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I think I need to use discriminant

 

Find the positive value of $n$ so that the quadratic equation $4x^2 + nx + 25 = 2x^2 + 5$ has exactly one solution in $x$.

 Aug 22, 2023

Best Answer 

 #1
avatar+177 
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The discriminant is absolutely useful here. The discriminant gives insight about the number of solutions for a particular quadratic equation in standard form. When the discriminant equals zero, then there is one and only one solution.

 

\(4x^2 + nx + 25 = 2x^2 + 5 \\ 2x^2 + nx + 20 = 0\)

 

Now, the equation is in standard form, so we can use the information from the discriminant to determine values of n that yield one and only solution in the x-variable.

 

\({\color{red}2}x^2 + {\color{blue}n}x + {\color{green}20} = 0 \\ \Delta = {\color{blue}b}^2 - 4{\color{red}a}{\color{green}c} \\ \Delta = {\color{blue}n}^2 - 4*{\color{red}2}*{\color{green}20} \\ \Delta = n^2 - 160\)

 

Set the discriminant to zero to find the n-values that lead to one and only one solution.

 

\(n^2 - 160 = 0 \\ n^2 = 160 \\ n = \pm \sqrt{160} \\ n = 4\sqrt{10} \text{ or } n = -4\sqrt{10}\)

 

The question asks for the positive value of n only, so we reject the negative answer. Therefore, \(n = 4 \sqrt{10}\)

 Aug 22, 2023
 #1
avatar+177 
+1
Best Answer

The discriminant is absolutely useful here. The discriminant gives insight about the number of solutions for a particular quadratic equation in standard form. When the discriminant equals zero, then there is one and only one solution.

 

\(4x^2 + nx + 25 = 2x^2 + 5 \\ 2x^2 + nx + 20 = 0\)

 

Now, the equation is in standard form, so we can use the information from the discriminant to determine values of n that yield one and only solution in the x-variable.

 

\({\color{red}2}x^2 + {\color{blue}n}x + {\color{green}20} = 0 \\ \Delta = {\color{blue}b}^2 - 4{\color{red}a}{\color{green}c} \\ \Delta = {\color{blue}n}^2 - 4*{\color{red}2}*{\color{green}20} \\ \Delta = n^2 - 160\)

 

Set the discriminant to zero to find the n-values that lead to one and only one solution.

 

\(n^2 - 160 = 0 \\ n^2 = 160 \\ n = \pm \sqrt{160} \\ n = 4\sqrt{10} \text{ or } n = -4\sqrt{10}\)

 

The question asks for the positive value of n only, so we reject the negative answer. Therefore, \(n = 4 \sqrt{10}\)

The3Mathketeers Aug 22, 2023

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