Example

3(r+2s)=2t-4
a⋅(b-2)=3b
3(r+2s)=2t-4
a⋅(b-2)=3b

Question

$$(4-k) { x }^{ 2 } +(2k+4)x+(8k+1)=0$$

Answer

x=-(k+2-3*sqrt(k*(k-3)))/(4-k),-(k+2+3*sqrt(k*(k-3)))/(4-k)

Solution


Remove parentheses.
\[(4-k){x}^{2}+(2k+4)x+8k+1=0\]
Regroup terms.
\[{x}^{2}(4-k)+(2k+4)x+8k+1=0\]
Regroup terms.
\[{x}^{2}(4-k)+x(2k+4)+8k+1=0\]
Use the Quadratic Formula.
\[x=\frac{-2k-4+6\sqrt{k(k-3)}}{2(4-k)},\frac{-2k-4-6\sqrt{k(k-3)}}{2(4-k)}\]
Simplify solutions.
\[x=-\frac{k+2-3\sqrt{k(k-3)}}{4-k},-\frac{k+2+3\sqrt{k(k-3)}}{4-k}\]
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