Jtdylktuy

I do not know how to solve this questions. Can somebody please help? A car averages 27 miles per gallon. If gas costs $4.04 per gallon, which of the following is closest to how much the gas would cost for this car to travel 2,727 typical miles?

To begin with, notice that

begin{align*}
(x-a)(x-b) = h^{2} Longleftrightarrow x^{2} - (a+b)x + ab - h^{2} = 0 Longleftrightarrow x = frac{a+b pmsqrt{(a-b)^{2}+4h^{2}}}{2}
end{align*}

Once $(a-b)^{2} + 4h^{2} geq 0$, both solutions are real.

Note that

$(x-a)(x-b)=h^2 iff x^2-(a+b)x+ab-h^2=0.$

And note that, for the quadratic equation to have real roots, we need $$D=(a+b)^2-4(ab-h^2)geq 0$$

And observe that

$$D=a^2+b^2-2ab+4h^2=(a-b)^2+h^2geq 0$$

since $r^2geq 0$ for any real number $r$.

Therefore, that quadratic equation has always real roots.

From a geometric point of view,
$$y = (x-a)(x-b)$$
is a parabola that crosses the $x$-axis at $a$ and $b$. Since it opens upward, it also crosses any horizontal line above the $x$-axis.

The graph of the function on the left side is a parabola with zeroes at $a$ and $b$ which opens upward. The vertex of the parabola is at the midpoint between $a$ and $b$ and is on or below the x-axis.

Since $h^2geq 0$, the horizontal line $y=h^2$ cuts the parabola on or above the x-axis at points whose x-coordinates are precisely the solutions to this equation.

If $a$ and $b$ are distinct, there are two solutions. The solutions will coincide if and only if $a=b$ and $h=0$, in which case the horizontal line is the x-axis and the vertex is on the x-axis.

Given equation is $(x-a)(x-b)=h^2$
$$x^2-(a+b)x+ab=h^2$$
$$x^2-(a+b)x+(ab-h^2)=0$$

Now, discriminant $implies b^2-4ac$

where $a=1, b=-(a+b),c=ab-h^2$
$$b^2-4ac=[-(a+b)]^2-4cdot1cdot(ab-h^2)$$
$$=(a-b)^2+4h^2ge0$$
Therefore, the roots of $(x-a)(x-b)=h^2$ are always real.

$(x-a)(x-b)=h^2implies x^2-(a+b)x+ab-h^2=0implies x=frac{(a+b)pmsqrt{(a+b)^2-4cdot (ab-h^2)}}2$.

So we just need to check that the discriminant $Delta =(a+b)^2-4(ab-h^2)ge0$. But $Delta =a^2+b^2-2ab+4h^2=(a-b)^2+4h^2$ is indeed nonnegative.