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Chapter 7: Problem 2

Verify that \(x=-2\) and \(x=-3\) are both solutions of \(x^{2}+5 x+6=0\)

Short Answer

Expert verified
Answer: Yes, both -2 and -3 are solutions to the given quadratic equation.

Step by step solution

01

Substitute x = -2 into the equation

Substitute x = -2 into the given equation: \((-2)^{2} + 5(-2) + 6\)
02

Simplify and solve for x = -2

Solve the equation: \((4) + (-10) + 6 = 0\) \(-6 + 6 = 0\) \(0 = 0\) The equation holds true when x = -2.
03

Substitute x = -3 into the equation

Substitute x = -3 into the given equation: \((-3)^{2} + 5(-3) + 6\)
04

Simplify and solve for x = -3

Solve the equation: \((9) + (-15) + 6 = 0\) \(-6 + 6 = 0\) \(0 = 0\) The equation holds true when x = -3. As both x = -2 and x = -3 satisfy the equation \(x^{2}+5x+6=0\), we can verify that \(x=-2\) and \(x=-3\) are both solutions of the given quadratic equation.

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Most popular questions from this chapter

Rewrite each of the statements without using a modulus sign: $$ |x| \leq 7.5 $$

Express in partial fractions $$ C(s)=\frac{K}{s(1+\tau s)} $$ where \(K\) and \(\tau\) are constants.

Factorise \(x^{3}+6 x^{2}+6 x+5\) given that \(x+5\) is a factor.

Rewrite each of the statements without using a modulus sign: $$ |x|<5 $$

In each case verify that the given values satisfy (b) \(x=4, y=3\) satisfy \(x+y=7\) and the given simultaneous equations: (a) \(x=2, y=-2\) satisfy \(7 x+y=12\) and (c) \(x=-3, y=2\) satisfy \(8 x-y=-26\) \(-3 x-y=-4\) and \(9 x+2 y=-23\)
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