Let f(x) = Ax^2 + Bx and g(x) = Lx^2 + Mx + N | vedclass

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(A) Let $h(x) = f(x) - g(x) = (A - L)x^2 + (B - M)x - N$. Given: $h(2) = 4(A - L) + 2(B - M) - N = 1$ ... $(i)$ $h(3) = 9(A - L) + 3(B - M) - N = 4$ .

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(A) Let $h(x) = f(x) - g(x) = (A - L)x^2 + (B - M)x - N$. Given: $h(2) = 4(A - L) + 2(B - M) - N = 1$ ... $(i)$ $h(3) = 9(A - L) + 3(B - M) - N = 4$ ... $(ii)$ $h(4) = 16(A - L) + 4(B - M) - N = 9$ ... $(iii)$ Subtracting $(i)$ from $(ii)$: $5(A - L) + (B - M) = 3$ ... $(iv)$ Subtracting $(ii)$ from $(iii)$: $7(A - L) + (B - M) = 5$ ... $(v)$ Subtracting $(iv)$ from $(v)$: $2(A - L) = 2 \Rightarrow A - L = 1$. Substituting $A - L = 1$ in $(iv)$: $5(1) + (B - M) = 3 \Rightarrow B - M = -2$. Substituting $A - L = 1$ and $B - M = -2$ in $(i)$: $4(1) + 2(-2) - N = 1$ $\Rightarrow 4 - 4 - N = 1$ $\Rightarrow N = -1$. Thus, $h(x) = (1)x^2 + (-2)x - (-1) = x^2 - 2x + 1$. Setting $h(x) = 0$: $x^2 - 2x + 1 = 0$ $\Rightarrow (x - 1)^2 = 0$ $\Rightarrow x = 1$.

Let $f(x) = Ax^2 + Bx$ and $g(x) = Lx^2 + Mx + N$. Given that $f(2) - g(2) = 1$, $f(3) - g(3) = 4$, and $f(4) - g(4) = 9$. Then a root of $f(x) - g(x) = 0$ is

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