Question

For what values of k, the roots of the equation x square + 4x + k= 0 are real

Answer 1

$\large\bf\underline{Given:-}$

• equation :- x² + 4x + k = 0

$\large\bf\underline {To \: find:-}$

• Value of k

$\huge\bf\underline{Solution:-}$

≫ Equation = x² + 4x + k

where,

• a = 1
• b = 4
• c = k

Condition for real roots:-

• Discriminant = 0

✝️≫ Discriminant = b² - 4ac

»» b² - 4ac = 0

»» 4² - 4 × 1 × k = 0

»» 16 - 4k = 0

»» -4k = -16

»» 4k = 16

»» k = 16/4

• ≫ k = 4.

$\rule{200}3$

$\bf\boxed{ \begin{minipage}{6cm}Discriminant\:| \:Roots\\\bf\:b^2-4ac>0 -\:\:two\: real \:roots \\\bf\:b^2-4ac<0- \:\:imaginary \:roots\\\bf\:b^2-4ac=0-\:\:real\: roots \end{minipage}}$

Answer 2

${ \huge{ \bold{ \underline{ \underline{ \blue{Question:-}}}}}}$

For what values of k, the roots of the equation x square + 4x + k= 0 are real ..

${ \huge{ \bold{ \underline{ \underline{ \orange{Answer:-}}}}}}$

✒Given : -

• Equation = x² + 4x + k = 0

✒To Find : -

• Value of k

✒Solution : -

$\dashrightarrow\sf{{x}^{2}+4x+k}$

✝ Here : -

• a = 1
• b = 4
• c = k

✒Formula Used : -

$\leadsto\sf{{ \small{ \bold{ \bold{ \bold{ \red{{b}^{2}-4ac}}}}}}}$

✒On Substituting : -

$\dashrightarrow\sf{{(4)}^{2}-4(1) (k) =0}$

$\dashrightarrow\sf{16-4k=0}$

$\dashrightarrow\sf{-4k = -16}$

$\dashrightarrow\sf{k=\cancel\dfrac{-16}{-4}}$

$\leadsto\sf{{ \large{ \boxed{ \bold{ \bold{ \green{k=4}}}}}}}$

✒So, the value of k is 4 ..