we'll start off by grouping some
so we have a missing guy at the end in order to get the a perfect square trinomial from that group, hmmm, what is it anyway?
well, let's recall that a perfect square trinomial is
so we know that the middle term in the trinomial, is really 2 times the other two without the exponent, well, in our case, the middle term is just "x", well is really -x, but we'll add the minus later, we only use the positive coefficient and variable, so we'll use "x" to find the last term.
so, there's our fellow, however, let's recall that all we're doing is borrowing from our very good friend Mr Zero, 0, so if we add (1/2)², we also have to subtract (1/2)²
![\bf \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2-\left[ \cfrac{1}{2} \right]^2 \right)=6\implies \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2 \right)-\left[ \cfrac{1}{2} \right]^2=6 \\\\\\ \left(x-\cfrac{1}{2} \right)^2=6+\cfrac{1}{4}\implies \left(x-\cfrac{1}{2} \right)^2=\cfrac{25}{4}\implies x-\cfrac{1}{2}=\sqrt{\cfrac{25}{4}} \\\\\\ x-\cfrac{1}{2}=\cfrac{\sqrt{25}}{\sqrt{4}}\implies x-\cfrac{1}{2}=\cfrac{5}{2}\implies x=\cfrac{5}{2}+\cfrac{1}{2}\implies x=\cfrac{6}{2}\implies \boxed{x=3}](https://tex.z-dn.net/?f=%5Cbf%20%5Cleft%28%20x%5E2%20-x%20%2B%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%5D%5E2-%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%5D%5E2%20%5Cright%29%3D6%5Cimplies%20%5Cleft%28%20x%5E2%20-x%20%2B%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%5D%5E2%20%5Cright%29-%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%5D%5E2%3D6%20%5C%5C%5C%5C%5C%5C%20%5Cleft%28x-%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2%3D6%2B%5Ccfrac%7B1%7D%7B4%7D%5Cimplies%20%5Cleft%28x-%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2%3D%5Ccfrac%7B25%7D%7B4%7D%5Cimplies%20x-%5Ccfrac%7B1%7D%7B2%7D%3D%5Csqrt%7B%5Ccfrac%7B25%7D%7B4%7D%7D%20%5C%5C%5C%5C%5C%5C%20x-%5Ccfrac%7B1%7D%7B2%7D%3D%5Ccfrac%7B%5Csqrt%7B25%7D%7D%7B%5Csqrt%7B4%7D%7D%5Cimplies%20x-%5Ccfrac%7B1%7D%7B2%7D%3D%5Ccfrac%7B5%7D%7B2%7D%5Cimplies%20x%3D%5Ccfrac%7B5%7D%7B2%7D%2B%5Ccfrac%7B1%7D%7B2%7D%5Cimplies%20x%3D%5Ccfrac%7B6%7D%7B2%7D%5Cimplies%20%5Cboxed%7Bx%3D3%7D)
Answer: x=26
Step-by-step explanation:
m x H = ![\left[\begin{array}{ccc}-25&37.5&-12.5\\\9\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-25%2637.5%26-12.5%5C%5C%5C9%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
Step 1; Multiply 5 with this matrix ![\left[\begin{array}{ccc}-1&2\\4&8\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%262%5C%5C4%268%5C%5C%5Cend%7Barray%7D%5Cright%5D)
and we get a matrix ![\left[\begin{array}{ccc}-5&10\\20&40\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-5%2610%5C%5C20%2640%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Multiply the fraction 
with the matrix and we get ![\left[\begin{array}{ccc}-\frac{2m}{5} &\frac{4m}{5} \\\frac{8m}{5} &\frac{16m}{5} \\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-%5Cfrac%7B2m%7D%7B5%7D%20%26%5Cfrac%7B4m%7D%7B5%7D%20%5C%5C%5Cfrac%7B8m%7D%7B5%7D%20%26%5Cfrac%7B16m%7D%7B5%7D%20%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Step2; Now equate corresponding values of the matrices with each other.
-5 = 
and so on. By equating we get the value of m as 
Step 3; Add the matrices to get the value of matrix m.
Adding the three matrices on the RHS we get ![\left[\begin{array}{ccc}2&9&-9\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%269%26-9%5C%5C%5Cend%7Barray%7D%5Cright%5D)
.
Step 4; Adding the matrices on the LHS we get the resulting matrix as H +
![\left[\begin{array}{ccc}4&6&-8\\\9\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%266%26-8%5C%5C%5C9%5Cend%7Barray%7D%5Cright%5D)
. Equating the matrices from step 3 and 4 we get the value of H as ![\left[\begin{array}{ccc}-2&3&-1\\\9\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%263%26-1%5C%5C%5C9%5Cend%7Barray%7D%5Cright%5D)
Step 5; Now to find the value of m x H we need to multiply the value of with the matrix
Step 6; Multiplying we get the matrix m x H = [ -25 
]
Step-by-step explanation:
factor 4 out of the variable terms, as this helps.
but my approach is simply to define the target and then calculate "backwards".
we want to find
(ax + b)² = a²x² + 2abx + b²
and now we compare with the original equation :
a²x² = 4x²
a² = 4
a = 2
2abx = 16x
2×2×bx = 16x
4b = 16
b = 4
b² = 16, but we have only 3, so we need to subtract 16-3 = 13 from the completed square.
so, our equation is
(2x + 4)² - 13 = 0
(2x + 4)² = 13
2x + 4 = sqrt(13)
2x = sqrt(13) - 4
x = sqrt(13)/2 - 2
