Question Video: Forming Quadratic Equations in the Simplest Form Using the Relation between Them and Their Roots | Nagwa Question Video: Forming Quadratic Equations in the Simplest Form Using the Relation between Them and Their Roots | Nagwa

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

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Given that πΏ and π are roots of the equation π₯Β² + π₯ β 2 = 0, find, in its simplest form, the quadratic equation whose roots are πΏΒ² + π and πΒ² + πΏ.

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### Video Transcript

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

For our first step, weβll need to find what πΏ and π are. We know that theyβre the roots of this equation, π₯ squared plus π₯ minus two equals zero. We can solve this by factoring. π₯ squared can be broken into π₯ and π₯. We need factors of two, and there are only two of them: one and two. We know that two is negative. And that means one of these factors is a plus and one is a minus.

Our middle term is positive, and that means the largest constant needs to be positive. The factors of this equation can be found by π₯ minus one and π₯ plus two equals zero, where π₯ minus one equals zero and π₯ plus two equals zero. We solved for π₯ to find the roots. Adding one to both sides of the equation, π₯ equals one. We can call our πΏ one. Weβll do the same thing to find our π value: subtract two from both sides of the equation. π₯ equals negative two. Our π value is negative two. And really weβre finished with the first equation now.

Now we use these πΏ and π values to help us create a new equation. This new equation has roots πΏ squared plus π and π squared plus πΏ. Their π₯ is equal to πΏ squared plus π or π squared plus πΏ. πΏ squared, one squared, plus negative two, one squared plus negative two equals negative one. π₯ equals negative one. π squared, negative two squared, plus πΏ, plus one, negative two squared equals four plus one equals five. Now that we have our new π₯ values, π₯ equals negative one and π₯ equals 15, weβre kind of going to work in reverse.

With the first equation, we started with something equal to zero and the end result was π₯ equals something. But now we want these equations to be set to zero. We want to say π₯ plus or minus something equals zero. And we do that by adding one to both sides of this equation. Minus one plus one equals zero. π₯ plus one equals zero. And for the equation on the right, subtract five from both sides. Plus five minus five equals zero. π₯ minus five equals zero. And then we take these two expressions and multiply them together.

π₯ plus one times π₯ minus five equals zero. Weβll FOIL this out. π₯ times π₯ equals π₯ squared. π₯ times negative five equals negative five π₯. One times π₯ equals π₯. And one times negative five equals negative five. Then we bring down our equal zero. Our two middle terms, negative five π₯ plus π₯, can be simplified. Negative five π₯ plus π₯ equals negative four π₯. Then weβll bring down the remaining parts of the equation: π₯ squared minus four π₯ minus five equals zero.

The first part, we started with an equation set to zero, and we found the roots. And in the second set, we took those roots, modified them, and then found an equation, our new equation being π₯ squared minus four π₯ minus five equals zero.

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