Question

Find the value of k for which the following quadratic equation x2-5x+k=0 has real roots

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Value\:of\:k=\frac{-25}{4}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}}\\ \tt: \implies {x}^{2} - 5x + k = 0 \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies Value \: of \: k =?$

• According to given question :

$\tt \circ \: {x}^{2} - 5x + k = 0 \\ \\ \tt \circ \: a = 1 \: \: \: \: \: b = - 5 \: \: \: \: \: \: c = k\\ \\ \bold{As \: we \: know \: that} \\ \\ \text{For \: real \: roots} \to \\ \tt: \implies Discriminant = 0 \\ \\ \tt: \implies {b}^{2} - 4ac = 0 \\ \\ \tt: \implies {( - 5)}^{2} - 4 \times 1 \times k = 0 \\ \\ \tt: \implies 25 - 4k = 0 \\ \\ \tt: \implies 4k = - 25 \\ \\ \green{\tt: \implies k = \frac{ - 25}{4} } \\ \\ \blue{\bold{Some \: related \: formula }} \\ \orange{ \tt \circ \: x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} }$

Answer 2

Step-by-step explanation:

$\sf \bf We \: have,$

$\sf x^2 -5x+k=0$

$\sf \bf We \: know \: that,$

$\sf For \: real \: roots,$

$\sf \underline{\boxed{\sf \bf b^2-4ac≥0}}$

$\sf : \implies (-5)^2-4(1)(k)≥0$

$\sf : \implies 25-4k≥0$

$\sf : \implies 4k-25≤0$

$\sf : \implies \underline{\boxed{\sf \bf{k ≤\dfrac{25}{4}}}}$