \sqrt{11} \pm 3 \)
\( \Rightarrow x = \sqrt{11} + 3 \) or \( x = \sqrt{11} - 3 \)
Hence, roots of the given equation are \( \sqrt{11} + 3 \) and \( \sqrt{11} - 3 \).

Question. Find the roots of the following quadratic equations by the factorisation method:
(A) \( 2x^2 + \frac{5}{3}x – 2 = 0 \)
(B) \( \frac{2}{5}x^2 – x – \frac{3}{5} = 0 \)
(C) \( 3\sqrt{2}x^2 – 5x – \sqrt{2} = 0 \)
(D) \( 3x^2 + 5\sqrt{5}x – 10 = 0 \)
(E) \( 21x^2 – 2x + \frac{1}{21} = 0 \)
Answer:
(A) The given equation is: \( 2x^2 + \frac{5}{3}x – 2 = 0 \)
Multiplying both the sides by 3, we get
\( 6x^2 + 5x – 6 = 0 \)
Splitting the middle term, we have
\( 6x^2 + 9x – 4x – 6 = 0 \)
\( 3x(2x + 3) – 2(2x + 3) = 0 \)
\( \Rightarrow (2x + 3)(3x – 2) = 0 \)
\( \Rightarrow 2x + 3 = 0 \) or \( 3x – 2 = 0 \)
\( x = – \frac{3}{2} \) or \( x = \frac{2}{3} \)
Hence, the roots of the given equation are \( – \frac{3}{2} \) and \( \frac{2}{3} \)
(B) The given equation is: \( \frac{2}{5}x^2 – x – \frac{3}{5} = 0 \)
Multiplying both the sides by 5, we get
\( \Rightarrow 2x^2 – 5x – 3 = 0 \)
Splitting the middle term, we have
\( \Rightarrow 2x^2 – 6x + x – 3 = 0 \)
\( \Rightarrow 2x(x – 3) + 1(x – 3) = 0 \)
\( \Rightarrow (2x + 1)(x – 3) = 0 \)
\( \Rightarrow 2x + 1 = 0 \) or \( x – 3 = 0 \)
\( \Rightarrow x = – \frac{1}{2} \) or \( x = 3 \)
Hence, the roots of the equation are \( – \frac{1}{2} \) and 3.
(C) The given equation is: \( 3\sqrt{2}x^2 – 5x – \sqrt{2} = 0 \)
Splitting the middle term, we have
\( 3\sqrt{2}x^2 – 6x + x – \sqrt{2} = 0 \)
\( \Rightarrow 3\sqrt{2}x^2 – 3\sqrt{2} \cdot \sqrt{2}x + x – \sqrt{2} = 0 \)
\( \Rightarrow 3\sqrt{2}x(x – \sqrt{2}) + 1(x – \sqrt{2}) = 0 \)
\( \Rightarrow (x – \sqrt{2})(3\sqrt{2}x + 1) = 0 \)
\( \Rightarrow x – \sqrt{2} = 0 \)
or \( 3\sqrt{2}x + 1 = 0 \)
\( \Rightarrow x = \sqrt{2} \) or \( x = -\frac{1}{3\sqrt{2}} = -\frac{\sqrt{2}}{6} \)
Hence, the roots of the equation are \( \sqrt{2} \) and \( -\frac{\sqrt{2}}{6} \).
(D) The given equation is: \( 3x^2 + 5\sqrt{5}x – 10 = 0 \)
Splitting the middle term, we have
\( 3x^2 + 6\sqrt{5}x – \sqrt{5}x – 10 = 0 \)
\( \Rightarrow 3x^2 + 6\sqrt{5}x – \sqrt{5}x – (2\sqrt{5})(\sqrt{5})= 0 \)
\( \Rightarrow 3x(x + 2\sqrt{5}) – \sqrt{5}(x + 2\sqrt{5}) = 0 \)
\( \Rightarrow (x + 2\sqrt{5})(3x – \sqrt{5}) = 0 \)
\( \Rightarrow (x + 2\sqrt{5}) = 0 \) or \( (3x – \sqrt{5}) = 0 \)
\( \Rightarrow x = -2\sqrt{5} \) or \( x = \frac{\sqrt{5}}{3} \)
Hence, roots of the given equation: \( 3x^2 + 5\sqrt{5}x – 10 \) are \( -2\sqrt{5} \) and \( \frac{\sqrt{5}}{3} \).
(E) The given equation is: \( 21x^2 – 2x + \frac{1}{21} = 0 \)
Multiplying both the sides by 21, we get
\( (21)(21)x^2 – (21)(2x) + \frac{1}{21} (21) = 0 \)
\( 441x^2 – 42x + 1 = 0 \)
Splitting the middle term, we have
\( 441x^2 – 21x – 21x +1 = 0 \)

Question. The sum of the areas of two squares is 157 m\(^2\). If the sum of their perimeters is 68 m, find the sides of the two squares.
Answer: Let 'x' and 'y' be the length of the sides of the two squares.
The, area of first square = (side)\(^2\) = \( x^2 \)
area of second square = (side)\(^2\) = \( y^2 \)
According to the question,
\( x^2 + y^2 = 157 \) ...(i)
Now, the perimeter of the first square = 4 × side = 4x
Perimeter of the second square = 4 × side = 4y
According to the question:
\( 4x + 4y = 68 \)
or \( x + y = 17 \)
\( y = 17 - x \) ...(ii)
Put the value of 'y' from equation (ii), in equation (i).
\( x^2 + (17 - x)^2 = 157 \)
\( x^2 + 289 + x^2 - 34x - 157 = 0 \)
\( 2x^2 - 34x + 132 = 0 \)
\( x^2 - 17x + 66 = 0 \)
\( (x - 6) (x - 11) = 0 \)
\( \therefore x = 6 \) or \( 11 \)
When, \( x = 6 \), then \( y = 11 \)
when, \( x = 11 \), then \( y = 6 \)
Hence, the sides of the squares are 6 m and 11 m.