Question

Find the values of k for which the given equation has real and equal roots.
$2 {x}^{2} - 10x + k = 0$​

Answer 1

Answer: k = 12.5

Explanation:-)

We know that an equation which is of the form

$a {x}^{2} + bx + c = 0$

have real and equal roots if

Its discriminant is equals to zero .

i.e.

${ b}^{2} - 4ac = 0$

So,

Here

a = 2

b = -10

and

c = k

Now putting in above Relation we get

$( - 10 {) {}^{2} - 4 \times 2 \times k} = 0 \\ \\ 100 - 8k = 0 \\ \\ 8k = 100 \\ \\ k = \large \frac{ \cancel{100}}{ \cancel8} \\ \\ k = \frac{25}{2} = 12.5$

Which is the required answer

Answer 2

Question :-

Find the value of K for which the given equation has real and equal roots

2x^2 - 10x + k = 0

Solution :-

b^2 - 4ac determines whether the quadratic equation ax^2 + bx + c = 0 has real roots or not, b^2 - 4ac is called discriminant of the quadratic equation.

So, a quadratic equation

ax^2 + bx + c = 0 has

1) Two distinct real roots, if b^2 - 4ac > 0

2) Two equal real roots, if b^2 - 4ac = 0

3) No real roots, if b^2 - 4ac < 0

Now, come to the solution,

2x^2 - 10x + k = 0

Here, a = 2 , b = -10 , c = k

By using discriminant,

b^2 - 4ac

Put the required values,

( - 10)^2 - 4 * 2 * k

100 - 8k

-8k = -100

k = 100/8

k = 50/ 4

k = 25/2

Hence, The value of K = 25/2 $.$