x square-5x+3=0 by perfect square method
Answers
Answer:
do you mean completing square method then here is your answer
The expression {x}^{2}-5x+3x
2
−5x+3 fits the form a{x}^{2}+bx+cax
2
+bx+c. Let's complete the square, where:
\begin{aligned}&a=1\\&b=-5\\&c=3\end{aligned}
a=1
b=−5
c=3
2 Introduce the constant kk, which is \frac{25}{4}
4
25
in our case.
{x}^{2}-5x+\frac{25}{4}-\frac{25}{4}+3x
2
−5x+
4
25
−
4
25
+3
3 Use Square of Difference: {(a-b)}^{2}={a}^{2}-2ab+{b}^{2}(a−b)
2
=a
2
−2ab+b
2
.
{(x-\frac{5}{2})}^{2}-\frac{25}{4}+3(x−
2
5
)
2
−
4
25
+3
4 Simplify.
{(x-\frac{5}{2})}^{2}-\frac{13}{4}(x−
2
5
)
2
−
4
13
5 Substitute the above back into the original equation.
{(x-\frac{5}{2})}^{2}-\frac{13}{4}=0(x−
2
5
)
2
−
4
13
=0
6 Add \frac{13}{4}
4
13
to both sides.
{(x-\frac{5}{2})}^{2}=\frac{13}{4}(x−
2
5
)
2
=
4
13
7 Take the square root of both sides.
x-\frac{5}{2}=\pm \sqrt{\frac{13}{4}}x−
2
5
=±√
4
13
8 Simplify \sqrt{\frac{13}{4}}√
4
13
to \frac{\sqrt{13}}{\sqrt{4}}
√
4
√
13
.
x-\frac{5}{2}=\pm \frac{\sqrt{13}}{\sqrt{4}}x−
2
5
=±
√
4
√
13
9 Since 2\times 2=42×2=4, the square root of 44 is 22.
x-\frac{5}{2}=\pm \frac{\sqrt{13}}{2}x−
2
5
=±
2
√
13
10 Break down the problem into these 2 equations.
x-\frac{5}{2}=\frac{\sqrt{13}}{2}x−
2
5
=
2
√
13
x-\frac{5}{2}=-\frac{\sqrt{13}}{2}x−
2
5
=−
2
√
13
11 Solve the 1st equation: x-\frac{5}{2}=\frac{\sqrt{13}}{2}x−
2
5
=
2
√
13
.
x=\frac{\sqrt{13}+5}{2}x=
2
√
13
+5
12 Solve the 2nd equation: x-\frac{5}{2}=-\frac{\sqrt{13}}{2}x−
2
5
=−
2
√
13
.
x=\frac{-\sqrt{13}+5}{2}x=
2
−√
13
+5
13 Collect all solutions.
x=\frac{\sqrt{13}+5}{2},\frac{-\sqrt{13}+5}{2}x=
2
√
13
+5
,
2
−√
13
+5