Question Video: Solving Quadratic Equations Using the Quadratic Formula Mathematics

Given that π‘₯ = βˆ’2 is a root of the equation π‘₯Β² βˆ’ 4π‘šπ‘₯ βˆ’ (π‘šΒ² βˆ’ 6) = 0, find the set of all possible values of π‘š.

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Video Transcript

Given that π‘₯ equals negative two is a root of the equation π‘₯ squared minus four π‘šπ‘₯ minus π‘š squared minus six equals zero, find the set of all possible values of π‘š.

In this question, we’re given a quadratic equation π‘₯ squared minus four π‘šπ‘₯ minus π‘š squared minus six equals zero and are told that a root or solution to this equation is π‘₯ equals negative two. This means that when we substitute π‘₯ equals negative two into the left-hand side, we get an answer of zero. Negative two squared minus four multiplied by π‘š multiplied by negative two minus π‘š squared minus six is equal to zero.

Squaring negative two gives us four. Multiplying negative four π‘š by negative two gives us eight π‘š. And distributing the parentheses by multiplying negative one by π‘š squared minus six gives us negative π‘š squared plus six. Four plus eight π‘š minus π‘š squared plus six equals zero. This can be rewritten as negative π‘š squared plus eight π‘š plus 10 equals zero.

We now have a quadratic equation in the form π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐 equals zero, where π‘Ž, 𝑏, and 𝑐 are constants and π‘Ž is nonzero. We can solve an equation of this type using the quadratic formula, which states that π‘₯ is equal to negative 𝑏 plus or minus the square root of 𝑏 squared minus four π‘Žπ‘ all divided by two π‘Ž. In this question, we can use the quadratic formula to find the set of possible values of π‘š. We have values of π‘Ž, 𝑏, and 𝑐 of negative one, eight, and 10, respectively.

Substituting in these values, we have π‘š is equal to negative eight plus or minus the square root of eight squared minus four multiplied by negative one multiplied by 10 all divided by two multiplied by negative one. Eight squared is equal to 64. And negative four multiplied by negative one multiplied by 10 is 40. π‘š is therefore equal to negative eight plus or minus the square root of 104 divided by negative two. Using our laws of radicals or surds, we can rewrite the square root of 104 as the square root of four multiplied by the square root of 26. And since the square root of four is two, this is equal to two root 26.

There are two possible values of π‘š. Either π‘š is equal to negative eight plus two root 26 over negative two, or π‘š is equal to negative eight minus two root 26 over negative two. In both cases, we can divide through by negative two, giving us π‘š is equal to four minus the square root of 26 and π‘š is equal to four plus the square root of 26. These are the two possible values of π‘š such that π‘₯ equals negative two is a root of the equation π‘₯ squared minus four π‘šπ‘₯ minus π‘š squared minus six equals zero.

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