CBSE Class 10 Maths HOTs Quadratic Equations Set J

Very Short Answer Type Questions

Question. Solve the following quadratic equation for \(x\): \(4\sqrt{3}x^2 + 5x - 2\sqrt{3} = 0\)
Answer: We have \(4\sqrt{3}x^2 + 5x - 2\sqrt{3} = 0\)
\(4\sqrt{3}x^2 + 8x - 3x - 2\sqrt{3} = 0\)
\(4x(\sqrt{3}x + 2) - \sqrt{3}(\sqrt{3}x + 2) = 0\)
\((\sqrt{3}x + 2)(4x - \sqrt{3}) = 0\)
Thus \(x = -\frac{2}{\sqrt{3}}, \frac{\sqrt{3}}{4}\)

Question. Solve for \(x : x^2 - (\sqrt{3} + 1)x + \sqrt{3} = 0\)
Answer: We have \(x^2 - (\sqrt{3} + 1)x + \sqrt{3} = 0\)
\(x^2 - \sqrt{3}x - 1x + \sqrt{3} = 0\)
\(x(x - \sqrt{3}) - 1(x - \sqrt{3}) = 0\)
\((x - \sqrt{3})(x - 1) = 0\)
Thus \(x = \sqrt{3}, x = 1\)

Question. Find the roots of the following quadratic equation: \((x + 3)(x - 1) = 3(x - \frac{1}{3})\)
Answer: We have \((x + 3)(x - 1) = 3(x - \frac{1}{3})\)
\(x^2 + 2x - 3 = 3x - 1\)
\(x^2 - x - 2 = 0\)
\(x^2 - 2x + x - 2 = 0\)
\(x(x - 2) + 1(x - 2) = 0\)
\((x - 2)(x + 1) = 0\)
Thus \(x = 2, -1\)

Question. Find the roots of the following quadratic equation: \(\frac{2}{5}x^2 - x - \frac{3}{5} = 0\)
Answer: We have \(\frac{2}{5}x^2 - x - \frac{3}{5} = 0\)
\(\frac{2x^2 - 5x - 3}{5} = 0\)
\(2x^2 - 5x - 3 = 0\)
\(2x^2 - 6x + x - 3 = 0\)
\(2x(x - 3) + 1(x - 3) = 0\)
\((2x + 1)(x - 3) = 0\)
Thus \(x = -\frac{1}{2}, 3\)

Question. Solve the following quadratic equation for \(x : 4x^2 - 4a^2x + (a^4 - b^4) = 0\)
Answer: We have \(4x^2 - 4a^2x + (a^4 - b^4) = 0\)
Compare with \(Ax^2 + Bx + C = 0\) we have
\(A = 4, B = -4a^2, C = (a^4 - b^4)\)
\(x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\)
\(= \frac{4a^2 \pm \sqrt{(-4a^2)^2 - 4 \times 4 \times (a^4 - b^4)}}{2 \times 4}\)

Question. [Continuation of previous problem]
Answer: \( = \frac{4a^2 \pm \sqrt{16a^4 - 16a^4 + 16b^4}}{8} \)
\( = \frac{4a^2 \pm \sqrt{16b^4}}{8} \)
or, \( x = \frac{4a^2 \pm 4b^2}{8} = \frac{a^2 \pm b^2}{2} \)
Thus either \( x = \frac{a^2 + b^2}{2} \) or \( x = \frac{a^2 - b^2}{2} \)

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

Question. Solve the following quadratic equation for \( x \):
\( x^2 - 2ax - (4b^2 - a^2) = 0 \)
Answer: We have \( x^2 - 2ax - (4b^2 - a^2) = 0 \)
\( x^2 - 2ax + a^2 - 4b^2 = 0 \)
\( (x - a)^2 - (2b)^2 = 0 \)
\( (x - a + 2b)(x - a - 2b) = 0 \)
Thus \( x = a - 2b, x = a + 2b \)

Question. Solve for \(x : \frac{x+1}{x-1} + \frac{x-2}{x+2} = 4 - \frac{2x+3}{x-2}; x \neq 1, -2, 2\)
Answer: We have \(\frac{x+1}{x-1} + \frac{x-2}{x+2} = 4 - \frac{2x+3}{x-2}\)
\(\frac{x^2 + 3x + 2 + x^2 - 3x + 2}{x^2 + x - 2} = \frac{4x - 8 - 2x - 3}{x - 2}\)
\(\frac{2x^2 + 4}{x^2 + x - 2} = \frac{2x - 11}{x - 2}\)
\((2x^2 + 4)(x - 2) = (2x - 11)(x^2 + x - 2)\)
\(5x^2 + 19x - 30 = 0\)
\((5x - 6)(x + 5) = 0\)
\(x = -5, \frac{6}{5}\)

Question. Solve for \(x : \frac{2x}{x-3} + \frac{1}{2x+3} + \frac{3x+9}{(x-3)(2x+3)} = 0, x \neq 3, -\frac{3}{2}\)
Answer: We have \(2x(2x+3) + (x-3) + (3x+9) = 0\)
\(4x^2 + 6x + x - 3 + 3x + 9 = 0\)
\(4x^2 + 10x + 6 = 0\)
\(2x^2 + 5x + 3 = 0\)
\((x+1)(2x+3) = 0\)
Thus \(x = -1, x = -\frac{3}{2}\)

Short Answer Type Questions

Question. For what value of \( k \), the roots of the quadratic equation \( kx(x - 2\sqrt{5}) + 10 = 0 \) are equal ?
Answer: We have \( kx(x - 2\sqrt{5}) + 10 = 0 \)
or, \( kx^2 - 2\sqrt{5}kx + 10 = 0 \)
Compare with \( ax^2 + bx + c = 0 \) we get
\( a = k, b = -2\sqrt{5}k, c = 10 \)
Since, roots are equal, \( D = b^2 - 4ac = 0 \)
\( (-2\sqrt{5}k)^2 - 4 \times k \times 10 = 0 \)
\( 20k^2 - 40k = 0 \)
\( 20k(k - 2) = 0 \)
\( k(k - 2) = 0 \)
Since \( k \neq 0 \), we get \( k = 2 \)

Question. Find that non-zero value of \(k\), for which the quadratic equation \(kx^2 + 1 - 2(k - 1)x + x^2 = 0\) has equal roots. Hence find the roots of the equation.
Answer: We have \(kx^2 + 1 - 2(k - 1)x + x^2 = 0\)
\((k + 1)x^2 - 2(k - 1)x + 1 = 0\)
Compare with \(ax^2 + bx + c = 0\) we get
\(a = k + 1, b = -2(k - 1), c = 1\)
For real and equal roots, \(b^2 - 4ac = 0\)
\(4(k - 1)^2 - 4(k + 1) \times 1 = 0\)
\(4k^2 - 8k + 4 - 4k - 4 = 0\)
\(4k^2 - 12k = 0\)
\(4k(k - 3) = 0\)
As \(k\) can’t be zero, thus \(k = 3\).

Question. Find the value of \(k\) for which the quadratic equation \((k - 2)x^2 + 2(2k - 3)x + (5k - 6) = 0\) has equal roots.
Answer: We have \((k - 2)x^2 + 2(2k - 3)x + (5k - 6) = 0\)
Compare with \(ax^2 + bx + c = 0\) we get
\(a = k - 2, b = 2(2k - 3), c = (5k - 6)\)
For real and equal roots, \(b^2 - 4ac = 0\)
\(\{2(2k - 3)\}^2 - 4(k - 2)(5k - 6) = 0\)
\(4(4k^2 - 12k + 9) - 4(k - 2)(5k - 6) = 0\)
\(4k^2 - 12k + 9 - (5k^2 - 6k - 10k + 12) = 0\)
\(k^2 - 4k + 3 = 0\)
\(k^2 - 3k - k + 3 = 0\)
\(k(k - 3) - 1(k - 3) = 0\)
\((k - 3)(k - 1) = 0\)
Thus \(k = 1, 3\)

Question. If the roots of the quadratic equation \((a - b)x^2 + (b - c)x + (c - a) = 0\) are equal, prove that \(2a = b + c\).
Answer: We have \((a - b)x^2 + (b - c)x + (c - a) = 0\)
Compare with \(ax^2 + bx + c = 0\) we get
\(a = (a - b), b = (b - c), c = (c - a)\)
For real and equal roots, \(b^2 - 4ac = 0\)
\((b - c)^2 - 4(a - b)(c - a) = 0\)
\(b^2 + c^2 - 2bc - 4(ac - a^2 - bc + ab) = 0\)
\(b^2 + c^2 - 2bc - 4ac + 4a^2 + 4bc - 4ab = 0\)
\(4a^2 + b^2 + c^2 + 2bc - 4ab - 4ac = 0\)
Using \(a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = (a + b + c)^2\) we have
\((-2a + b + c)^2 = 0\)
\(-2a + b + c = 0\)
Hence, \(b + c = 2a\)

Question. If the quadratic equation, \((1 + a^2)b^2x^2 + 2abcx + (c^2 - m^2) = 0\) in \(x\) has equal roots, prove that \(c^2 = m^2(1 + a^2)\).
Answer: We have \((1 + a^2)b^2x^2 + 2abcx + (c^2 - m^2) = 0\)
Compare with \(Ax^2 + Bx + C = 0\) we get
\(A = (1 + a^2)b^2, B = 2abc, C = (c^2 - m^2)\)
If roots are equal, \(B^2 - 4AC = 0\)
\((2abc)^2 - 4(1 + a^2)b^2(c^2 - m^2) = 0\)
\(4a^2b^2c^2 - (4b^2 + 4a^2b^2)(c^2 - m^2) = 0\)
\(4a^2b^2c^2 - [4b^2c^2 - 4b^2m^2 + 4a^2b^2c^2 - 4a^2b^2m^2] = 0\)
\(4a^2b^2c^2 - 4b^2c^2 + 4b^2m^2 - 4a^2b^2c^2 + 4a^2b^2m^2 = 0\)
\(4b^2[a^2m^2 + m^2 - c^2] = 0\)
\(c^2 = a^2m^2 + m^2\)
\(c^2 = m^2(1 + a^2)\)

Question. If \(-3\) is a root of quadratic equation \(2x^2 + px - 15 = 0\), while the quadratic equation \(x^2 - 4px + k = 0\) has equal roots. Find the value of \(k\).
Answer: Given \(-3\) is a root of quadratic equation \(2x^2 + px - 15 = 0\). Since \(-3\) is a root of above equation, it must satisfy it. Substituting \(x = -3\) in above equation we have
\(2(-3)^2 + p(-3) - 15 = 0\)
\(2 \times 9 - 3p - 15 = 0\)
\(p = 1\)
Since \(x^2 - 4px + k = 0\) has equal roots,
or \(x^2 - 4x + k = 0\) has equal roots,
\(b^2 - 4ac = 0\)
\(4^2 - 4k = 0\)
\(16 - 4k = 0\)
\(4k = 16\)
\(k = 4\)

Question. If \(ad \neq bc\), then prove that the equation \((a^2 + b^2)x^2 + 2(ac + bd)x + (c^2 + d^2) = 0\) has no real roots.
Answer: We have \((a^2 + b^2)x^2 + 2(ac + bd)x + (c^2 + d^2) = 0\)
Compare with \(Ax^2 + Bx + C = 0\) we get
\(A = (a^2 + b^2), B = 2(ac + bd)\) and \(C = (c^2 + d^2)\)
For no real roots, \(D = B^2 - 4AC < 0\)
\(D = [2(ac + bd)]^2 - 4(a^2 + b^2)(c^2 + d^2)\)
\(= 4[a^2c^2 + 2abcd + b^2d^2] - 4[a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2]\)
\(= 4[a^2c^2 + 2abcd + b^2d^2 - a^2c^2 - a^2d^2 - b^2c^2 - b^2d^2]\)
\(= -4[a^2d^2 + b^2c^2 - 2abcd]\)
\(= -4(ad - bc)^2\)
Since \(ad \neq bc\), therefore \(D \neq 0\) and always negative.
Hence the equation has no real roots.

Question. Find the value of \(c\) for which the quadratic equation \(4x^2 - 2(c + 1)x + (c + 1) = 0\) has equal roots.
Answer: We have \(4x^2 - 2(c + 1)x + (c + 1) = 0\)
Compare with \(Ax^2 + Bx + C = 0\) we get
\(A = 4, B = -2(c + 1), C = (c + 1)\)
If roots are equal, \(B^2 - 4AC = 0\)
\([-2(c + 1)]^2 - 4 \times 4(c + 1) = 0\)
\(4(c^2 + 2c + 1) - 16(c + 1) = 0\)
\(4(c^2 + 2c + 1 - 4c - 4) = 0\)
\(c^2 - 2c - 3 = 0\)
\(c^2 - 3c + c - 3 = 0\)
\(c(c - 3) + 1(c - 3) = 0\)
\((c - 3)(c + 1) = 0\)
\(c = 3, -1\)
Hence for equal roots \(c = 3, -1\).

Question. Show that if the roots of the following equation are equal that \(ad = bc\) or \(\frac{a}{b} = \frac{c}{d}\).
\(x^2(a^2 + b^2) + 2(ac + bd)x + c^2 + d^2 = 0\)
Answer: We have \(x^2(a^2 + b^2) + 2(ac + bd)x + c^2 + d^2 = 0\)
Compare with \(Ax^2 + Bx + C = 0\) we get
\(A = a^2 + b^2, B = 2(ac + bd), C = c^2 + d^2\)
If roots are equal, \(B^2 - 4AC = 0\)
\([2(ac + bd)]^2 - 4(a^2 + b^2)(c^2 + d^2) = 0\)
\(4(a^2c^2 + 2abcd + b^2d^2) - 4(a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2) = 0\)
\(4(a^2c^2 + 2abcd + b^2d^2 - a^2c^2 - a^2d^2 - b^2c^2 - b^2d^2) = 0\)
\(4(2abcd - a^2d^2 - b^2c^2) = 0\)
\((ad - bc)^2 = 0\)
\(ad = bc\)
\(\frac{a}{b} = \frac{c}{d}\) Hence Proved.

Question. If roots of the quadratic equation \(x^2 + 2px + mn = 0\) are real and equal, show that the roots of the quadratic equation \(x^2 - 2(m + n)x + (m^2 + n^2 + 2p^2) = 0\) are also equal.
Answer: We have \(x^2 + 2px + mn = 0\)
Compare with \(Ax^2 + Bx + C = 0\) we get
\(A = 1, B = 2p, C = mn\)
If roots are equal, \(B^2 - 4AC = 0\)
\(4p^2 - 4mn = 0\)
or, \(p^2 = mn\) ... (1)
Now we have \(x^2 - 2(m + n)x + (m^2 + n^2 + 2p^2) = 0\)
Compare with \(Ax^2 + Bx + C = 0\) we get
\(A = 1, B = -2(m + n), C = (m^2 + n^2 + 2p^2)\)
If roots are equal, \(B^2 - 4AC = 0\)
\(4(m + n)^2 - 4(m^2 + n^2 + 2p^2) = 0\)
\(m^2 + n^2 + 2mn - m^2 - n^2 - 2p^2 = 0\)
\(2mn - 2p^2 = 0\)
\(p^2 = mn\)
Thus if roots of \(x^2 + 2px + mn = 0\) are equal then those of \(x^2 - 2(m + n)x + (m^2 + n^2 + 2p^2) = 0\) are also equal.

Question. Find the positive values of \(k\) for which quadratic equations \(x^2 + kx + 64 = 0\) and \(x^2 - 8x + k = 0\) both will have the real roots.
Answer: (i) For \(x^2 + kx + 64 = 0\) to have real roots
\(k^2 - 256 \ge 0\)
\(k^2 \ge 256\)
\(k \ge 16\) or \(k \le -16\)
(ii) For \(x^2 - 8x + k = 0\) to have real roots
\(64 - 4k \ge 0\)
\(16 - k \ge 0\)
\(16 \ge k\)
For (i) and (ii) to hold simultaneously
\(k = 16\)

Question. Find the positive value of \(k\) for which \(x^2 - 8x + k = 0\), will have real roots.
Answer: We have \(x^2 - 8x + k = 0\)
Compare with \(Ax^2 + Bx + C = 0\) we get
\(A = 1, B = -8, C = k\)
If roots are equal, \(B^2 - 4AC = 0\)
Since the given equation has real roots, \(B^2 - 4AC \ge 0\)
\((-8)^2 - 4(1)(k) \ge 0\)
\(64 - 4k \ge 0\)
\(16 - k \ge 0\)
\(16 \ge k\)
Thus \(k \le 16\)

Question. Find the values of \(k\) for which the quadratic equations \((k + 4)x^2 + (k + 1)x + 1 = 0\) has equal roots. Also, find the roots.
Answer: We have \((k + 4)x^2 + (k + 1)x + 1 = 0\)
Compare with \(Ax^2 + Bx + C = 0\) we get \(A = (k + 4), B = (k + 1), C = 1\)
If roots are equal, \(B^2 - 4AC = 0\)
\((k + 1)^2 - 4(k + 4)(1) = 0\)
\(k^2 + 1 + 2k - 4k - 16 = 0\)
\(k^2 - 2k - 15 = 0\)
\((k - 5)(k + 3) = 0\)
\(k = 5, -3\)
For \(k = 5\), equation becomes
\(9x^2 + 6x + 1 = 0\)
\((3x + 1)^2 = 0\)
or \(x = -\frac{1}{3}\)
For \(k = -3\), equation becomes
\(x^2 - 2x + 1 = 0\)
\((x - 1)^2 = 0\)
\(x = 1\)
Hence roots are \(1\) and \(-\frac{1}{3}\).

Question. If \(x = -2\) is a root of the equation \(3x^2 + 7x + p = 0\), find the value of \(k\) so that the roots of the equation \(x^2 + k(4x + k - 1) + p = 0\) are equal.
Answer: We have \(3x^2 + 7x + p = 0\)
Since \(x = -2\) is the root of above equation. It must satisfy it.
Thus \(3(-2)^2 + 7(-2) + p = 0\)
\(p = 2\)
Since roots of the equation \(x^2 + 4kx + k^2 - k + 2 = 0\) are equal.
\(16k^2 - 4(k^2 - k + 2) = 0\)
\(16k^2 - 4k^2 + 4k - 8 = 0\)
\(12k^2 + 4k - 8 = 0\)
\(3k^2 + k - 2 = 0\)
\((3k - 2)(k + 1) = 0\)
\(k = \frac{2}{3}, -1\)
Hence, roots = \(\frac{2}{3}, -1\)

Question. If \(x = -4\) is a root of the equation \(x^2 + 2x + 4p = 0\), find the values of \(k\) for which the equation \(x^2 + px(1 + 3k) + 7(3 + 2k) = 0\) has equal roots.
Answer: We have \(x^2 + 2x + 4p = 0\)
Since \(x = -4\) is the root of above equation. It must satisfy it.
\((-4)^2 + (2 \times -4) + 4p = 0\)
\(p = -2\)
Since equation \(x^2 - 2(1 + 3k)x + 7(3 + 2k) = 0\) has equal roots.
\(4(1 + 3k)^2 - 28(3 + 2k) = 0\)
\(9k^2 - 8k - 20 = 0\)
\((9k + 10)(k - 2) = 0\)
\(k = -\frac{10}{9}, 2\)
Hence, the value of \(k\) are \(-\frac{10}{9}\) and \(2\).

Question. Find the value of \(p\) for which the quadratic equation \((p + 1)x^2 - 6(p + 1)x + 3(p + 9) = 0, p \neq -1\) has equal roots. Hence find the roots of the equation.
Answer: We have \((p + 1)x^2 - 6(p + 1)x + 3(p + 9) = 0\)
Compare with \(ax^2 + bx + c = 0\) we get
\(a = p + 1, b = -6(p + 1), c = 3(p + 9)\)
For real and equal roots, \(b^2 - 4ac = 0\)
\(36(p + 1)^2 - 4(p + 1) \times 3(p + 9) = 0\)
\(3(p^2 + 2p + 1) - (p + 1)(p + 9) = 0\)
\(3p^2 + 6p + 3 - (p^2 + 9p + p + 9) = 0\)
\(2p^2 - 4p - 6 = 0\)
\(p^2 - 2p - 3 = 0\)
\(p^2 - 3p + p - 3 = 0\)
\(p(p - 3) + 1(p - 3) = 0\)
\((p - 3)(p + 1) = 0\)
\(p = -1, 3\)
Neglecting \(p \neq -1\) we get \(p = 3\)
Now the equation becomes
\(4x^2 - 24x + 36 = 0\)
or \(x^2 - 6x + 9 = 0\)
or, \((x - 3)(x - 3) = 0\)
\(x = 3, 3\)
Thus roots are \(3\) and \(3\).

Question. If the equation \((1 + m^2)x^2 + 2mcx + (c^2 - a^2) = 0\) has equal roots, prove that \(c^2 = a^2(1 + m^2)\)
Answer: We have \((1 + m^2)x^2 + 2mcx + (c^2 - a^2) = 0\)
Compare with \(Ax^2 + Bx + C = 0\) we get
\(A = 1 + m^2, B = 2mc, C = (c^2 - a^2)\)
If roots are equal, \(B^2 - 4AC = 0\)
\((2mc)^2 - 4(1 + m^2)(c^2 - a^2) = 0\)
\(4m^2c^2 - 4(1 + m^2)(c^2 - a^2) = 0\)
\(m^2c^2 - (c^2 - a^2 + m^2c^2 - m^2a^2) = 0\)
\(m^2c^2 - c^2 + a^2 - m^2c^2 + m^2a^2 = 0\)
\(-c^2 + a^2 + m^2a^2 = 0\)
\(c^2 = a^2(1 + m^2)\)
Hence Proved.

Question. If \((-5)\) is a root of the quadratic equation \(2x^2 + px - 15 = 0\) and the quadratic equation \(p(x^2 + x) + k = 0\) has equal roots, then find the values of \(p\) and \(k\).
Answer: We have \(2x^2 + px - 15 = 0\)
Since \(x = -5\) is the root of above equation. It must satisfy it.
\(2(-5)^2 + p(-5) - 15 = 0\)
\(50 - 5p - 15 = 0\)
\(5p = 35 \Rightarrow p = 7\)
Now \(p(x^2 + x) + k = 0\) has equal roots
or \(7x^2 + 7x + k = 0\)
Taking \(b^2 - 4ac = 0\) we have
\(7^2 - 4 \times 7 \times k = 0\)
\(7 - 4k = 0\)
\(k = \frac{7}{4}\)
Hence \(p = 7\) and \(k = \frac{7}{4}\).