Question

Please answer. I will mark as brainliest if I'm satisfied. ​

Answer 1

Answer:

K=2

Step-by-step explanation:

Hope it helps uh!

Answer 2

Solution :

We have, Equation

$\tt\dagger \: \: \: \: \: {x}^{2} + kx - \dfrac{5}{4} = 0.$

Let's One Root be,

$\tt \dagger \: \: \: \alpha = \dfrac{1}{2}$

What is meaning of given sentence ?

• When, Putting the given roots as place of x then our Equation become 0.

$\rule{90}1$

Taking Equation,

$\tt\longmapsto {x}^{2} + kx - \dfrac{4}{5} = 0.$

$\tt\longmapsto \dfrac{4 {x}^{2} + 4kx - 5}{4} = 0.$

$\tt\longmapsto 4 {x}^{2} + 4kx - 5 = 0.$

Now, Putting the value of Alpha.

$\tt\longmapsto 4 {( \dfrac{1}{2} )}^{2} + 4k( \dfrac{1}{2} ) - 5 = 0.$

$\tt\longmapsto 4 \times \dfrac{1}{4} + 4k \times \dfrac{1}{2} - 5 = 0.$

$\tt \longmapsto 1 + 2k = 5.$

$\tt\longmapsto 2k = 4.$

$\tt \longmapsto k = 2.$

Therefore, the value of k be 2.

$\rule{200}3$

Let Find Both Roots Of given Quadratic equation,

$\tt \longmapsto 4 {x}^{2} + 8x - 5 = 0.$

Compare with General Equation,

$\tt \dagger \: \: \: a {x}^{2} + bx + c = 0.$

Where,

$\tt \red{ a \neq0}.$

$\tt \longmapsto 4 {x}^{2} + 8x - 5 = 0.$

$\tt\longmapsto 4 {x}^{2} + 10x - 2x - 5 = 0.$

$\tt\longmapsto 4 {x}^{2} - 2x + 10x - 5 = 0.$

$\tt\longmapsto 2x(2x - 1) + 5(2x - 1) =0 .$

$\tt\longmapsto (2x + 5)(2x - 1) = 0.$

$\rule{90}1$

Either,

$\tt\mapsto 2x + 5 = 0.$

$\tt\mapsto x = \dfrac{-5}{2}.$

$\rule{90}1$

Or,

$\tt\mapsto 2x - 1 = 0.$

$\tt\mapsto x = \dfrac{1}{2}$

$\rule{90}1$

Both Roots Of given Equation be,

$\tt \dagger \: \: \: \alpha = \dfrac{1}{2} \: \: \beta = \dfrac{-5}{2}$