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Chapter 1: Problem 89

Show that if \(a x^{2}+b x+c=0\) for all \(x,\) then \(a=b=c=0\)

Short Answer

Expert verified
If the equation \(ax^{2}+bx+c=0\) holds for all values of \(x\), then \(a=b=c=0\).

Step by step solution

01

Set the equation

The given equation is \(ax^{2}+bx+c=0\). This equation holds for all \(x\).
02

Check the degree of the equation

Notice that this is a quadratic equation i.e., its degree is 2, and hence its polynomial can have at most two roots.
03

Apply the fundamental theorem

According to the fundamental theorem of algebra, a second-degree polynomial can have at most 2 roots. If our given equation is valid for all \(x\), every real number could be a root, which contradicts the fundamental theorem of algebra.
04

Identify the only possibility

The only way the quadratic expression could be equal to zero for all \(x\) is when all coefficients are zero i.e., when \(a = 0\), \(b = 0\), and \(c = 0\).

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Most popular questions from this chapter

Describe all possible \(2 \times 2\) reduced row-echelon matrices. Support your answer with examples.

Determine conditions on \(a, b, c,\) and \(d\) such that the matrix $$\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]$$ will be row-equivalent to the given matrix. $$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$

Determine whether the matrix is in row-echelon form. If it is, determine whether it is also in reduced row-echelon form. $$\left[\begin{array}{llll} 0 & 1 & 0 & 0 \\ 1 & 0 & 2 & 1 \end{array}\right]$$

Determine the value(s) of \(k\) such that the system of linear equations has the indicated number of solutions. No solution \\[ \begin{aligned} x+2 y+k z &=6 \\ 3 x+6 y+8 z &=4 \end{aligned} \\]

Solve the system using either Gaussian elimination with back-substitution or Gauss-Jordan elimination. $$\begin{array}{rr} 2 x+y-z+2 w= & -6 \\ 3 x+4 y+w= & 1 \\ x+5 y+2 z+6 w= & -3 \\ 5 x+2 y-z-w= & 3 \end{array}$$
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