Find the value of K if the equation has equal roots 1.x2-2x(1+3k)+7(3+2k)
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
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}