Find the value of K if …

If the quadratic equation has two equal roots then D=0.

⇒ b²-4ac=0

Here, 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: {tex}{x^2} - 2x\left( {1 + 3k} \right) + 7\left( {3 + 2k} \right) = 0{/tex}

Here, {tex}a = 1,b =- 2\left( {1 + 3k} \right),c = 7\left( {3 + 2k} \right){/tex}

Since, for equal roots, the condition is {tex}{b^2} - 4ac = 0{/tex}

Therefore, {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} or {tex}9k+10=0{/tex}

=>     {tex}k=2{/tex} or {tex}k = {{ - 10} \over 9}{/tex}