-3, 1, and 1 maybe? That's my guess.
Answer:
two solutions : (-3;4) and (-5;6)
Step-by-step explanation:
hello :
X²+y²+18x+29=0 ..(1)
X+y=1 ...(2)
by (2) : y = 1 - x
put this value in (1) : x² +(1-x)² +18x+29 = 0
x² +1 +x² -2x+18x +29 =0
2x²+16x +30 = 0
x²+8x+15 =0
delta = b²-4ac a=1 b=8 c = 15
delta = (8)²-4(1)(15)=64-60 =4 = 2²
X1=(-8+2)/2 = - 3
X2=(-8-2)/2 = - 5
case 1 : x = -3 y = 1 - (-3) = 4
case 2 : x = -5 y = 1 - (-5) = 6
The formula for quadratic equation is given by:
x² - (sum)x + (product) = 0
so if we have real roots of 3 and 5
<span> x² - (3 + 5)x + (3*5) = 0
</span> x² - 8<span>x + 15 = 0
</span>
so if we have real roots of -3 and -6
<span> x² - (-3 - 6)x + (-2*-6) = 0
</span> x² - (-9<span>x) + 12 = 0
</span> x² + 9<span>x + 12 = 0</span>
P would equal -3/7
Hope it helps!
