Answer:
[tex]h(x)=2x^2+14x-60[/tex]
Step-by-step explanation:
This question can be solved by two methods
Method 1: Substitute x=3 and x=-10 in all the equations and determine which equals to zero (ie., check h(3)=0 and h(-10)=0 for all the equations)
Equation 1
[tex]h(x)=x^2-13x-30[/tex]
[tex]h(3)=3^2-13(3)-30[/tex]
[tex]h(3)=-60[/tex]
As h(3)≠0, Equation 1 is discounted
Equation 2
[tex]h(x)=x^2-7x-30[/tex]
[tex]h(3)=3^2-7(3)-30[/tex]
[tex]h(3)=-42[/tex]
As h(3)≠0, Equation 2 is discounted
Equation 3
[tex]h(x)=2x^2+26x-60[/tex]
[tex]h(3)=2(3)^2+26(3)-60[/tex]
[tex]h(3)=36[/tex]
As h(3)≠0, Equation 3 is discounted
Equation 4
[tex]h(x)=2x^2+14x-60[/tex]
[tex]h(3)=2(3)^2+14(3)-60[/tex]
[tex]h(3)=0[/tex]
[tex]h(x)=2x^2+14x-60[/tex]
[tex]h(-10)=2(-10)^2+14(-10)-60[/tex]
[tex]h(-10)=0[/tex]
As h(3)=0 and h(-10)=0, Equation 4 represents h(x)
Method 2: Solve to find the roots of each equation where h(x)=0 using the quadratic formula. Roots should be x=3,x=-10
The quadratic formula is:
[tex]x=\frac{-b\±\sqrt{b^2-4ac}}{2a}[/tex]
where a, b and c are as below
[tex]h(x)=ax^2+bx+c=0[/tex]
Equation 1
[tex]h(x)=x^2-13x-30=0[/tex]
[tex]x=\frac{-b\±\sqrt{b^2-4ac}}{2a}[/tex]
[tex]x=\frac{13\±\sqrt{(-13)^2-4(1)(-30)}}{2(1)}[/tex]
[tex]x=15,x=-2[/tex]
As roots are not x=3 and x=-10, Equation 1 is discounted
Equation 2
[tex]h(x)=x^2-7x-30[/tex]
[tex]x=\frac{-b\±\sqrt{b^2-4ac}}{2a}[/tex]
[tex]x=\frac{-(-7)\±\sqrt{(-7)^2-4(1)(-30)}}{2(1)}[/tex]
[tex]x=10,x=-3[/tex]
As roots are not x=3 and x=-10, Equation 2 is discounted
Equation 3
[tex]h(x)=2x^2+26x-60[/tex]
[tex]x=\frac{-b\±\sqrt{b^2-4ac}}{2a}[/tex]
[tex]x=\frac{-(26)\±\sqrt{(26)^2-4(2)(-60)}}{2(2)}[/tex]
[tex]x=2,x=-15[/tex]
As roots are not x=3 and x=-10, Equation 3 is discounted
Equation 4
[tex]h(x)=2x^2+14x-60[/tex]
[tex]x=\frac{-b\±\sqrt{b^2-4ac}}{2a}[/tex]
[tex]x=\frac{-(14)\±\sqrt{(14)^2-4(2)(-60)}}{2(2)}[/tex]
[tex]x=3,x=-10[/tex]
As roots are x=3 and x=-10, Equation 4 represents h(x)
