Question

1. If the roots of 2x²- 6x + k = 0 are real and equal,
find k.
​

Answer 1

Answer:

4

Step-by-step explanation:

If roots are real and equal, discriminant must be 0. Discriminant is ax² +bx + c = 0 is given by b² - 4ac

Here,

a = 2, b = - 6, c = k. Since roots are equal.

=> (-6)² - 4(2)(k) = 0

=> 36 = 8k

=> 36/8 = k

=> 4 = k

Required value of k is 4

Answer 2

Here,

$p(x) = {2x}^{2} - 6x + k$

And,

• It is as ax² + bx + c

Also,

$\alpha = \beta \small{ \: \: \: \: \: \: \: \bf \: \: \: \: \: \: \: \: \: \: \{ these \: \: are \: \: roots\}}$

Then,

Applying the relationship of zeroes & coefficients, we get,

$\alpha + \beta = \frac{ - b}{a} \\ \\ 2\alpha = \frac{ - ( - 6)}{2} = \frac{6}{2} \\ \\ \alpha = \frac{ \frac{6}{2} }{2} = \frac{6}{4} = \frac{3}{2}$

And,

$\alpha \times \beta = \frac{c}{a} \\ \\ {(\alpha)}^{2} = \frac{k}{2} \\ \\ { \bigg( \frac{3}{2} \bigg)}^{2} = \frac{k}{2} \\ \\ \frac{9}{4} = \frac{k}{2} \\ \\ k = \frac{9 \times 2}{4} \\ \\ k = \pink{\frac{9}{2} }$