Question

Find the value of k for which the quadritic equation 2x² - (k+2)x + k = 0 has real and equal roots

Answer 1

Answer:

As the given quadratic equation is:-

$2 {x}^{2} - (k + 2)x + k = 0$

has real and equal roots so its

$Discriminant \:D( {b}^{2} - 4ac) = 0$

By putting this restriction on the given quadratic equation which can easily calculate the value of k as shown below :-

$As \: \\ {( - (k + 2))}^{2} - 4 \times 2 \times k = 0 \: \\\implies {k}^{2} + 2 + 4k - 8k = 0 \\ \implies \: {k}^{2} - 4k + 2 = 0 \\ \implies {(k - 2)}^{2} = 0 \\ \implies \: k - 2 = 0 \\ \implies \: k = 2 \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \boxed{ANSWER}}$

We can also verify our answer by putting ke equals to 2 in that quadratic equation as:-)

By putting k=2,

equation becomes

$2 {x}^{2} - 4x + 2 = 0 \\ \implies {x}^{2} - 2x - 2x + 2 = 0 \\ \implies \: x(x - 2) - 2(x - 2) = 0 \\ \implies (x - 2)(x - 2) = 0 \\ \implies x = 2$

As after putting K = 2 it gives only one root it means that our answer is correct.

Answer 2

Answer:

As the given quadratic equation is:-

$2 {x}^{2} - (k + 2)x + k = 0$

has real and equal roots so its

$D( {b}^{2} - 4ac) = 0$

By putting this restriction on the given quadratic equation which can easily calculate the value of k as shown below :-

$As \: \\ {( - (k + 2))}^{2} - 4 \times 2 \times k = 0 \: \\\implies {k}^{2} + 2 + 4k - 8k = 0 \\ \implies \: {k}^{2} - 4k + 2 = 0 \\ \implies {(k - 2)}^{2} = 0 \\ \implies \: k - 2 = 0 \\ \implies \: k = 2 \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \boxed{ANSWER}}$