A quadratic equation is a second-order polynomial. This means that the highest exponent of x is 2.
A typical quadratic equation is in the form:
$$Ax^2+Bx+C=0$$
They are three ways of solving the quadratic equation:
1. Factorization method
2. Completing the square method
3. Quadratic formula method
Today, we shall be solving quadratic equations using the factorization method.
Example 1
Solve $x^2+4x-12=0$ by factorization
Solution:
$x^2+4x-12=0$
$x^2+6x-2x-12=0$
$x(x+6)-2(x+6)=0$
$(x-2)(x+6)=0$
Now, we equate each factor to zero
$x-2=0$ or $x+6=0$
$x=2$ or $x=-6$
Example 2
Factorize the quadratic equation
$5x^2+7x-6=0$
Solution
By factorization
$5x^2+7x-6=0$
$5x^2+10x-3x-6=0$
$5x(x+2)-3(x+2)=0$
$(5x-3)(x+2)=0$
This translates to:
$5x=3$ or $x=-2$
$x=\frac{3}{5}$ or $x=-2$
In some cases, you need to substitute before solving the equation quadratically.
Here is an example:
Example 3
Solve $x^4+4x^2-12=0$
Solution.$x^4+4x^2-12=0$
$x^2(x^2)+4(x^2)-12=0$
Let $x^2$ be y
$y(y)+4(y)-12=0$
$y^2+4y-12=0$
By factorization
$y^2+4y-12=0$
$y^2+6y-2y-12=0$
$y(y+6)-2(y+6)=0$
$(y-2)(y+6)=0$
This translate to
$y-2=0$ or $y+6=0$
$y=2$ or $y=-6$
Recalled that $x^2$ is equal to y
So, when $x^2=2$ $$\sqrt{x^2}=\sqrt{2}$$
$$x=\sqrt{2}$$
Note: the square of a whole number cannot be negative, thus, $-6$ if extraneous
In other cases(as you will see below), the quadratic equation may be re-arranged.
Example 4
Solve $-25=4x^2+20x$ by factorization
Solution:
Re-arranging
$0=4x^2+20x+25$
$4x^2+20x+25=0$
Solving quadratically
$4x^2+20x+25=0$
$4x^2+10x+10x+25=0$
$2x(2x+5)+5(2x+5)=0$
$(2x+5)(2x+5)=0$
This translate to
$2x=-5$
Divide both sides by 2
$\frac{\require{cancel}\bcancel{2}x}{ \require{cancel}\bcancel{2}}=\frac{-5}{2}$
$x=\frac{-5}{2}$
In some cases, you may be given an incomplete quadratic equation
The next example will illustrate an incomplete quadratic equation
Example 5
Solve $4x^2-25=0$ by factorization
Solution:
This is a difference of two states $(2x)^2$, $(5)$, hence
$(2x+5)(2x-5)=0$
Hence, either
$(2x+5)=0$ or $(2x-5)=0$
$2x=-5$ or $2x=5$
This translate to
$x=\frac{-5}{2}$ or $x=\frac{5}{2}$
Related post
Example 6
Find the solution to $x^2+\frac{14x}{4}+\frac{49}{16}=0$
Solution:
First, let take L.C.M
$x^2+\frac{14x}{4}+\frac{49}{16}=0$
$\frac{16x^2+56x+49}{16}=0$
Cross multiply
$16x^2+56x+49=0\times16$
$16x^2+56x+49=0$
By factorization
$16x^2+28x+28x+49=0$
$4x(4x+7)+7(4x+7)=0$
$(4x+7)^2$=0
This translates to
$4x=-7$
$x=\frac{-7}{4}$
Example 7
Solve this: $9x^2+18x+19=26$
Solution:
First, let's rearrange
$9x^2+18x+19-26=0$
$9x^2+18x-7=0$
By factorization
$9x^2-3x+21x-7=0$$
3x(3x-1)+7(3x-1)=0$
$(3x+7)(3x-1)=0$
$3x+7=0$ or $3x-1=0$
$3x=-7$ or $3x=1$
$\frac{\require{cancel}\bcancel{3}x}{\require{cancel}\bcancel{3}}=\frac{-7}{3}$ or $\frac{\require{cancel}\bcancel{3}x}{\require{cancel}\bcancel{3}}=\frac{1}{3}$
$x=\frac{-7}{3}$ or $\frac{1}{3}$.
There you have it! You should be able to solve the quadratic equation by now. If not, then you can always go back to the examples above.
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