Chapter Review
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To solve the equation ax^2+bx+c=0, use the Quadratic Formula.
About (4.87,14.75) and about (- 2.87, - 0.75).
We want to solve the given system of equations using the Substitution Method. y=x^2-9 & (I) y=2x+5 & (II) The y-variable is isolated in Equation (II). This allows us to substitute its value, 2x+5, for y in Equation (I).
(I): y= 2x+5
(I): LHS-5=RHS-5
(I): LHS-2x=RHS-2x
(I): Rearrange equation
Notice that in Equation (I), we have a quadratic equation in terms of only the x-variable. x^2-2x-14=0 ⇔ 1x^2+( - 2)x+( - 14)=0
Substitute values
- (- a)=a
Calculate power
a * 1=a
- a(- b)=a* b
Add terms
Split into factors
sqrt(a* b)=sqrt(a)*sqrt(b)
Calculate root
Factor out 2
Cancel out common factors
This result tells us that we have two solutions for x. One of them will use the positive sign and the other one will use the negative sign. Therefore, we have that x_1=1+sqrt(15) and x_2=1-sqrt(15). Now, consider Equation (II). y=2x+5 We can substitute x_1=1+sqrt(15) and x_2=1-sqrt(15) in the above equation to find the values for y. Let's start with x_1=1+sqrt(15).
x= 1+sqrt(15)
Distribute 2
Add terms
We found that y= 7+2sqrt(15) when x=1+sqrt(15). One solution to the system is (1+sqrt(15),7+2sqrt(15)). To find the other solution, we will substitute 1-sqrt(15) for x in Equation (II) again.
x= 1-sqrt(15)
Distribute 2
Add terms
We found that y= 7-2sqrt(15) when x=1-sqrt(15). Therefore, our second solution is (1-sqrt(15),7-2sqrt(15)). Finally, if desired, we can approximate our solutions to two decimal places. (1+sqrt(15),7+2sqrt(15)) &≈ (4.87,14.75) (1-sqrt(15),7-2sqrt(15)) &≈ (- 2.87, - 0.75)