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For what value of m does the equation (x + 4)(x + 1) = m + 3x have exactly one real solution?

 
 Nov 22, 2020
 #1
avatar+10581 
+1

For what value of m does the equation (x + 4)(x + 1) = m + 3x have exactly one real solution?

 

Hello Guest!

 

\((x + 4)(x + 1) = m + 3x\\ m=(x + 4)(x + 1) -3x\\ m=x^2+x+4x+4-3x\)

 

\(m(x)=x^2+2x+4\)

\(\frac{dm(x)}{dx}=2x+2=0 \)

\(x=-1\\ m=3\)

 

 \(\color{BrickRed}(x + 4)(x + 1) = m + 3x\\ x^2+x+4x+4=3+3x\\ x^2+2x+1=0\\ x=-1\pm\sqrt{(-1)^2-1}\\ \color{blue}x=-1\)

                         

laugh  !

 
 Nov 22, 2020
edited by asinus  Nov 22, 2020
edited by asinus  Nov 22, 2020
 #2
avatar+10581 
+1

solution without calculus

 

\(\color{BrickRed}m=x^2+x+4x+4-3x\\ x^2+2x+4-m=0\\ x=-\frac{p}{2}\pm \sqrt{(\frac{p}{2})^2-q}\\ x=-1\pm\sqrt{1^2-4+m}\)

\(\color{blue}1^2-4+m=0\\ m=4-1\)

\(m=3\\ x=-1\)

laugh  !

 
asinus  Nov 22, 2020
edited by asinus  Nov 22, 2020

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