Find all values of b for which the equation has a unique solution.x−2=2(b−1)x+1I found the roots of the equation.x=b+1±b2+2b−2As I understand, there is one solution: b=1, then x=3. But I don’t understand how to find it?

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Find all values ​​of b for which the equation has a unique solution.x−2=2(b−1)x+1I found the roots of the equation.x=b+1±b2+2b−2As I understand, there is one solution: b=1, then x=3. But I don’t understand how to find it?

bondsk384r · Accepted Answer

Step 1I suppose what is required is a real solution. Nowx−2=2(b−1)x+1⇔(x−2)2=2(b−1)x+1=0&amp;nbsp;andx−2≥0K⇔p(x)=defx2-2(b+1)x+3=0&amp;nbsp;&amp;nbsp;and&amp;nbsp;&amp;nbsp;x≥2So there&#039;s a single root if and only if the reduced discriminantΔ′=(b+1)2−3=b2+2b−2≥0and 2 separates the roots.These conditions are easy to translate: they mean thatb≤−1−3&amp;nbsp;or&amp;nbsp;b≥−1+3&amp;nbsp;for the first condition,p(2)=3−4b&amp;lt;0⇔b&amp;gt;34&amp;nbsp;for the second condition.The latter condition implies&amp;nbsp;b≻1+3&amp;nbsp;Therefore it is this latter condition which ensures there&#039;s a unique solution.

Nico Buck · Accepted Answer

Step 1You need to find for which values of b, the equation has only ONE solution. For example, when&amp;nbsp;b=1&amp;nbsp;you found there was only one solution which is&amp;nbsp;x=3. For unknown bx−2=2(b−1)x+1⇒(x−2)2=2(b−1)x+1Hence x must be a solution ofx2−4x+4=2bx−2x+1⇔x2−2(b+1)x+3=0ThenΔ=4(b+1)2−12=4(b2+2b−2)Can you discuss whether this kind of expression has no, one or two solutions ? What conditions on b do you need on the solution x so it makes sense ? (Think about the original equation containing a&amp;nbsp;&amp;nbsp;).

Find all values of b for which the

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