Question

If the roots are real of equation
x² + 4x + k = 0 then k is​
explain with steps

Answer 1

Answer:

4

Step-by-step explanation:

x² −4x+k=0 is of the form ax

x² −4x+k=0 is of the form ax 2

x² −4x+k=0 is of the form ax 2 +bx+c=0 where a=1b=−4c=k

x² −4x+k=0 is of the form ax 2 +bx+c=0 where a=1b=−4c=kRoots are equal

x² −4x+k=0 is of the form ax 2 +bx+c=0 where a=1b=−4c=kRoots are equal⇒ Discriminant=0

x² −4x+k=0 is of the form ax 2 +bx+c=0 where a=1b=−4c=kRoots are equal⇒ Discriminant=0⇒b

x² −4x+k=0 is of the form ax 2 +bx+c=0 where a=1b=−4c=kRoots are equal⇒ Discriminant=0⇒b 2

x² −4x+k=0 is of the form ax 2 +bx+c=0 where a=1b=−4c=kRoots are equal⇒ Discriminant=0⇒b 2 −4ac=0

x² −4x+k=0 is of the form ax 2 +bx+c=0 where a=1b=−4c=kRoots are equal⇒ Discriminant=0⇒b 2 −4ac=0⇒(−4)

x² −4x+k=0 is of the form ax 2 +bx+c=0 where a=1b=−4c=kRoots are equal⇒ Discriminant=0⇒b 2 −4ac=0⇒(−4) 2

x² −4x+k=0 is of the form ax 2 +bx+c=0 where a=1b=−4c=kRoots are equal⇒ Discriminant=0⇒b 2 −4ac=0⇒(−4) 2 −4×1×k=0

x² −4x+k=0 is of the form ax 2 +bx+c=0 where a=1b=−4c=kRoots are equal⇒ Discriminant=0⇒b 2 −4ac=0⇒(−4) 2 −4×1×k=0⇒16−4k=0

x² −4x+k=0 is of the form ax 2 +bx+c=0 where a=1b=−4c=kRoots are equal⇒ Discriminant=0⇒b 2 −4ac=0⇒(−4) 2 −4×1×k=0⇒16−4k=0⇒−4k=−16

x² −4x+k=0 is of the form ax 2 +bx+c=0 where a=1b=−4c=kRoots are equal⇒ Discriminant=0⇒b 2 −4ac=0⇒(−4) 2 −4×1×k=0⇒16−4k=0⇒−4k=−16∴k=

x² −4x+k=0 is of the form ax 2 +bx+c=0 where a=1b=−4c=kRoots are equal⇒ Discriminant=0⇒b 2 −4ac=0⇒(−4) 2 −4×1×k=0⇒16−4k=0⇒−4k=−16∴k= 4

x² −4x+k=0 is of the form ax 2 +bx+c=0 where a=1b=−4c=kRoots are equal⇒ Discriminant=0⇒b 2 −4ac=0⇒(−4) 2 −4×1×k=0⇒16−4k=0⇒−4k=−16∴k= 416

x² −4x+k=0 is of the form ax 2 +bx+c=0 where a=1b=−4c=kRoots are equal⇒ Discriminant=0⇒b 2 −4ac=0⇒(−4) 2 −4×1×k=0⇒16−4k=0⇒−4k=−16∴k= 416

x² −4x+k=0 is of the form ax 2 +bx+c=0 where a=1b=−4c=kRoots are equal⇒ Discriminant=0⇒b 2 −4ac=0⇒(−4) 2 −4×1×k=0⇒16−4k=0⇒−4k=−16∴k= 416 =4