Do I look quadratic to you?

The first equation means that $x-10=\pm 7$ . We can square this equation and keep all solutions: $x^2-20x+100=49$ $x^2-20x+51=0$ The answer is $-20+51=\boxed{31}$ .

NOTE: There will be the same number of solutions since an absolute value equation has 2 solutions if the RHS is positive

|x-10| = 7

tells that x = 3 , x = 17

making it as a quadratic equation

(x - 17) (x - 3) gives

x^2 - 20x + 51

a = -20 b = 51

a+b = 31

Absolute value of (x-10)=7 is also equal to (x-10)^2 = 49. Now the problem can be expanded so x^2-20x+100=49. The equation simplifies to x^2-20x+51. So your a and b are -20 and 51, which gets you answer of 31

|x-10|=7 |17-10|=7 --> 1 solution |3-10|=7 ---> 2 Solution

{Since the next equation is quadratic , we need two solutions}

Substituting , we get 289+17a+b=0 and 9-3a+b=0

Solving them , we arrive at a=-21 and b =50

therefore -21+50 = 31 .

quite easy,,,,, first u had to find out the possible values of x from equation 1 which were - 3 and 17. then , as its given that the equations are equivalent, put the same values of x one by one in equation 2, then u will obtain a pair of linear equations in 2 variables. after solving them the values of a and b come out to be - (-20) and 51 respectively . therefore the answer comes out to be 31

Equivalent equations have equivalent roots, so start by finding roots of the first equation. Splitting the absolute value function into separate equations, $x-10=7$ and $x-10=-7$ , so $x=17,3$ . These roots can be expressed as $(x-17)=0$ and $(x-3)=0$ . Multiplying together yields $(x-17)(x-3)=0$ , the factored form of a quadratic. The problem asks for coefficients in standard form though, so expanding the latter form yields $x^2-20x+51=0$ . The coefficients in question are $-20$ and $51$ , so the answer is $-20+51=31$ .

$|x - 10| = 7$ Squaring the equation gets rid of the absolute value, giving: $\begin{aligned} x^2 - 20x + 100 &= 49 \\ x^2 - 20x + 51 &= 0 \end{aligned}$ The sum of the coefficients of the $x$ and constant terms is $-20 + 51 = \boxed{31}$ .

Therefore, the answer is 31 .

from the first equation....x=3,17 substitute x in second equation..... 289 + 17a+b=0 ........(1) 9+3a+b=0.......(2) subtract ....(1) minus (2)....we will get a= -20 substitute a in any one of equation....we will get b = 51 .....so a+b=31

Very easy, $|x-10|=7$ has two solution , which is $3$ OR $17$ .

Cause $|x-10|=7$ equivalent $x^2+ax+b=0$ so,

we can substitute $x=3$ to $x^2+ax+b=0$ and we got

$3a+b=-9$

now subtitute $x=17$ , we got

$17a+b=-289$

by system equation two variable , we got

$a=-20$ , $b=51$ and the answer is $-20+51$ --> $31$