Question Video: Forming Quadratic Equations in the Simplest Form Using the Relation between a Quadratic Equation and Its Roots | Nagwa Question Video: Forming Quadratic Equations in the Simplest Form Using the Relation between a Quadratic Equation and Its Roots | Nagwa

# Question Video: Forming Quadratic Equations in the Simplest Form Using the Relation between a Quadratic Equation and Its Roots Mathematics • First Year of Secondary School

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Given that πΏ and π are the roots of the equation π₯Β² β 9π₯ β 7 = 0, find, in its simplest form, the quadratic equation whose roots are πΏ β π and π β πΏ.

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### Video Transcript

Given that πΏ and π are the roots of the equation π₯ squared minus nine π₯ minus seven equals zero, find, in its simplest form, the quadratic equation whose roots are πΏ minus π and π minus πΏ.

We begin by recalling some facts about the roots of any quadratic equation in the form ππ₯ squared plus ππ₯ plus π equals zero, where π, π, and π are constants and π is nonzero. If π sub one and π sub two are the two roots of the equation, their sum, π sub one plus π sub two, is equal to negative π over π. The product of the two roots, π one multiplied by π two, is equal to π over π.

In this question, we are given the equation π₯ squared minus nine π₯ minus seven equals zero. This means that π is equal to one, π is equal to negative nine, and π is equal to negative seven. As the two roots of this equation are πΏ and π, πΏ plus π is equal to negative negative nine over one. This is equal to nine. The product of the roots πΏ multiplied by π is equal to negative seven over one. This is equal to negative seven.

We are asked to find the equation whose roots are πΏ minus π and π minus πΏ. Letβs begin by considering the sum of these roots. We have πΏ minus π plus π minus πΏ. Removing the parentheses, we see that πΏ minus πΏ is equal to zero. Likewise, negative π plus π is also equal to zero. This means that the sum of the roots equals zero. And we know for this new equation this must be equal to negative π over π. For the product of the roots, we need to multiply πΏ minus π and π minus πΏ. Distributing the parentheses or expanding the brackets using the FOIL method gives us πΏπ minus πΏ squared minus π squared plus πΏπ. This can be simplified to two πΏπ minus πΏ squared plus π squared.

We now recall the property of the sum of two squares. π₯ squared plus π¦ squared is equal to π₯ plus π¦ all squared minus two π₯π¦. This means that we can rewrite πΏ squared plus π squared as πΏ plus π all squared minus two πΏπ. We need to subtract all of this from two πΏπ. Collecting our like terms, this simplifies to four πΏπ minus πΏ plus π all squared. We have already calculated the values of πΏ plus π and πΏ multiplied by π. Substituting these into our expression gives us four multiplied by negative seven minus nine squared. Four multiplied by negative seven is negative 28, and nine squared is 81. This simplifies to negative 28 minus 81, which in turn is equal to negative 109. As this is the product of our two roots, this is equal to π over π.

We now have two equations: negative π over π equals zero and π over π is equal to negative 109. From the first equation, we see that π must be equal to zero. And if we let π equal one, then π will be equal to negative 109. Substituting these values into the general form of the quadratic equation gives us π₯ squared plus zero π₯ minus 109 is equal to zero. In its simplest form, the quadratic equation whose roots are πΏ minus π and π minus πΏ is π₯ squared minus 109 equals zero.

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