Question

Find the value of n such that x^2-3x+n is a perfect square trinomial

Answer 1

so a perfect square trinomial like
`\bf \begin{array}{cccccllllll} {{ a}}^2& + &2{{ a}}{{ b}}&+&{{ b}}^2\\ \downarrow && &&\downarrow \\ {{ a}}&& &&{{ b}}\\ &\to &({{ a}} + {{ b}})^2&\leftarrow \end{array}\qquad % perfect square trinomial, negative middle term \begin{array}{cccccllllll} {{ a}}^2& - &2{{ a}}{{ b}}&+&{{ b}}^2\\ \downarrow && &&\downarrow \\ {{ a}}&& &&{{ b}}\\ &\to &({{ a}} - {{ b}})^2&\leftarrow \end{array}`

has its middle term, by multiplying 2 * the left guy * the right guy

so.. .let's see your equation now x² - 3x +n

so.. the middle term is 3x, neverminding the sign

`\bf 2\cdot √(x^2)\cdot √(n)=3x\implies 2x√(n)=3x\implies √(n)=\cfrac{3x}{2x} \\\\\\ √(n)=\cfrac{3}{2}\implies n=\left( \cfrac{3}{2} \right)^2\implies n=\cfrac{3^2}{2^2}\implies n=\cfrac{9}{4}`