CBSE Class 10 Mathematics Quadratic Equations VBQs Set 16

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
\( \implies \) \( k(2)^2 + 2(2) - 3 = 0 \)
\( \implies \) \( 4k = -1 \)
\( \implies \) \( k = -\frac{1}{4} \)
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 \( c = 3 \)
\(\therefore\) \( b^2 - 4ac = (k)^2 - 4 \times 3 \times 3 = 0 \)
\( \implies \) \( k^2 = 36 \)
\( \implies \) \( k = \pm 6 \)
Hence, the value of \( k \) is \(\pm 6\).

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, \( D = b^2 - 4ac \)
\( = (-4)^2 - 4 \times 2 \times 3 \)
\( = 16 - 24 = -8 < 0 \)
\( \because D < 0 \)
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. \( D \geq 0 \)
\( b^2 - 4ac \geq 0 \)
Here, \( a = 1, b = 4, c = k \)
\( \implies \) \( (4)^2 - 4 \times 1 \times k \geq 0 \)
\( \implies \) \( 16 - 4k \geq 0 \)
\( \implies \) \( k \leq 4 \)
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, \( 3(2)^2 - 2k(2) + 2(3) = 0 \)
\( \implies \) \( 12 - 4k + 6 = 0 \)
\( \implies \) \( -4k + 18 = 0 \)
\( \implies \) \( k = \frac{9}{2} \).

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, \( 6(\frac{2}{3})^2 - \frac{2}{3} - k = 0 \)
\( \implies \) \( 6 \times \frac{4}{9} - \frac{2}{3} - k = 0 \)
\( \implies \) \( \frac{8}{3} - \frac{2}{3} - k = 0 \)
\( \implies \) \( k = 2 \)
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 \( c = 8 \)
\( \implies \) \( k^2 - 4 \times 2 \times 8 = 0 \)
\( \implies \) \( k^2 = 64 \)
\( \implies \) \( k = \pm 8 \)
Hence, the value of \( k \) is 8 or – 8.

SHORT ANSWER (SA-I) 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, \( D < 0 \)
Then \( k^2 - 36 < 0 \)
\( \implies \) \( k^2 < 36 \)
\( \implies \) \( k < \pm 6 \)
\( \implies \) \( k > -6 \) or \( k < 6 \)
Hence, the value of \( k < 6 \) (positive value) for no real roots.

Question. Solve for \( x : 6x^2 + 11x + 3 = 0 \)
Answer: \( 6x^2 + 11x + 3 = 0 \)
\( \implies \) \( 6x^2 + 9x + 2x + 3 = 0 \)
\( \implies \) \( 3x (2x + 3) + 1 (2x + 3) = 0 \)
\( \implies \) \( (2x + 3) (3x + 1) = 0 \)
\( \implies \) \( 2x + 3 = 0 \) 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: \( 8x^2 - 2x - 3 = 0 \)
\( \implies \) \( 8x^2 - 6x + 4x - 3 = 0 \)
\( \implies \) \( 2x(4x - 3) + 1(4x - 3) = 0 \)
\( \implies \) \( (4x - 3) (2x + 1) = 0 \)
\( \implies \) \( 4x - 3 = 0 \) or \( 2x + 1 = 0 \)
i.e., \( x = \frac{3}{4} \) or \( x = -\frac{1}{2} \)

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, discriminant, \( D = B^2 - 4AC \)
\( = (-7ab)^2 - 4 \times 6a^2 \times (-3b^2) \)
\( = 49a^2b^2 + 72a^2b^2 \)
\( = 121 a^2b^2 \)
\( = (11 ab)^2 \)
Then, roots of the equation, \( x = \frac{-B \pm \sqrt{D}}{2A} \)
\( = \frac{-(-7ab) \pm 11ab}{2 \times 6a^2} \)
\( = \frac{18ab}{12a^2} \) 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, \( 8x^2 - 4x + 2x - 1 = 0 \)
\( 4x(2x - 1) + 1(2x - 1) = 0 \implies (4x + 1)(2x - 1) = 0 \)
If \( 4x + 1 = 0 \implies x = -\frac{1}{4} \)
\( 2x - 1 = 0 \implies x = \frac{1}{2} \)
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, \( D = b^2 - 4ac = (-6)^2 - 4(1)(7) \implies D = 36 - 28 = 8 \)
Since, discriminant is not a perfect square, therefore it will have irrational roots.
The roots will be \( \frac{-b \pm \sqrt{D}}{2a} = \frac{6 \pm \sqrt{8}}{2} = \frac{6 \pm 2\sqrt{2}}{2} = 3 \pm \sqrt{2} \)
The roots will be \( 3 + \sqrt{2} \) and \( 3 - \sqrt{2} \), are both irrational.

Question. Solve for \( x : \frac{x+3}{x+2} = \frac{3x-7}{2x-3} \)
Answer: \( \implies (x + 3) (2x - 3) = (x + 2) (3x - 7) \)
\( \implies 2x^2 + 6x - 3x - 9 = 3x^2 + 6x - 7x - 14 \)
\( \implies 2x^2 - 3x^2 + 3x + x - 9 + 14 = 0 \)
\( \implies - x^2 + 4x + 5 = 0 \implies x^2 - 4x - 5 = 0 \)
\( \implies x^2 - 5x + x - 5 = 0 \) (on splitting the middle term)
\( \implies x(x - 5) + 1(x - 5) = 0 \implies (x + 1) (x - 5) = 0 \implies x = - 1, 5 \)
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 \( ax^2 + bx + c = 0 \), 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 \( \frac{-5}{\sqrt{2}} = -\sqrt{2} \) and \( -\frac{5}{\sqrt{2}} \)
Hence, the roots of the given equation are \( -\sqrt{2} \) and \( -\frac{5}{\sqrt{2}} \).

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 \( x^2 + 0 \cdot x + c = 0 \implies x^2 + c = 0 \implies x^2 = -c \)
Here, \( c < 0 \implies -c > 0 \implies x = \pm \sqrt{-c} \)
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., \( b^2 - 4ac = 0 \)
\( (2k)^2 - 4 \times 1 \times (k^2 - k + 2) = 0 \)
\( 4k^2 - 4k^2 + 4k - 8 = 0 \implies 4k - 8 = 0 \implies k = 2 \)
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: \( \frac{2}{3} + (-3) = -\frac{7}{a} \implies -\frac{7}{3} = -\frac{7}{a} \implies a = 3 \).
Product of roots: \( \frac{2}{3} \times (-3) = \frac{b}{a} \implies - 2 = \frac{b}{3} \) [\(\because a = 3\)] \( \implies b = - 6 \)
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 \( = ab = \frac{\text{constant term}}{\text{Coefficient of } x^2} = -b \)
So, \( a + b = -a \implies b = -2a \)
and, \( ab = -b \implies a = -1 \)
Putting the value of \( a \), we get \( b = -2 \times (-1) = 2 \). Hence, \( a = -1 \) and \( b = 2 \).

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

SHORT ANSWER (SA-II) Type Questions

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) \( 5x^2 + 13x + 8 = 0 \)
Answer: (A) Given: \( 2x^2 - 3x - 5 = 0 \)
Comparing with \( ax^2 + bx + c = 0 \), \( a = 2, b = -3, c = -5 \)
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-5)}}{2(2)} = \frac{3 \pm \sqrt{9 + 40}}{4} = \frac{3 \pm 7}{4} \)
\( \implies x = \frac{10}{4} = \frac{5}{2} \) or \( x = \frac{-4}{4} = -1 \).
(B) Given: \( 5x^2 + 13x + 8 = 0 \)
\( a = 5, b = 13, c = 8 \).
\( x = \frac{-13 \pm \sqrt{(13)^2 - 4(5)(8)}}{2(5)} = \frac{-13 \pm \sqrt{169 - 160}}{10} = \frac{-13 \pm 3}{10} \)
\( \implies x = \frac{-10}{10} = -1 \) or \( x = \frac{-16}{10} = -\frac{8}{5} \).

Question. Find the roots of the following quadratic equations by the factorisation method: (A) \( 2x^2 + \frac{5}{3}x - 2 = 0 \), (B) \( \frac{2}{5}x^2 - x - \frac{3}{5} = 0 \)
Answer: (A) \( 2x^2 + \frac{5}{3}x - 2 = 0 \). Multiplying by 3, \( 6x^2 + 5x - 6 = 0 \).
\( 6x^2 + 9x - 4x - 6 = 0 \implies 3x(2x + 3) - 2(2x + 3) = 0 \implies (2x + 3)(3x - 2) = 0 \implies x = -\frac{3}{2}, \frac{2}{3} \).
(B) \( \frac{2}{5}x^2 - x - \frac{3}{5} = 0 \). Multiplying by 5, \( 2x^2 - 5x - 3 = 0 \).
\( 2x^2 - 6x + x - 3 = 0 \implies 2x(x - 3) + 1(x - 3) = 0 \implies (2x + 1)(x - 3) = 0 \implies x = -\frac{1}{2}, 3 \).

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

Question. Determine the condition for one root of the quadratic equation \( ax^2 + bx + c = 0 \) to be thrice the other.
Answer: Let roots be \( \alpha \) and \( 3\alpha \).
Sum \( = 4\alpha = -\frac{b}{a} \implies \alpha = -\frac{b}{4a} \)
Product \( = 3\alpha^2 = \frac{c}{a} \)
\( 3(-\frac{b}{4a})^2 = \frac{c}{a} \implies \frac{3b^2}{16a^2} = \frac{c}{a} \implies 3b^2 = 16ac \).

Question. The sum of the areas of two squares is 157 m². If the sum of their perimeters is 68 m, find the sides of the two squares.
Answer: Let sides be \( x \) and \( y \). \( x^2 + y^2 = 157 \) and \( 4x + 4y = 68 \implies x + y = 17 \implies y = 17 - x \).
\( x^2 + (17 - x)^2 = 157 \implies x^2 + 289 + x^2 - 34x - 157 = 0 \implies 2x^2 - 34x + 132 = 0 \implies x^2 - 17x + 66 = 0 \)
\( (x - 6)(x - 11) = 0 \implies x = 6, 11 \).
Sides are 6 m and 11 m.

Question. Show that if the roots of the quadratic equation are equal, then \( ad = bc \): \( x^2(a^2 + b^2) + 2(ac + bd)x + (c^2 + d^2) = 0 \).
Answer: For equal roots, \( B^2 - 4AC = 0 \).
\( [2(ac + bd)]^2 - 4(a^2 + b^2)(c^2 + d^2) = 0 \)
\( \implies 4(a^2c^2 + b^2d^2 + 2abcd) - 4(a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2) = 0 \)
\( \implies 2abcd - a^2d^2 - b^2c^2 = 0 \implies -(ad - bc)^2 = 0 \implies ad = bc \).