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Ludmilka [50]
3 years ago
11

For what value of k, the zeroes of x2 + kx + 12 will differ by 1?

Mathematics
1 answer:
asambeis [7]3 years ago
8 0

Let <em>a</em> and <em>b</em> be the zeroes of <em>x</em>² + <em>kx</em> + 12 such that |<em>a</em> - <em>b</em>| = 1.

By the factor theorem, we can write the quadratic in terms of its zeroes as

<em>x</em>² + <em>kx</em> + 12 = (<em>x</em> - <em>a</em>) (<em>x</em> - <em>b</em>)

Expand the right side and equate the coefficients:

<em>x</em>² + <em>kx</em> + 12 = <em>x</em>² - (<em>a</em> + <em>b</em>) <em>x</em> + <em>ab</em>

Then

<em>a</em> + <em>b</em> = -<em>k</em>

<em>ab</em> = 12

The condition that |<em>a</em> - <em>b</em>| = 1 has two cases, so without loss of generality assume <em>a</em> > <em>b</em>, so that |<em>a</em> - <em>b</em>| = <em>a</em> - <em>b</em>.

Then if <em>a</em> - <em>b</em> = 1, we get <em>b</em> = <em>a</em> - 1. Substitute this into the equations above and solve for <em>k</em> :

<em>a</em> + (<em>a</em> - 1) = -<em>k</em>   →   2<em>a</em> = 1 - <em>k</em>   →   <em>a</em> = (1 - <em>k</em>)/2

<em>a</em> (<em>a</em> - 1) = 12   →   (1 - <em>k</em>)/2 • ((1 - <em>k</em>)/2 - 1) = 12

→   (1 - <em>k</em>)²/4 - (1 - <em>k</em>)/2 = 12

→   (1 - <em>k</em>)² - 2 (1 - <em>k</em>) = 48

→   (1 - 2<em>k</em> + <em>k</em>²) - 2 (1 - <em>k</em>) = 48

→   <em>k</em>² - 1 = 48

→   <em>k</em>² = 49

→   <em>k</em> = ± √(49) = ±7

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