Question

HEY MATH LOVERS, PLSE HELP ME OUT!
If the quadratic equation
$2 {x}^{2} + kx + 4 = 0$
Has 1 root double of the other , find the value of k.​

Answer 1

For what value(s) of k will 2x^2+kx+4=0 have real roots?

Answer



Follo

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4 ANSWERS


Answered Jan 16, 2018

For what values(s) of k will 2x^2+kx+4=0 have real roots?

Let a = 2, b = k, and c = 4 in the quadratic formula

x=−b±b2−4ac−−−−−−−√2a.x=−b±b2−4ac2a.

x=−k±k2−4(2)(4)−−−−−−−−−−√2(2)x=−k±k2−4(2)(4)2(2)

=−k±k2−32−−−−−−√4=−k±k2−324

=−k4±14k2−32−−−−−−√.=−k4±14k2−32.

For xx to be real, must have

k2−32≥0.k2−32≥0.

Solving the equation k2−32=0k2−32=0, get that k2=32⟹k=±32−−√k2=32⟹k=±32

=±16(2)−−−−√=±16(2)

=±16−−√2–√=±162

=±42–√≈±5.656854.=±42≈±5.656854.

First choose a point to the left of −42–√−42 on the number line, say −8.−8.

This implies that k2−32=(−8)2−32k2−32=(−8)2−32

=64−32=32>0.=64−32=32>0.

So k<−42–√k<−42 is a piece of the solution.

Then choose kk so that −42–√<k<42–√.−42<k<42.Using k=0⟹k=0⟹

k2−32=02−32k2−32=02−32

=0−32=−32<0.=0−32=−32<0.

So −42–√<k<42–√−42<k<42

is not a part of the solution.

Next, pick a value of k>42–√.k>42. Letting k=10⟹k2−32k=10⟹k2−32

=102−32=100−32=102−32=100−32

=68>0.=68>0. This is a part of the solution.

Lastly, when k=±42–√k=±42 get k2−32k2−32

=(±42–√)2−32=(±42)2−32

=42(2–√)2−32=42(2)2−32

=16(2)−32=32−32=0.=16(2)−32=32−32=0. So k=±42–√k=±42 are in the solution.

∴∴ the values of kk that give real roots for 2x2+kx+4=02x2+kx+4=0 are

k≤−42–√ or k≥42–√.
THANKS TO BOL DO YRR

Answer 2

$10 \sqrt{2}$