

InterviewSolution
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Determine the nature of the roots of the following quadratic equations:(i) \((x-2a)(x-2b)=4ab\) (ii) \(9a^2b^2x^2-24abcdx+16c^2d^2=0,a\neq0,b\neq0\) (iii) \(2(a^2+b^2)x^2+2(a+b)x+1=0\)(iv) \((b+c)x^2-(a+b+c)x+a=0\) |
Answer» (i) \((x-2a)(x-2b)=4ab\) For a quadratic equation, ax2 + bx + c = 0, D = b2 – 4ac If D < 0, roots are not real If D > 0, roots are real and unequal If D = 0, roots are real and equal \((x-2a)(x-2b)=4ab\) ⇒ x2 – (2a + 2b)x + 4ab = 4ab ⇒ x2 – (2a + 2b)x = 0 D = (2a + 2b)2 – 0 = (2a + 2b)2 Roots are real and distinct (ii) \(9a^2b^2x^2-24abcdx+16c^2d^2=0,a\neq0,b\neq0\) For a quadratic equation, ax2 + bx + c = 0, D = b2 – 4ac If D < 0, roots are not real If D > 0, roots are real and unequal If D = 0, roots are real and equal \(9a^2b^2x^2-24abcdx+16c^2d^2=0,a\neq0,b\neq0\) ⇒ D = 576a2b2c2d2 – 4 × 16 × 9 × a2b2c2d2 = 0 Roots are real and equal (iii) \(2(a^2+b^2)x^2+2(a+b)x+1=0\) For a quadratic equation, ax2 + bx + c = 0, D = b2 – 4ac If D < 0, roots are not real If D > 0, roots are real and unequal If D = 0, roots are real and equal \(2(a^2+b^2)x^2+2(a+b)x+1=0\) ⇒ D = 4(a + b)2 – 4 × 2 × (a2 + b2) ⇒ D = 4(a + b)2 – 4 × 2 × (a2 + b2) Roots are not real (iv) \((b+c)x^2-(a+b+c)x+a=0\) For a quadratic equation, ax2 + bx + c = 0, D = b2 – 4ac If D < 0, roots are not real If D > 0, roots are real and unequal If D = 0, roots are real and equal \((b+c)x^2-(a+b+c)x+a=0\) ⇒ D = (a + b + c)2 – 4a(b + c) ⇒ D = a2 + b2 + c2 – 2ab – 2ac + 2bc ⇒ D = (a – b – c)2 Thus, roots are real and unequal |
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