Answer:
\(h(x)=2x^2+14x-60\)
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
\(h(x)=x^2-13x-30\)
\(h(3)=3^2-13(3)-30\)
\(h(3)=-60\)
As h(3)≠0, Equation 1 is discounted
Equation 2
\(h(x)=x^2-7x-30\)
\(h(3)=3^2-7(3)-30\)
\(h(3)=-42\)
As h(3)≠0, Equation 2 is discounted
Equation 3
\(h(x)=2x^2+26x-60\)
\(h(3)=2(3)^2+26(3)-60\)
\(h(3)=36\)
As h(3)≠0, Equation 3 is discounted
Equation 4
\(h(x)=2x^2+14x-60\)
\(h(3)=2(3)^2+14(3)-60\)
\(h(3)=0\)
\(h(x)=2x^2+14x-60\)
\(h(-10)=2(-10)^2+14(-10)-60\)
\(h(-10)=0\)
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:
\(x=\frac{-b\±\sqrt{b^2-4ac}}{2a}\)
where a, b and c are as below
\(h(x)=ax^2+bx+c=0\)
Equation 1
\(h(x)=x^2-13x-30=0\)
\(x=\frac{-b\±\sqrt{b^2-4ac}}{2a}\)
\(x=\frac{13\±\sqrt{(-13)^2-4(1)(-30)}}{2(1)}\)
\(x=15,x=-2\)
As roots are not x=3 and x=-10, Equation 1 is discounted
Equation 2
\(h(x)=x^2-7x-30\)
\(x=\frac{-b\±\sqrt{b^2-4ac}}{2a}\)
\(x=\frac{-(-7)\±\sqrt{(-7)^2-4(1)(-30)}}{2(1)}\)
\(x=10,x=-3\)
As roots are not x=3 and x=-10, Equation 2 is discounted
Equation 3
\(h(x)=2x^2+26x-60\)
\(x=\frac{-b\±\sqrt{b^2-4ac}}{2a}\)
\(x=\frac{-(26)\±\sqrt{(26)^2-4(2)(-60)}}{2(2)}\)
\(x=2,x=-15\)
As roots are not x=3 and x=-10, Equation 3 is discounted
Equation 4
\(h(x)=2x^2+14x-60\)
\(x=\frac{-b\±\sqrt{b^2-4ac}}{2a}\)
\(x=\frac{-(14)\±\sqrt{(14)^2-4(2)(-60)}}{2(2)}\)
\(x=3,x=-10\)
As roots are x=3 and x=-10, Equation 4 represents h(x)
