if the sum of two numbers is four and the sum of their squares minus three times their product is 76, find the numbers
1 answer:
Answer:
Numbers =(6,-2 )
Step-by step explanation:
Given
the sum of two numbers is four and the sum of their squares minus three times their product is 76,
Let the first term =X
Second term will be=4-x
So
X^2+(4-x)^2-3x(4-x)=76
x^2+16+x^2-8x-12x+3x^2=76
5x^2-20x+16=76
5x^2-20x-60=0
5(x^2-4x-12)=0
So x^2-4x-12=0
x^2-(6-2)x-12=0
x^2-6x+2x-12=0
X(x-6)+2(x-6)=0
(X-6)(X+2)=0
X=6,-2
Let X=6 and -2 then numbers will be 6 and (4-6)=-2
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27 divided by 3 (j) is 9. If j=3 then plug in 3 for the equation
4x^2+23x+15=0
Δ=23^2-4*4*15=529-240=289
x1= (-23+17)/8=-6/8= -3/4
x2= (-23-17)/8=-5
4x^2+23x+15= 4(x+3/4)(x+5)=(4x+3)(x+5)
Answer:
12.4
Step-by-step explanation:
take the 310 and divide it by 5 and divide the 62 and you get 12.4 you can check by multiplying 12.4 times 5
