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# Algebra

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

Jul 4, 2024

#1
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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

Jul 4, 2024
#3
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Yes, it is correct.  I was still typing when your answer was

posted, but we both solved it the same way.  Good work.

,

Bosco  Jul 4, 2024
#2
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Find the constant k such that the quadratic 2x^2 + 3x + 8x - x^2 + 4x + k has a double root.

2x2 + 3x + 8x – x2 + 4x + k

combine like terms                                         x2 + 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 ax2 + bx +c

its discriminant is the quantity b2 – 4ac            b2 – 4ac

152 – (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)

.

Jul 4, 2024