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Find the value of k so that the quadratic equation x^2-2x(1+3k)+7(3+2k)=0 has two equal roots.please give ans |
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Answer» Answer: Hello we have given equation is x² + 2x(1 + 3k) +7(3 + 2K) = 0 or x² + 2(1 + 3k)X +7(3 + 2k) = 0 roots are EQUAL we have to find the value of k =? now we know that an equation have equal roots if and only if D = 0 => b² -4ac = 0 here comparing the equation with ax² + bx + c = 0 here a= 1, b = 2(1 + 3k) and c= 7(3 + 2k) now b² - 4ac = 0
= [ 2(1 + 3k) ]² - 4 ×1×7(3 + 2k) = 0 = 4×(1 + 3k)² - 4×7(3 + 2k) = 0 = 4(1 + 9k² + 6k) - 4(21 +14K) =0 = (1 + 9k² + 6k) - (21 +14k) =0 => 9k² - 8k -20 = 0 now solving the equation we get = 9k² - 18k + 10k - 20 = 0 = 9k( k- 2) + 10(k -2) =0 => 9k +10=0 and k -2 = 0 => k = -10/9 and 2 answer ✯✯ hope it helps ✯✯ |
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