Find the value of k if the discriminant = 0?

Question

x $x^{2} - 3 x + 2 + \frac{1}{k} = 0$ $x^2 -3x+2+1/k=0$

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Answer 1 · 2017-11-10T12:31:12

$k = 4$ $k=4$

The equation is $x^{2} - 3 x + 2 + \frac{1}{k} = 0$ $x^2-3x+2+1/k=0$

Here,

$a = 1$ $a=1$ , $b = - 3$ $b=-3$ , $c = 2 + \frac{1}{k}$ $c=2+1/k$

According to the question the discriminant is equal to $0$ $0$

We know the determinant of a quadratic equation is $b^{2} - 4 a c$ $b^2-4ac$

Therefore,

$b^{2} - 4 a c = 0$ $b^2-4ac=0$

$b^{2} = 4 a c$ $b^2=4ac$

Substitute the values of $a$ $a$ , $b$ $b$ and $c$ $c$

${(- 3)}^{2} = 4 \times 1 \times (2 + \frac{1}{k})$ $(-3)^2=4 xx 1 xx (2+1/k)$

$9 = 8 + \frac{4}{k}$ $9=8+4/k$

$9 - 8 = \frac{4}{k}$ $9-8=4/k$

$1 = \frac{4}{k}$ $1=4/k$

$k = 4$ $k=4$