Question

(06.02)

What are the solutions to the following system of equations?
y = x2 + 12x + 30
8x − y = 10

Answer 1

sub x²+12x+30 for y

8x-y=10
8x-(x²+12x+30)=10
8x-x²-12x-30=10
-x²-4x-30=10
times -1 both sides
x²+4x+30=-10
add 10 both sides
x²+4x+40=0
factor
what 2 numbers multiply to get 40 and add to get 4?
none
ok, erm
how many real roots?
use discrimiant,
b²-4ac so
for ax²+bx+c=0
4²-4(1)(40)=16-160<0, no real roots

no solutions

Answer 2

No solution

Explanation:

We are given that two equations

`y=x^2+12x+30`

`8x-y=10`

We have to find the solutions of equations

Substitute the value of y in second equation

`8x-(x^2+12x+30)=10`

`8x-x^2-12x-30=10`

`-x^2-4x-30-10=0`

`-x^2-4x-40=0`

`x^2+4x+40=0`

To find the nature of roots by finding discriminant

`D=b^2-4ac`

`D=4^2-4*1* 40=16-160=-144 <0`

Hence , no real roots exist .Therefore, no solution of the system of equations .