Week 6 – Pre Calc 11 – Manroop's Blog

This week in Pre-Calculus 11 we continued the quadratics unit. This week we learned about different ways to solve quadratic equations and how to factor perfect square trinomials.

What is a Perfect Square Trinomial? A perfect square trinomial is when the first term and the third term are perfect squares. A perfect square trinomial cannot have a third term that is negative.

Ex. $4x^2+20x+25$

$(2x+5)^2$

The example above is a perfect square trinomial because both the first and third term are perfect squares. To factor the perfect square trinomial I found the square root of the first term and the square root of the third term.

Ex. $x^2-9x-36$

$(x-12)(x+3)$

Although the first and third term are perfect squares, the expression is NOT considered a perfect square trinomial because the last term is negative. The expression is not a perfect square trinomial but it can still be factored.

What is Completing the Square? Completing the square is when the second or third term in a perfect square trinomial is missing. When the second term is missing you have to take the third term find the square root of it and then multiply the square root by 2.

Ex. $x^2$ +___+100

$\sqrt100=10$

$10*2=20$

$x^2+20x+100$

$(x+10)^2$

When the third term of a perfect square trinomial is missing you can take the second term divide it by 2 and then square it to find the third term.

Ex. $x^2+7x$ + ___

$\frac{7}{1}\times\frac{1}{2}=\frac{7}{2}$

$(\frac{7}{2})^2=\frac{49}{4}$

$x^2+7x+\frac{49}{4}$

$(x+\frac{7}{2})^2$

What is Solving a Quadratic Equation? Solving a quadratic equation is when you have to find a value of x. When solving a quadratic formula there will always be an equals sign at the end of the equation. After factoring the equation you have to isolate the variable in order to find the value of x.

Ex. $x^2-6x-27=0$

$(x-9)(x+3)=0$

$x-9=0$

$x=9$

$x+3=0$

$x=-3$

$x=9,-3$

Solving a Quadratic Equation By Completing the Square: Another way to solve quadratic equations is by completing the square. Although this strategy can be time consuming it’s easy and it works.

Ex. $x^2+6x+23=0$

$3+3=6$

$3^2=9$

$x^2+6x+9-9+23=0$

$(x+3)^2-9+23=0$

$(x+3)^2+14=0$

$(x+3)^2=-14$

$\sqrt{(x+3)^2}=\pm\sqrt{14}$

$x+3=\pm\sqrt{14}$

$x=-3\pm\sqrt{14}$

In the example above in front of $\sqrt{14}$ is the plus or minus sign ( $\pm)$ . The $\pm$ is there because the square root of 14 can be -7 or 7. I did not calculate the value of $\sqrt{14}$ because by leaving it in the square root makes it an exact value.

The Quadratic Formula: The quadratic formula is another method to solve a quadratic equation. The formula is $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ . By plugging in the values of a, b and c, you can simply calculate the value of x.