Given the straight line y=0 is the tangent line to the curve f(x)=2x²+(h-1)x+2k², where h and k are constants Express h in terms of k.

When a line/curve is tangent to another line/curve, the discriminant (Δ) of their combined function will equal to 0. To find the h in terms of k, we can use this way:

First, we combine both equations:

[tex]\left\{\begin{aligned}y&=0\\y&=2x^2+(h-1)x+2k^2\end{aligned}[/tex]

Since y = y, then:

[tex]2x^2+(h-1)x+2k^2=0[/tex]

Next, we find the discriminant of the equation by using this formula:

[tex]\boxed{\Delta=b^2-4ac}[/tex]

Given:

[tex]a=\bf 2[/tex]
[tex]b= \bf (h - 1)[/tex]
[tex]c = \bf 2k^2[/tex]

Hence:

[tex]\Delta=0\ \Longleftrightarrow \text{the line is tangent to the curve}[/tex]

[tex]b^2-4ac=0[/tex]

[tex](h-1)^2-4(2)(2k^2)=0[/tex]

[tex](h-1)^2=16k^2[/tex]

[tex]h-1=\sqrt{16k^2}[/tex]

[tex]h-1=\pm4k[/tex]

[tex]h=\pm4k+1[/tex]

Therefore:

[tex]\bf h=4k+1\ or\ -4k+1[/tex]

Given the straight line y=0 is the tangent line to the curve f(x)=2x²+(h-1)x+2k², where h and k are constants Express h in terms of k.​

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Given the straight line y=0 is the tangent line to the curve f(x)=2x²+(h-1)x+2k², where h and k are constants Express h in terms of k.