7/6iws terminating decimal or not with solution
Answers
Answer:
\begin{gathered}\\\;\underbrace{\underline{\sf{Understanding\;the\;Question\;:-}}}\end{gathered}
UnderstandingtheQuestion:−
Here the Concept of Quadratic Equations has been used. Here we see that we are given a quadratic equation. We know that a Quadratic Equation has 2 roots either can be equal, imaginary or distinct but here we have to go with equal case since, we need to find the value of a such that there are only one roots that is only one value of x.
Let's do it !!
_______________________________________________
★ Formula Used :-
\begin{gathered}\\\;\boxed{\sf{For\;equal\;roots\;:\;b^{2}\;-\;4ac\;=\;\bf{0}}}\end{gathered}
Forequalroots:b
2
−4ac=0
_______________________________________________
★ Solution :-
Given,
✒ ax² - 5x + 2 = 0
Here, a = a, b = -5 and c = 2.
We know that,
\begin{gathered}\\\;\;\sf{:\mapsto\;\;For\;equal\;roots\;:\;b^{2}\;-\;4ac\;=\;\bf{0}}\end{gathered}
:↦Forequalroots:b
2
−4ac=0
\begin{gathered}\\\;\;\sf{:\mapsto\;\;b^{2}\;-\;4ac\;=\;\bf{0}}\end{gathered}
:↦b
2
−4ac=0
By applying values, we get,
\begin{gathered}\\\;\;\sf{:\mapsto\;\;(-\:5)^{2}\;-\;4\;\times\;a\;\times\;2\;=\;\bf{0}}\end{gathered}
:↦(−5)
2
−4×a×2=0
\begin{gathered}\\\;\;\sf{:\mapsto\;\;25\;-\;8\;\times\;a\;=\;\bf{0}}\end{gathered}
:↦25−8×a=0
\begin{gathered}\\\;\;\sf{:\mapsto\;\;\;-\;8\;\times\;a\;=\;\bf{-\:25}}\end{gathered}
:↦−8×a=−25
\begin{gathered}\\\;\;\sf{:\mapsto\;\;\;8\;\times\;a\;=\;\bf{25}}\end{gathered}
:↦8×a=25
\begin{gathered}\\\;\;\underline{\underline{\bf{:\mapsto\;\;\;a\;=\;\bf{\dfrac{25}{8}}}}}\end{gathered}
:↦a=
8
25
Also, when a = 0 , we get,
✒ ax² - 5x + 2 = 0
✒ 0x² - 5x + 2 = 0
✒ -5x = -2
✒ 5x = 2
✒ x = 2/5
This also, gives single root of equation since it forms linear equation. Thus,
a = 0
is also correct answer.
\begin{gathered}\\\;\underline{\boxed{\tt{Hence\;\;for\;\;single\;\;roots,\;a\;=\;\bf{\dfrac{25}{8}\;\;\tt{or}\;\;\bf{0}}}}}\end{gathered}
Henceforsingleroots,a=
8
25
or0
_______________________________________________
★ More to know :-
• Verification ::
We need to verify, if the equation has single roots or not for the value of a we got. Then,
✒ ax² - 5x + 2 = 0
✒ (25/8)x² - 5x + 2 = 0
✒ 25x² - 40x + 8 = 0
Now using the Quadratic Formula, we get,
\begin{gathered}\\\;\tt{\Rightarrow\;\;x\;=\;\dfrac{-\:b\;\pm\;\sqrt{b^{2}\;-\;4ac}}{2a}}\end{gathered}
⇒x=
2a
−b±
b
2
−4ac
\begin{gathered}\\\;\tt{\Rightarrow\;\;x\;=\;\dfrac{-\:40\;\pm\;\sqrt{(-\:40)^{2}\;-\;4\;\times\;25\;\times\;8}}{2\;\times\;25}}\end{gathered}
⇒x=
2×25
−40±
(−40)
2
−4×25×8
\begin{gathered}\\\;\tt{\Rightarrow\;\;x\;=\;\dfrac{-\:(-\:40)\;\pm\;\sqrt{1600\;-\;1600}}{50}}\end{gathered}
⇒x=
50
−(−40)±
1600−1600
\begin{gathered}\\\;\tt{\Rightarrow\;\;x\;=\;\dfrac{40\;\pm\;\sqrt{0}}{50}}\end{gathered}
⇒x=
50
40±
0
\begin{gathered}\\\;\tt{\Rightarrow\;\;x\;=\;\dfrac{40}{50}\;=\;\dfrac{4}{5}}\end{gathered}
⇒x=
50
40
=
5
4
Clearly,
\begin{gathered}\\\;\bf{\Rightarrow\;\;x\;=\;\dfrac{4}{5}\;or\;\dfrac{4}{5}}\end{gathered}
⇒x=
5
4
or
5
4
This is because, this is Quadratic equation so it must have two equal roots.
This gives that this equation has only single root that is 4/5 when a = 25/8
So our answer is correct.
Hence, Verified.
_______________________________________________
★ Supplementary Counsel :-
\begin{gathered}\\\;\sf{\leadsto\;\;For\;imaginary\;roots\;:\;b^{2}\;-\;4ac\; < \;0}\end{gathered}
⇝Forimaginaryroots:b
2
−4ac<0
\begin{gathered}\\\;\sf{\leadsto\;\;For\;distinct\;roots\;:\;b^{2}\;-\;4ac\; > \;0}\end{gathered}
⇝Fordistinctroots:b
2
−4ac>0