X×x -2(1+3k)x +7(3+2k)=0 find k

1.	X×x -2(1+3k)x +7(3+2k)=0 find k
Answer» The given equation is x² -\xa02x(1 + 3k) +7(3 + 2k) = 0\xa0or x² - 2(1 + 3k)x +7(3 + 2k) = 0\xa0It is given that the given quadratic equation has equal roots,\xa0⇒ b² -4ac = 0In the given equation we have,a= 1, b = - 2(1 + 3k) and c= 7(3 + 2k)\xa0Now b² - 4ac = 0 ⇒ [ - 2(1 + 3k)\xa0]² - 4\xa0×1×7(3 + 2k) = 0\xa0⇒ 4×(1 + 3k)² - 4×7(3 + 2k) = 0⇒ 4(1 + 9k² + 6k) - 4(21 +14k) =0⇒ (1 + 9k² + 6k) - (21 +14k) = 0\xa0⇒ 1+ 9{tex}k^2{/tex}+ 6k - 21 - 14k = 0⇒ 9k² - 8k - 20 = 0\xa0\xa0Factorize\xa0above equation\xa0we get\xa0 9k² - 18k + 10k - 20 = 0⇒ 9k( k- 2) + 10(k -2) =0\xa0Therefore, either 9k +10=0 and k -2 = 0⇒ k= -10/9and k = 2Hence the values of k are -10/9, 2

Answer»

The given equation is x² -\xa02x(1 + 3k) +7(3 + 2k) = 0\xa0or x² - 2(1 + 3k)x +7(3 + 2k) = 0\xa0It is given that the given quadratic equation has equal roots,\xa0⇒ b² -4ac = 0In the given equation we have,a= 1, b = - 2(1 + 3k) and c= 7(3 + 2k)\xa0Now b² - 4ac = 0 ⇒ [ - 2(1 + 3k)\xa0]² - 4\xa0×1×7(3 + 2k) = 0\xa0⇒ 4×(1 + 3k)² - 4×7(3 + 2k) = 0⇒ 4(1 + 9k² + 6k) - 4(21 +14k) =0⇒ (1 + 9k² + 6k) - (21 +14k) = 0\xa0⇒ 1+ 9{tex}k^2{/tex}+ 6k - 21 - 14k = 0⇒ 9k² - 8k - 20 = 0\xa0\xa0Factorize\xa0above equation\xa0we get\xa0 9k² - 18k + 10k - 20 = 0⇒ 9k( k- 2) + 10(k -2) =0\xa0Therefore, either 9k +10=0 and k -2 = 0⇒ k= -10/9and k = 2Hence the values of k are -10/9, 2

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X×x -2(1+3k)x +7(3+2k)=0 find k

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