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Find the value of K if the equation has equal roots 1.x2-2x(1+3k)+7(3+2k) |
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Answer» If the quadratic equation has two equal roots then D=0.⇒ b2-4ac=0Here, a=1, b=-2(1+3k), c=7(3+2k)⇒(-2(1+3k))2-4×1×7(3+2k) =0⇒4(1+2×1×3k+9k2)-28(3+2k) =0⇒4+24k+36 k2 -84-56k=0⇒36 k2 -32k-80=0⇒9 k2 -8k-20=0⇒9 k2 -18k+10k-20=0⇒ 9k(k-2)+10(k-2)=0⇒(9k+10) (k-2)=0⇒(9k+10)=0 (k-2)=0⇒ k=-10/9 k=2. Given:\xa0{tex}{x^2} - 2x\\left( {1 + 3k} \\right) + 7\\left( {3 + 2k} \\right) = 0{/tex}Here,\xa0{tex}a = 1,b =- 2\\left( {1 + 3k} \\right),c = 7\\left( {3 + 2k} \\right){/tex}Since, for equal roots, the condition is\xa0{tex}{b^2} - 4ac = 0{/tex}Therefore,\xa0{tex}{\\left\\{ {-2\\left( {1 + 3k} \\right)} \\right\\}^2} - 4 \\times 1 \\times 7\\left( {3 + 2k} \\right) = 0{/tex}=> {tex}4\\left( {1 + 9{k^2} + 6k} \\right) - 84 - 56k = 0{/tex}=> {tex}4 + 36{k^2} + 24k - 84 - 56k = 0{/tex}=> {tex}36{k^2} - 32k - 80 = 0{/tex}=> {tex}9{k^2} - 8k - 20 = 0{/tex}=> {tex}9{k^2} - 18k + 10k - 20 = 0{/tex}=> {tex}9k\\left( {k - 2} \\right) + 10\\left( {k - 2} \\right) = 0{/tex}=> {tex}\\left( {k - 2} \\right)\\left( {9k + 10} \\right) = 0{/tex}=> {tex}k-2=0{/tex}\xa0or\xa0{tex}9k+10=0{/tex}=> {tex}k=2{/tex}\xa0or\xa0{tex}k = {{ - 10} \\over 9}{/tex} |
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