Find the constant k such that the quadratic 2x^2 + 3x + 8x - x^2 + 4x + k has a double root.

Rangcr897 Jul 4, 2024

#1**+1 **

for a quadratic to have a double root we want the variables of b^2 -4ac = 0

we can combine like terms in this quadratic via simplification and get

x^2+11x+4x + k

==> x^2 + 15x + k = 0

now we just plug this into the discriminant equation ( b^2 - 4ac = 0 )

== > 15^2 - 4 x 1 x c =0

simplfy == > 225 - 4c = 0

thus c = 56.25

please check if this is right

breadstickim01 Jul 4, 2024

#2**+1 **

*Find the constant k such that the quadratic 2x^2 + 3x + 8x - x^2 + 4x + k has a double root.*

2x^{2} + 3x + 8x – x^{2} + 4x + k

combine like terms x^{2} + 15x + k

for a quadratic to have a double root,

which is also called a repeated root,

its discriminant must equal zero

when a quadratic is in the form ax^{2} + bx +c

its discriminant is the quantity b^{2} – 4ac b^{2} – 4ac

15^{2} – (4)(1)(k)

225 – 4k

solve for k k = (225 / 4)

**k = 56.25**

by the way, the quadratic in this problem factors to (x + 7**.**5)(x + 7**.**5)

_{.}

Bosco Jul 4, 2024