Question

The value of K so that the quadratic equations x^2-2(1+3k)x+7(3+2k)=0 has equal roots.
Please someone help me out ASAP.
Thank you for your help.​

Answer 1

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Answer 2

Given equation :

• x² - 2(1+3k)x + 7(3+2k) = 0
• Roots are equal.

To Find :

• Value of k.

Solution :

Compare the given quadratic equation by the general form of the quadratic equation.

General Form :

• ax² + bx + c = 0

We now get the values of variable,

• a = 1
• b = -2(1+3k) = -2-6k
• c = 7(3+2k) = 21 + 14k

We know that for roots to be equal, the value of discrimant, D should be equal to zero.

D = b² - 4a

D = 0

Block in the values,

$\sf{0\:=\:b^2-4ac}$

$\sf{(2+6k)^2\:-\:4(1)(21+14k)}$

$\sf{\big[Divided\:value\:of\:b\:by\:minus\:\big]}$

$\sf{0\:=\:2^2\:\:+(\:2\:\times\:2\:\times\:6k)\:+\:6k^2\:-\:4(21+14k)}$

$\sf{\big[Using\:the\:identity\::(a+b)^2\:=\:a^2\:+\:2ab\:+\:b^2\big]}$

$\sf{0\:=\:4\:+\:2\:\times\:12k\:+\:36k^2\:-84\:-\:56k}$

$\sf{0\:=\:4\:+\:24k\:+36k^2-84-56k}$

$\sf{0\:=\:36k^2\:+\:24k\:-\:56k\:-84+4}$

$\longrightarrow$ $\sf{0\:=\:36k^2\:-\:32k\:-80}$

Divide throughout by 4,

$\longrightarrow$ $\sf{0\:=\:9k^2\:-\:8k\:-\:20}$

$\longrightarrow$ $\sf{0\:=\:9k^2\:-\:18k+10k-20}$

$\longrightarrow$ $\sf{0=9k(k-2)+10(k-2)}$

$\longrightarrow$ $\sf{0\:=\:(k-2)(9k+10) }$

$\longrightarrow$ $\sf{k-2=0\:\:or\:\:9k+10=0}$

$\longrightarrow$ $\sf{k=2\:\:or\:\:9k=-10}$

$\longrightarrow$ $\sf{k=2\:\:or\:\:k=\dfrac{-10}{9}}$

$\large{\boxed{\bold{\purple{Value\:of\:k\:=\:2\:\:or\:\:\dfrac{-10}{9}}}}}$