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 • First Year of Secondary School

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Solve the equation 4π₯Β² + 40π₯ + 40 = β60 by factoring.

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### Video Transcript

Solve the equation four π₯ squared plus 40π₯ plus 40 equals negative 60 by factoring.

Weβll copy down our equation four π₯ squared plus 40π₯ plus 40 equals negative 60. As Iβm writing this down, I notice that each of these coefficients is divisible by four. So I want to divide the entire equation by four. Four π₯ squared divided by four equals π₯ squared. 40π₯ divided by four equals 10π₯. 40 divided by four equals 10. And negative 60 divided by four equals negative 15.

Now, we have a much simpler equation to work with. Solving the equation means find the places where itβs equal to zero. That means we wanna take this negative 15 and move it to the other side. We add 15 to both sides of our equation. And on the right side, weβre left with zero. Our left side becomes π₯ squared plus 10π₯ plus 25 equals zero.

Now, we can factor. Because weβre dealing with π₯ squared, our first term for each of the factors will be π₯ by itself. We know that the second terms must be multiplied together to equal 25. The factors of 25, one times 25; two is not a factor; three is not a factor; four is not a factor; five is a factor. Five times five equals 25.

Not only do our factors need to multiply together to equal 25, they must add together to equal 10. That means that one and 25 would not work. Our missing factors here are five. π₯ plus five times π₯ plus five would equal π₯ squared plus 10π₯ plus 25. π₯ plus five squared equals zero. We take the square root of the left and the square root of the right. π₯ plus five is equal to zero. If we subtract five from both sides of our equation, on the left we have π₯ and on the right we have negative five.

The only solution for this equation is negative five.

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