#1**0 **

let the roots be x and y. we know that x+y=-b/a and xy=c/a. the sum of the squares of the roots is^{ }x^{2 }+ y^{2} and we want to rearrange this to be a comination of (x+y) and xy. (x+y)^{2} = x^{2 }+ y^{2 }+ 2xy. therefore, we can write x^{2 }+ y^{2 }as (x+y)^{2 }-2xy. subtitute in the other answers from the sum of the roots and the product of the roots gives you (-b/a)^{2} -2(c/a). in the quadratic 18x^{2 }+21x-400, a=18,b=21,c=400. substitute those values in and you get (-21/18)^{2} -2(-400/18). hope that helps.

nadja.birch1
Apr 9, 2018

#2**0 **

Solve for x:

18 x^2 + 21 x - 400 = 0

The left hand side factors into a product with two terms:

(3 x + 16) (6 x - 25) = 0

Split into two equations:

3 x + 16 = 0 or 6 x - 25 = 0

Subtract 16 from both sides:

3 x = -16 or 6 x - 25 = 0

Divide both sides by 3:

x = -16/3 or 6 x - 25 = 0

Add 25 to both sides:

x = -16/3 or 6 x = 25

Divide both sides by 6:

**x = -16/3 or x = 25/6 Sum of the squares =[-16/3]^2 + [25/6]^2 =1649 / 36**

Guest Apr 9, 2018