Question

find the value of k 3k²=4(kx-1) ,if the root is equal​

Answer 1

Solution :

3k² = 4(kx - 1)

for quadratic equation are

b² + 4ac = 0

so, now

3kx² - 4 kx + 4 = 0

b² + 4ac = (-4k)² + 4 × 3k × 4 = 0

16k² + 12k × 4 = 0

16k² - 48k = 0

16k (k - 3) = 0

16k = 0 or k - 3 = 0

k = 0 or k = 3

∴ value of k is (0 , 3)

Answer 2

$\sf \underline{Given \: } :$

• 3k² = 4(kx - 1) ,

$\sf \underline{ To \: Find\: } :$

• find the value of k

$\sf \underline{ Solution \: \: } :$

$\sf : \implies {3k}^{2} = 4(kx - 1) \\ \\ \sf : \implies 3k {x}^{2} + 4kx - 4 = 0 \\ \\ \sf : \implies \: 0 = 0 \\ \\ \sf : \implies 0 = { b}^{2} - 4ac$

The roots are equal therefore  b²- 4ac = 0

Suthe value of b substitute values of a, b and c.

$\sf : \implies \: { - 4k}^{2} - 4 \times 3k \times 4 = 0 \\ \\ \sf : \implies \:16k - 48k = 0 \\ \\ \sf : \implies \:16k(k - 3) = 0 \\ \\ \sf : \implies \:16k = 0 \: or \: k - 3 \\ \\ \sf : \implies \:k = 0 \: or \: k = 3$

but k cannot be zero as 'a' term of quadratic equation is never zero.Therefore k = 3

$\rule{180}{1.5}$

$\boxed{\begin{minipage}{5.5 cm} { \bigstar\: \textsf{For a Quadratic Polynomial :}}\\\\ {\qquad\sf p(x) = ax ^\sf2 \sf + bx + c}\\\\\sf \: D= {b}^{2} - 4ac \\ \\ \sf \: x = \frac{ - b \pm \: \sqrt{ {b}^{2} - 4ac} \: }{2a} \\ \end{minipage}}$