The equation x^2-2kx+7k-12=0 has real and equal roots. Calculate k !

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

x^2 - 2kx + 7k - 12 = 0

if both roots are real and equals, then the equation has one solution. Then the discrminant must be zero.

==> delta = b^2 - 4ac = 0

==> 4k^2 - 4(1)(7k-12) = 0

==> 4k^2 - 28k + 48 = 0

Divide by 4:

==> k^2 - 7k + 12 = 0

==> (k -4)(k-3) = 0

Then there are two solutions:

k1 = 4

k2= 3

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

The roots of the equation  x^2-2kx+7k-12=0 are real and equal, when the discriminant of the equation, delta, is zero.

delta = b^2 - 4*a*c, where a,b,c are the coefficients of the equation.

Let's identify a,b,c:

a = 1

b = -2k

c = 7k-12

delta = (-2k)^2 - 4*1*( 7k-12)

delta = 0  => 4k^2 - 28k + 48 = 0

We'll solve the equation

4k^2 - 28k + 48 = 0

We'll divide by 4:

k^2 - 7k + 12 = 0

We'll solve the quadratic:

k1 = [7+sqrt(49-48)]/2

k1 = (7+1)/2

k1 = 4

k2 = (7-1)/2

k2 = 3

So, the equation will have 2 real equal roots when k = {3 ; 4}.

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