Does (3x+4)(x-4)=0 Have Two Solutions?

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It is true for all x ̸= 0. Since an algebraic identity is a statement about all numbers in a certain set, you can prove that a

The calculator computes exact solutions ... zeros and their multiplicities. ... The solutions of $ x^2-3x-4 = 0 $ are: $ x = -1 ~ \text{and} ~ x = 4$....

{\displaystyle x} ⁠ satisfying ... coefficients ⁠ ... {\displaystyle \Delta <0} ⁠, the equation has no real roots but has two distinct complex roots, which are complex conjugates of each other....

For each real number \(x\), \(\dfrac{1}{x(1 - x)} \ge 4\). When a statement is false, it is sometimes possible to add an assumption that will yield a true statement. This is usually done by using a conditional statement. So instead of working with the statement in (3), we will work with a related statement that is obtained by adding an assumption (or assumptions) to the hypothesis. For each real number \(x\), if \(0 < x < 1\), then \(\dfrac{1}{x(1 - x)} \ge 4\).

For the above equation to always be a true statement, that is \(0=0\), all the terms in the left side must be equal to \(0\).

The solutions to the equation (x+4)(x-4)=0 are x=-4,x=4

2.2 Solve : x+3 = 0 Subtract 3 from both sides of the equation : x = -3 · 2.3 Solve : x-4 = 0 Add 4 to both sides of the equation : x = 4

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