Find whether the following equations have real roots. If real roots exist, find them.
(i) \(8x^2 + 2x - 3 = 0\)
(ii) \(-2x^2 + 3x + 2 = 0\)
(iii) \(5x^2 - 2x - 10 = 0\)
(iv) \(\dfrac{1}{2x-3} + \dfrac{1}{x-5} = 1\), \(x \neq \dfrac{3}{2},\; 5\)
(v) \(x^2 + 5\sqrt{5}\,x - 70 = 0\)
(i) Real and distinct: \(x = -\dfrac{3}{4},\; \dfrac{1}{2}\).
(ii) Real and distinct: \(x = -\dfrac{1}{2},\; 2\).
(iii) Real and distinct: \(x = \dfrac{1 - \sqrt{51}}{5},\; \dfrac{1 + \sqrt{51}}{5}\).
(iv) Real and distinct: \(x = 4 \pm \dfrac{3\sqrt{2}}{2}\).
(v) Real and distinct: \(x = -7\sqrt{5},\; 2\sqrt{5}\).
General idea: For a quadratic equation of the form \(ax^2 + bx + c = 0\), we use the discriminant \(D = b^2 - 4ac\). - If \(D > 0\), we get two real and different roots. - If \(D = 0\), we get two equal real roots. - If \(D < 0\), we get no real roots.
Here, \(a = 8, b = 2, c = -3\).
Step 1: Find discriminant. \(D = b^2 - 4ac = 2^2 - 4(8)(-3) = 4 + 96 = 100\).
Step 2: Since \(D > 0\), there are 2 real and different roots.
Step 3: Use formula \(x = \dfrac{-b \pm \sqrt{D}}{2a}\).
\(x = \dfrac{-2 \pm 10}{16}\).
So, \(x = -\tfrac{3}{4}\) or \(x = \tfrac{1}{2}\).
First, multiply the whole equation by \(-1\) to make it easier: \(2x^2 - 3x - 2 = 0\).
Here, \(a = 2, b = -3, c = -2\).
Step 1: \(D = (-3)^2 - 4(2)(-2) = 9 + 16 = 25\).
Step 2: \(D > 0\), so two real and different roots.
Step 3: Formula: \(x = \dfrac{-b \pm \sqrt{D}}{2a} = \dfrac{3 \pm 5}{4}\).
So, \(x = -\tfrac{1}{2}\) or \(x = 2\).
Here, \(a = 5, b = -2, c = -10\).
Step 1: \(D = (-2)^2 - 4(5)(-10) = 4 + 200 = 204\).
Step 2: \(D > 0\), so two real and different roots.
Step 3: \(\sqrt{204} = 2\sqrt{51}\).
Step 4: Formula: \(x = \dfrac{2 \pm 2\sqrt{51}}{10} = \dfrac{1 \pm \sqrt{51}}{5}\).
Step 1: Take LCM on the left side:
\(\dfrac{(x-5) + (2x-3)}{(2x-3)(x-5)} = 1\).
So, numerator: \(3x - 8\).
Step 2: Cross multiply: \(3x - 8 = (2x-3)(x-5)\).
Expand: \(2x^2 - 10x - 3x + 15 = 2x^2 - 13x + 15\).
So the equation becomes: \(2x^2 - 16x + 23 = 0\).
Step 3: Find discriminant: \(D = (-16)^2 - 4(2)(23) = 256 - 184 = 72\).
\(\sqrt{72} = 6\sqrt{2}\).
Step 4: Formula: \(x = \dfrac{16 \pm 6\sqrt{2}}{4} = 4 \pm \tfrac{3\sqrt{2}}{2}\).
Here, \(a = 1, b = 5\sqrt{5}, c = -70\).
Step 1: \(D = (5\sqrt{5})^2 - 4(1)(-70) = 125 + 280 = 405\).
Step 2: \(D > 0\), so real and different roots.
Step 3: \(\sqrt{405} = 9\sqrt{5}\).
Step 4: Formula: \(x = \dfrac{-5\sqrt{5} \pm 9\sqrt{5}}{2}\).
Simplify: roots are \(-7\sqrt{5}\) and \(2\sqrt{5}\).