Ganit Prakash - Class-X - Let us work out 1.2
Let us work out 1.2 Solutions Step-by-Step Approach
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📗 Let us work out 1.2 (Q1 to Q3)
Basics(ii) \(8x^2 + 7x = 0\), \(0\) and \(-2\)
(iii) \(x + \frac{1}{x} = \frac{13}{6}\), \(\frac{5}{6}\) and \(\frac{4}{3}\)
(iv) \(x^2 - \sqrt{3}x - 6 = 0\), \(-\sqrt{3}\) and \(2\sqrt{3}\)
(i) \(x^2 + x - 1 = 0\)
Putting \(x = 1\) in the left side of the quadratic equation, we get,
\((1)^2 + 1 - 1 = 1 + 1 - 1 = 1 \neq 0\)
∴ \(x = 1\) does not satisfy the equation. So, \(1\) is not a root.
Putting \(x = -1\) in the left side, we get,
\((-1)^2 + (-1) - 1 = 1 - 1 - 1 = -1 \neq 0\)
∴ \(x = -1\) does not satisfy the equation. So, \(-1\) is not a root.
(ii) \(8x^2 + 7x = 0\)
Putting \(x = 0\) in the left side of the quadratic equation, we get,
\(8(0)^2 + 7(0) = 0 + 0 = 0\)
∴ \(x = 0\) satisfy the equation. So, \(0\) is a root.
Putting \(x = -2\) in the left side, we get,
\(8(-2)^2 + 7(-2) = 8(4) - 14 = 32 - 14 = 18 \neq 0\)
∴ \(x = -2\) does not satisfy the equation. So, \(-2\) is not a root.
(iii) \(x + \frac{1}{x} = \frac{13}{6}\)
Putting \(x = \frac{5}{6}\) in the left side, we get,
\(\frac{5}{6} + \frac{1}{\frac{5}{6}} = \frac{5}{6} + \frac{6}{5} = \frac{25 + 36}{30} = \frac{61}{30} \neq \frac{13}{6}\)
∴ \(x = \frac{5}{6}\) does not satisfy the equation. So, \(\frac{5}{6}\) is not a root.
Putting \(x = \frac{4}{3}\) in the left side, we get,
\(\frac{4}{3} + \frac{1}{\frac{4}{3}} = \frac{4}{3} + \frac{3}{4} = \frac{16 + 9}{12} = \frac{25}{12} \neq \frac{13}{6}\)
∴ \(x = \frac{4}{3}\) does not satisfy the equation. So, \(\frac{4}{3}\) is not a root.
(iv) \(x^2 - \sqrt{3}x - 6 = 0\)
Putting \(x = -\sqrt{3}\) in the left side, we get,
\((-\sqrt{3})^2 - \sqrt{3}(-\sqrt{3}) - 6 = 3 + 3 - 6 = 0\)
∴ \(x = -\sqrt{3}\) satisfy the equation. So, \(-\sqrt{3}\) is a root.
Putting \(x = 2\sqrt{3}\) in the left side, we get,
\((2\sqrt{3})^2 - \sqrt{3}(2\sqrt{3}) - 6 = 12 - 6 - 6 = 0\)
∴ \(x = 2\sqrt{3}\) satisfy the equation. So, \(2\sqrt{3}\) is a root.
(ii) For which \(-a\) will be a root of \(x^2 + 3ax + k = 0\).
(i) Since, \(\frac{2}{3}\) is a root of the quadratic equation \(7x^2 + kx - 3 = 0\),
therefore, \(7\left(\frac{2}{3}\right)^2 + k\left(\frac{2}{3}\right) - 3 = 0\)
or, \(7\left(\frac{4}{9}\right) + \frac{2k}{3} - 3 = 0\)
or, \(\frac{28}{9} + \frac{2k}{3} - 3 = 0\)
Multiplying both sides by 9,
or, \(28 + 6k - 27 = 0\)
or, \(6k + 1 = 0\)
or, \(6k = -1\)
∴ \(k = -\frac{1}{6}\)
(ii) Since, \(-a\) is a root of the quadratic equation \(x^2 + 3ax + k = 0\),
therefore, \((-a)^2 + 3a(-a) + k = 0\)
or, \(a^2 - 3a^2 + k = 0\)
or, \(-2a^2 + k = 0\)
or, \(k = 2a^2\)
∴ The required value of \(k\) is \(2a^2\).
Since \(2\) is a root of the equation \(ax^2 - 7x + b = 0\),
therefore, \(a(2)^2 - 7(2) + b = 0\)
or, \(4a - 14 + b = 0\) ......... (I)
Again, since \(-3\) is a root of the equation \(ax^2 - 7x + b = 0\),
therefore, \(a(-3)^2 - 7(-3) + b = 0\)
or, \(9a + 21 + b = 0\) ......... (II)
Subtracting equation (I) from equation (II), we get:
\((9a + 21 + b) - (4a - 14 + b) = 0\)
or, \(9a + 21 + b - 4a + 14 - b = 0\)
or, \(5a + 35 = 0\)
or, \(5a = -35\)
∴ \(a = -7\)
Putting the value of \(a = -7\) in equation (I), we get:
\(4(-7) - 14 + b = 0\)
or, \(-28 - 14 + b = 0\)
or, \(-42 + b = 0\)
∴ \(b = 42\)
Solution: \(a = -7\) and \(b = 42\).
📗 Let us solve Question 4
Detailed Solutionsor, \( 3y^2 + 2y^2 = 160 + 20 \)
or, \( 5y^2 = 180 \)
or, \( y^2 = 36 \)
or, \( y = \pm \sqrt{36} = \pm 6 \)
Either \( y = 6 \) or \( y = -6 \)
∴ The roots are \( 6, -6 \)
or, \( (4x^2 + 4x + 1) + (x^2 + 2x + 1) = 6x + 47 \)
or, \( 5x^2 + 6x + 2 = 6x + 47 \)
or, \( 5x^2 = 45 \)
or, \( x^2 = 9 \)
or, \( x = \pm \sqrt{9} = \pm 3 \)
Either \( x = 3 \) or \( x = -3 \)
∴ The roots are \( 3, -3 \)
or, \( x^2 - 9x - 7x + 63 = 195 \)
or, \( x^2 - 16x - 132 = 0 \)
or, \( x^2 - 22x + 6x - 132 = 0 \)
or, \( x(x - 22) + 6(x - 22) = 0 \)
or, \( (x - 22)(x + 6) = 0 \)
Either \( x - 22 = 0 \implies x = 22 \) or \( x + 6 = 0 \implies x = -6 \)
∴ The roots are \( 22, -6 \)
or, \( 3x - \frac{x}{3} = \frac{24}{x} \)
or, \( \frac{9x - x}{3} = \frac{24}{x} \)
or, \( \frac{8x}{3} = \frac{24}{x} \)
or, \( 8x^2 = 72 \)
or, \( x^2 = 9 \)
or, \( x = \pm \sqrt{9} = \pm 3 \)
Either \( x = 3 \) or \( x = -3 \)
∴ The roots are \( 3, -3 \)
or, \( \frac{x}{3} = \frac{15}{x} - \frac{3}{x} \)
or, \( \frac{x}{3} = \frac{12}{x} \)
or, \( x^2 = 36 \)
or, \( x = \pm \sqrt{36} = \pm 6 \)
Either \( x = 6 \) or \( x = -6 \)
∴ The roots are \( 6, -6 \)
or, \( \frac{10x^2 - 1}{x} = 3 \)
or, \( 10x^2 - 1 = 3x \)
or, \( 10x^2 - 3x - 1 = 0 \)
or, \( 10x^2 - 5x + 2x - 1 = 0 \)
or, \( 5x(2x - 1) + 1(2x - 1) = 0 \)
or, \( (2x - 1)(5x + 1) = 0 \)
Either \( 2x - 1 = 0 \implies x = \frac{1}{2} \) or \( 5x + 1 = 0 \implies x = -\frac{1}{5} \)
∴ The roots are \( \frac{1}{2}, -\frac{1}{5} \)
Let \( \frac{1}{x} = a \)
or, \( 2a^2 - 5a + 2 = 0 \)
or, \( 2a^2 - 4a - a + 2 = 0 \)
or, \( 2a(a - 2) - 1(a - 2) = 0 \)
or, \( (a - 2)(2a - 1) = 0 \)
Either \( a = 2 \) or \( a = \frac{1}{2} \)
When \( a = 2 \), \( \frac{1}{x} = 2 \implies x = \frac{1}{2} \)
When \( a = \frac{1}{2} \), \( \frac{1}{x} = \frac{1}{2} \implies x = 2 \)
∴ The roots are \( \frac{1}{2}, 2 \)
or, \( (x-2) \left[ \frac{1}{x+2} + \frac{6}{x-6} \right] = 1 \)
or, \( (x-2) \left[ \frac{x-6+6x+12}{(x+2)(x-6)} \right] = 1 \)
or, \( (x-2) \left[ \frac{7x+6}{x^2-4x-12} \right] = 1 \)
or, \( 7x^2 + 6x - 14x - 12 = x^2 - 4x - 12 \)
or, \( 6x^2 - 4x = 0 \)
or, \( 2x(3x - 2) = 0 \)
Either \( 2x = 0 \implies x = 0 \) or \( 3x - 2 = 0 \implies x = \frac{2}{3} \)
∴ The roots are \( 0, \frac{2}{3} \)
or, \( \frac{(x+5) - (x-3)}{(x-3)(x+5)} = \frac{1}{6} \)
or, \( \frac{8}{x^2 + 2x - 15} = \frac{1}{6} \)
or, \( x^2 + 2x - 15 = 48 \)
or, \( x^2 + 2x - 63 = 0 \)
or, \( x^2 + 9x - 7x - 63 = 0 \)
or, \( (x+9)(x-7) = 0 \)
Either \( x+9 = 0 \implies x = -9 \) or \( x-7 = 0 \implies x = 7 \)
∴ The roots are \( -9, 7 \)
Let \( \frac{x}{x+1} = a \)
or, \( a + \frac{1}{a} = \frac{5}{2} \)
or, \( \frac{a^2+1}{a} = \frac{5}{2} \)
or, \( 2a^2 - 5a + 2 = 0 \)
or, \( 2a^2 - 4a - a + 2 = 0 \)
or, \( 2a(a - 2) - 1(a - 2) = 0 \)
or, \( (2a-1)(a-2) = 0 \)
Either \( a = \frac{1}{2} \) or \( a = 2 \)
When \( a = \frac{1}{2} \), \( \frac{x}{x+1} = \frac{1}{2} \implies 2x = x + 1 \implies x = 1 \)
When \( a = 2 \), \( \frac{x}{x+1} = 2 \implies 2x + 2 = x \implies x = -2 \)
∴ The roots are \( 1, -2 \)
or, \( (ax+b)(c+dx) = (cx+d)(a+bx) \)
or, \( acx + adx^2 + bc + bdx = acx + bcx^2 + ad + bdx \)
or, \( adx^2 + bc = bcx^2 + ad \)
or, \( adx^2 - bcx^2 = ad - bc \)
or, \( x^2(ad - bc) = (ad - bc) \)
Since \( a \ne b, c \ne d \), assuming \( ad - bc \ne 0 \):
or, \( x^2 = 1 \)
Either \( x = 1 \) or \( x = -1 \)
∴ The roots are \( 1, -1 \)
Let \( 2x+1 = a \)
or, \( a + \frac{3}{a} = 4 \)
or, \( a^2 - 4a + 3 = 0 \)
or, \( a^2 - 3a - a + 3 = 0 \)
or, \( a(a - 3) - 1(a - 3) = 0 \)
or, \( (a-1)(a-3) = 0 \)
Either \( a = 1 \) or \( a = 3 \)
When \( a = 1 \), \( 2x+1 = 1 \implies x = 0 \)
When \( a = 3 \), \( 2x+1 = 3 \implies 2x = 2 \implies x = 1 \)
∴ The roots are \( 0, 1 \)
Let \( x+1 = a \)
or, \( \frac{a}{2} + \frac{2}{a} = \frac{a}{3} + \frac{3}{a} - \frac{5}{6} \)
or, \( \frac{a}{2} - \frac{a}{3} + \frac{5}{6} = \frac{3}{a} - \frac{2}{a} \)
or, \( \frac{3a - 2a + 5}{6} = \frac{1}{a} \)
or, \( \frac{a+5}{6} = \frac{1}{a} \)
or, \( a^2 + 5a - 6 = 0 \)
or, \( a^2 + 6a - a - 6 = 0 \)
or, \( a(a + 6) - 1(a + 6) = 0 \)
or, \( (a+6)(a-1) = 0 \)
Either \( a = -6 \) or \( a = 1 \)
When \( a = -6 \), \( x+1 = -6 \implies x = -7 \)
When \( a = 1 \), \( x+1 = 1 \implies x = 0 \)
∴ The roots are \( -7, 0 \)
or, \( \frac{(12x+17)(x+7) - (2x+15)(3x+1)}{(3x+1)(x+7)} = \frac{16}{5} \)
or, \( \frac{12x^2 + 84x + 17x + 119 - (6x^2 + 2x + 45x + 15)}{3x^2 + 21x + x + 7} = \frac{16}{5} \)
or, \( \frac{6x^2 + 54x + 104}{3x^2 + 22x + 7} = \frac{16}{5} \)
or, \( \frac{3x^2 + 27x + 52}{3x^2 + 22x + 7} = \frac{8}{5} \)
or, \( 15x^2 + 135x + 260 = 24x^2 + 176x + 56 \)
or, \( 9x^2 + 41x - 204 = 0 \)
or, \( 9x^2 + 68x - 27x - 204 = 0 \)
or, \( x(9x + 68) - 3(9x + 68) = 0 \)
or, \( (x - 3)(9x + 68) = 0 \)
Either \( x - 3 = 0 \implies x = 3 \) or \( 9x + 68 = 0 \implies x = -\frac{68}{9} \)
∴ The roots are \( 3, -7\frac{5}{9} \)
Let \( \frac{x+3}{x-3} = a \)
or, \( a + \frac{6}{a} = 5 \)
or, \( a^2 - 5a + 6 = 0 \)
or, \( a^2 - 3a - 2a + 6 = 0 \)
or, \( a(a - 3) - 2(a - 3) = 0 \)
or, \( (a-3)(a-2) = 0 \)
Either \( a = 3 \) or \( a = 2 \)
When \( a = 3 \), \( \frac{x+3}{x-3} = 3 \implies 3x - 9 = x + 3 \implies 2x = 12 \implies x = 6 \)
When \( a = 2 \), \( \frac{x+3}{x-3} = 2 \implies 2x - 6 = x + 3 \implies x = 9 \)
∴ The roots are \( 6, 9 \)
or, \( \frac{1}{a+b+x} - \frac{1}{x} = \frac{1}{a} + \frac{1}{b} \)
or, \( \frac{x - (a+b+x)}{x(a+b+x)} = \frac{b+a}{ab} \)
or, \( \frac{-(a+b)}{x^2+ax+bx} = \frac{a+b}{ab} \)
or, \( \frac{-1}{x^2+ax+bx} = \frac{1}{ab} \)
or, \( x^2+ax+bx = -ab \)
or, \( x(x+a) + b(x+a) = 0 \)
or, \( (x+a)(x+b) = 0 \)
Either \( x+a = 0 \implies x = -a \) or \( x+b = 0 \implies x = -b \)
∴ The roots are \( -a, -b \)
Let \( \frac{x+a}{x-a} = y \)
or, \( y^2 - 5y + 6 = 0 \)
or, \( y^2 - 3y - 2y + 6 = 0 \)
or, \( y(y - 3) - 2(y - 3) = 0 \)
or, \( (y-3)(y-2) = 0 \)
Either \( y = 3 \) or \( y = 2 \)
When \( y = 3 \), \( x+a = 3x - 3a \implies 2x = 4a \implies x = 2a \)
When \( y = 2 \), \( x+a = 2x - 2a \implies x = 3a \)
∴ The roots are \( 2a, 3a \)
or, \( \frac{x+b-x}{x(x+b)} = \frac{a+b-a}{a(a+b)} \)
or, \( \frac{b}{x^2+bx} = \frac{b}{a^2+ab} \)
or, \( x^2+bx = a^2+ab \)
or, \( x^2 - a^2 + bx - ab = 0 \)
or, \( (x-a)(x+a) + b(x-a) = 0 \)
or, \( (x-a)(x+a+b) = 0 \)
Either \( x - a = 0 \implies x = a \) or \( x + a + b = 0 \implies x = -(a+b) \)
∴ The roots are \( a, -(a+b) \)
\(\frac{1}{(x-1)(x-2)} + \frac{1}{(x-2)(x-3)} + \frac{1}{(x-3)(x-4)} = \frac{1}{6}\)
or, \( \left(\frac{1}{x-2} - \frac{1}{x-1}\right) + \left(\frac{1}{x-3} - \frac{1}{x-2}\right) + \left(\frac{1}{x-4} - \frac{1}{x-3}\right) = \frac{1}{6} \)
or, \( \frac{1}{x-4} - \frac{1}{x-1} = \frac{1}{6} \)
or, \( \frac{(x-1) - (x-4)}{(x-4)(x-1)} = \frac{1}{6} \)
or, \( \frac{3}{x^2-5x+4} = \frac{1}{6} \)
or, \( x^2-5x+4 = 18 \)
or, \( x^2-5x-14 = 0 \)
or, \( x^2-7x+2x-14 = 0 \)
or, \( x(x-7)+2(x-7) = 0 \)
or, \( (x-7)(x+2) = 0 \)
Either \( x = 7 \) or \( x = -2 \)
∴ The roots are \( 7, -2 \)
or, \( \frac{a}{x-a} - \frac{c}{x-c} = \frac{c}{x-c} - \frac{b}{x-b} \)
or, \( \frac{ax-ac-cx+ac}{(x-a)(x-c)} = \frac{cx-bc-bx+bc}{(x-c)(x-b)} \)
or, \( \frac{x(a-c)}{(x-a)(x-c)} = \frac{x(c-b)}{(x-c)(x-b)} \)
or, \( x \left[ \frac{a-c}{(x-a)(x-c)} - \frac{c-b}{(x-c)(x-b)} \right] = 0 \)
Either \( x = 0 \)
or, \( \frac{a-c}{(x-a)(x-c)} = \frac{c-b}{(x-c)(x-b)} \)
or, \( (a-c)(x-b) = (c-b)(x-a) \)
or, \( ax - ab - cx + bc = cx - ac - bx + ab \)
or, \( x(a + b - 2c) = 2ab - ac - bc \)
or, \( x = \frac{2ab - ac - bc}{a + b - 2c} \)
∴ The roots are \( 0, \frac{2ab - ac - bc}{a + b - 2c} \)
or, \( x^2 - \sqrt{3}x - 2x + 2\sqrt{3} = 0 \)
or, \( x(x - \sqrt{3}) - 2(x - \sqrt{3}) = 0 \)
or, \( (x - \sqrt{3})(x - 2) = 0 \)
Either \( x - \sqrt{3} = 0 \implies x = \sqrt{3} \)
or, \( x - 2 = 0 \implies x = 2 \)
∴ The roots are \( \sqrt{3}, 2 \)
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