Question

enter the correct answer in the box.
solve the equation by completing the square.
$x^2 - 10x + 26 = 8$
fill in the values of a and b to complete the solutions.
$x = a - \sqrt{b}$
$x = a + \sqrt{b}$

Explanation:

Step1: Isolate constant term

$x^2 - 10x = 8 - 26$
$x^2 -10x = -18$

Step2: Complete the square

Add $(\frac{-10}{2})^2 = 25$ to both sides:
$x^2 -10x +25 = -18 +25$
$(x -5)^2 = 7$? No, wait: 8-26=-18, -18+25=7? Wait no: original equation $x²-10x+26=8$ → $x²-10x=8-26=-18$ → add 25: $(x-5)²=7$? Wait no, wait 8-26 is -18? Wait 26-8=18, so moving 26 to right: $x²-10x=8-26=-18$? Wait no, wait let's check again: $x²-10x+26=8$ → subtract 26: $x²-10x=8-26=-18$ → then complete square: $x²-10x+25=-18+25=7$ → $(x-5)²=7$ → $x=5±√7$? Wait wait no, wait 8-26 is -18? Wait 26-8=18, so if we do $x²-10x+26-8=0$ → $x²-10x+18=0$ → then $x²-10x=-18$ → add 25: $(x-5)²=7$ → yes, so $a=5$, $b=7$? Wait wait no, wait the problem says fill a and b for $x=a±√b$. So yes, $a=5$, $b=7$. Wait wait I made a mistake earlier in calculation but corrected: original equation $x²-10x+26=8$ → $x²-10x=8-26=-18$ → complete square: $x²-10x+25=-18+25=7$ → $(x-5)^2=7$ → $x=5±√7$. So a=5, b=7.

Wait wait let's confirm: $(5+√7)^2 -10(5+√7)+26 = 25 +10√7 +7 -50 -10√7 +26 = (25+7+26-50)+(10√7-10√7)=8+0=8$. Correct. Similarly for $5-√7$: same result. So yes, a=5, b=7.

Step1: Rearrange the equation

$x^2 -10x = 8 - 26$
$x^2 -10x = -18$

Step2: Complete the square

Add $25$ to both sides:
$x^2 -10x +25 = -18 +25$
$(x-5)^2 = 7$

Step3: Solve for x

$x = 5 ± \sqrt{7}$
Thus, $a=5$, $b=7$

Answer:

a = 5, b = 3