Video: Solving Quadratic Equations by Factorization

Find the solution set of the equation (π‘₯ βˆ’ 5)Β² βˆ’ (π‘₯ βˆ’ 5) βˆ’ 3 = 0, giving values to one decimal place.

02:22

Video Transcript

Find the solution set of the equation π‘₯ minus five squared minus π‘₯ minus five minus three equals zero, giving values to one decimal place.

Now, what we’re not actually going to do is distribute our parentheses and simplify. Instead, we need to spot that we already have a quadratic equation of some from. If you need convincing of this, let’s perform a substitution. We’re going to let 𝑦 be equal to π‘₯ minus five. Then, π‘₯ minus five all squared is 𝑦 squared. Negative π‘₯ minus five becomes negative 𝑦. And then, we subtract three. So our equation becomes 𝑦 squared minus 𝑦 minus three equals zero.

Our job is to solve for 𝑦. Once we have values for 𝑦, we can use our earlier substitution to find the suitable values for π‘₯. So how do we solve this quadratic equation? Well, we can’t factor the expression 𝑦 squared minus 𝑦 minus three. So we have two options. We can either complete the square or we can use a quadratic formula. For a quadratic equation of the form π‘Žπ‘¦ squared plus 𝑏𝑦 plus 𝑐 equals zero, the solutions are given by 𝑦 equals negative 𝑏 plus or minus the square root of 𝑏 squared minus four π‘Žπ‘ over two π‘Ž. π‘Ž is the coefficient of 𝑦 squared, so it’s one. 𝑏 is the coefficient of 𝑦, so that’s actually negative one. And 𝑐 is the constant; it’s negative three.

We substitute everything we have into our quadratic formula. And we get 𝑦 is equal to negative negative one plus or minus the square root of negative one squared minus four times one times negative three all over two times one. This simplifies to 𝑦 equals one plus or minus the square root of 13 over two. So we have two solutions for 𝑦. We have one plus the square root of 13 over two and one minus the square root of 13 over two. But what are our solutions for π‘₯?

Well, we go back to our earlier substitution, we said 𝑦 was equal to π‘₯ minus five. By adding five to both sides, we find that π‘₯ is equal to 𝑦 plus five. And so for the solution 𝑦 equals one plus the square root of 13 over two, π‘₯ is that value plus five. And the other solution for π‘₯ is one minus the square root of 13 over two plus five. That gives us 7.3027 and so on or 3.6972 and so on. We round each number to one decimal place. And then, we use these little curly brackets to represent the solution set. And our solution set is 7.3 and 3.7.

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