What are the values of k such that 2x^2-12x+2k=0 has two solutions?

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What are the values of k such that 2x^2-12x+2k=0 has two solutions?

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Answer 1 · 2018-05-28T11:18:38

It must be $9 > k$ $9>k$

Explanation:

Dividing your equation by $2$ $2$
$x^{2} - 6 x + k = 0$ $x^2-6x+k=0$
using the quadratic formula
$x_{1, 2} = 3 \pm \sqrt{9 - k}$ $x_{1,2}=3pmsqrt{9-k}$
so we get two real Solutions for
$9 > k$ $9>k$

Answer 2 · 2018-05-28T11:22:52

$K \leq 9$ $K<=9$

Explanation:

For two Solution Discriminant(D) should be $D \geq 0$ $D>=0$
$D = b^{2} - 4 a c$ $D=b^2-4ac$
$\Rightarrow 12^{2} - 4 \cdot 2 \cdot 2 k \geq 0$ $rArr12^2-4*2*2k>=0$
$\Rightarrow 144 - 16 k \geq 0$ $rArr144-16k>=0$
$\Rightarrow 16 K \leq 144$ $rArr 16K<=144$
$\Rightarrow K \leq 9$ $rArr K<=9$

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What are the values of k such that 2x^2-12x+2k=0 has two solutions?

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