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If roots of the equation ax² + 2bx + c = 0 and bx² - 2 (ac) ½ × +b =0 are simultaneously real then prove that be =ac |
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Answer» D 1AND D2 be the discriminants of the TWO equations, then D1≥0 and D2≥0 ⇒4b2−4ac ≥0 and 4ac−4b2 ≥0 ⇒b2≥ac and ac≥b2 ⇒b2 =ac here b2=(b)2 (or) b whole SQUARE |
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