Find, in its simplest form, the quadratic equation whose roots are five plus root two and five minus root two.
We’ve been given the roots of a quadratic equation and asked to find the equation itself. In order to do this, let’s recall a general theorem. If the quadratic equation 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero, where 𝑎 must be nonzero, has roots 𝑥 sub one and 𝑥 sub two, then 𝑥 sub one plus 𝑥 sub two is equal to negative 𝑏 over 𝑎 and 𝑥 sub one multiplied by 𝑥 sub two is equal to 𝑐 over 𝑎. In other words, the sum of the two roots is equal to the negative of the coefficient of 𝑥 divided by the coefficient of 𝑥 squared and the product of the two roots is equal to the constant term divided by the coefficient of 𝑥 squared.
If, in fact, the value of 𝑎, the coefficient of 𝑥 squared, is one so the quadratic is of the form 𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero, then these equations simplify to 𝑥 sub one plus 𝑥 sub two equals negative 𝑏 and 𝑥 sub one multiplied by 𝑥 sub two is equal to 𝑐. When deriving a quadratic equation from its roots, it’s easier to start with this simple form in which the coefficient of 𝑥 squared is equal to one. If either of the coefficients 𝑏 or 𝑐 turn out to be fractions, then we can multiply the entire equation by the lowest common denominator of these two fractions to convert the quadratic equation to one with integer coefficients.
Starting with the quadratic equation 𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero then, we know that the sum of the roots is negative 𝑏 and the product of the roots is 𝑐. Let’s define 𝑥 sub one to be the root five plus root two and 𝑥 sub two to be five minus root two. Finding the sum of these two roots, we have the equation five plus root two plus five minus root two is equal to negative 𝑏. Now root two minus root two is equal to zero, so this equation simplifies to 10 is equal to negative 𝑏. We can solve for 𝑏 by multiplying or dividing both sides by negative one, and we find that 𝑏 is equal to negative 10.
So we found one of the unknowns in our quadratic equation. To find the value of 𝑐, we need to find the product of these two roots. We have five plus root two multiplied by five minus root two is equal to 𝑐. Distributing the parentheses using the FOIL method gives 25 minus five root two plus five root two minus root two squared is equal to 𝑐. The terms in the middle of this expansion cancel one another out. Root two squared is equal to two, so this simplifies to 25 minus two equals 𝑐. And we find that the value of 𝑐 is 23.
So we found the values of both 𝑏 and 𝑐. And all that remains is to substitute them into the quadratic equation, which gives 𝑥 squared minus 10𝑥 plus 23 is equal to zero. As each of these coefficients are integers, there’s no need to multiply through by any constants. And so we have our answer.
We can check this, though, by applying the quadratic formula. This tells us that the roots of the quadratic equation 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero, where 𝑎 is nonzero, are given by 𝑥 equals negative 𝑏 plus or minus the square root of 𝑏 squared minus four 𝑎𝑐 over two 𝑎. The value of 𝑎 is one, the value of 𝑏 is negative 10, and the value of 𝑐 is 23. So we have 𝑥 equals 10 plus or minus the square root of negative 10 squared minus four multiplied by one multiplied by 23 all over two multiplied by one. The expression under the square root evaluates to 100 minus 92, which is equal to eight. So this simplifies to 10 plus or minus root eight over two.
Now, the square root of eight is the square root of four multiplied by two. So using laws of surds or radicals, this simplifies to the square root of four multiplied by the square root of two, which is two root two. So this all simplifies to 10 plus or minus two root two all over two. If we then divide through by a common factor of two, we’re left with five plus or minus root two over one or simply five plus or minus root two. This is consistent with the roots being five plus root two and five minus root two. So this confirms that our answer is correct.
In its simplest form then, the quadratic equation whose roots are five plus root two and five minus root two is 𝑥 squared minus 10𝑥 plus 23 equals zero.