Question Video: Solving Quadratic Equations by Factoring Perfect Squares | Nagwa Question Video: Solving Quadratic Equations by Factoring Perfect Squares | Nagwa

Question Video: Solving Quadratic Equations by Factoring Perfect Squares Mathematics • Second Year of Preparatory School

Solve the equation 4𝑥² + 40𝑥 + 40 = −60 by factoring.

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Video Transcript

Solve the equation four 𝑥 squared plus 40𝑥 plus 40 equals negative 60 by factoring.

We’ve been given a quadratic equation and asked to solve it by factoring. Before we can do this, we need to collect every term on the same side of the equation. We’ll begin by adding 60 to each side of the equation. Doing so gives the equivalent quadratic equation four 𝑥 squared plus 40𝑥 plus 100 equals zero.

Before attempting to factor, we can observe that every coefficient in the equation is a multiple of four. Hence, we can simplify by dividing the entire equation by four. Zero divided by four is still zero. So, the right-hand side is unchanged, but the left-hand side becomes 𝑥 squared plus 10𝑥 plus 25. This is easier to work with because the coefficients are smaller. As the coefficient of 𝑥 squared in this simplified equation is one, and we’re told to solve by factoring, we know that the expression on the left-hand side can be written in the form 𝑥 plus 𝑎 multiplied by 𝑥 plus 𝑏, where 𝑎 and 𝑏 are constants that we need to determine.

The values 𝑎 and 𝑏 must satisfy two properties: their sum must be equal to the coefficient of 𝑥 in the quadratic equation, which is 10, and their product must be equal to the constant term, which is 25. With a bit of thought, or perhaps by listing the factor pairs of 25, we identify that the values of 𝑎 and 𝑏 are both five. Hence, the factored form of the quadratic equation is 𝑥 plus five multiplied by 𝑥 plus five is equal to zero.

So, we’ve written this quadratic equation in its fully factored form. In order to now solve the equation, we recall that if the product of two values is equal to zero, then at least one of those values must themselves be equal to zero. So, we set each of the two factors equal to zero and then solve the resulting equations. In this case, we have a repeated factor of 𝑥 plus five, and so we actually only need to solve this equation once. Subtracting five from each side of this equation gives 𝑥 equals negative five. The solution to the quadratic equation is therefore 𝑥 equals negative five.

A slightly alternative approach would have been to write the quadratic equation in the form 𝑥 plus five all squared equals zero. We would then argue that if something squared is equal to zero, that something must be equal to the square root of zero, which is also zero. So, we have 𝑥 plus five equals zero which leads to 𝑥 equals negative five as before. We can check our solution is correct by substituting this value of 𝑥 back into the original equation. Doing so gives 100 minus 200 plus 40, which is equal to negative 60 as required.

We’ve found that the solution to the equation four 𝑥 squared plus 40𝑥 plus 40 equals negative 60 is 𝑥 equals negative five.

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