In my previous post, we solved quadratic equations using the factorization method.
A quadratic equation can, however, be derived from a quadratic root.
A quadratic equation can, however, be derived from a quadratic root.
But first,
what is the root of a quadratic equation?
The root of a quadratic equation is simply the solution pair of a quadratic equation.
For example, $x^2+x-6=0$ has a solution pair of $(2$, $-3$)
Hence $2$ and $-3$ are said to be roots of $x^2+x-6=0$.
To derive the quadratic equation from roots,
You will use the formula:
$x^2-$(sum of roots)x+(product of roots)$=0$
👉This post is the extension of our series on quadratic equation👈
Example 1
If the roots of a quadratic equation are $(3$,$5)$, find the equation.
solution:
Remember that the formula is
$x^2-(sum of roots)x+(product of roots)=0$
The Sum of roots is $3+5=8$
Product of roots is $3\times5=15$
Hence,
$x^2-(8)x+(15)=0$
$x^2-8x+15=0$
Example 2
If the roots of a quadratic equation are
$(\frac{-1}{2}$, $2$), find the equation.
Solution:
Sum of roots is $\frac{-1}{2}+2=\frac{3}{2}$
Product of roots is $\frac{-1}{2}\times2=-1$
$x^2-(\frac{3}{2})x+(-1)=0$
$x^2-\frac{3x}{2}-1=0$
Taking the L.C.M
$\frac{2x^2-3x-2}{2}=0$
$2x^2-3x-2=0\times2$
$2x^2-3x-2=0$
Example 3
Find the quadratic equation whose roots is $\frac{2}{5}$ and $\frac{-1}{2}$
Solution:
Sum of roots is $\frac{2}{5}+\frac{-1}{2}=\frac{-1}{10}$
Product of roots is $\frac{2}{5}\times\frac{-1}{2}=\frac{-1}{5}$
Inserting the value in the formula
$x^2-(\frac{-1}{10})x+(\frac{-1}{5})=0$
$x^2+\frac{1x}{10}-\frac{1}{5}=0$
Taking the L.C.M
$\frac{10x^2+1x-2}{10}=0$
Cross multiply
$10x^2+1x-2=0\times10$
$10x^2+1x-2=0$
Example 4
Given that the roots of a quadratic equation are $18$ and $12$, find the equation.Solution:
The Sum of roots is $18+12=30$
Product if roots is $18\times12=216$
$x^2-(sum of roots)x+(product of roots)=0$.
$x^2-(30)x+(216)=0$
$x^2-30x+216=0$
Example 5
Find the quadratic equation whose roots are $-6$ and $4$.
Solution:
The Sum of roots is $-6+4=-2$
Product of roots is $-6\times4=-24$
$x^2-(-2)x+(-24)=0$
$x^2+2x-24=0$
Example 6
Find the quadratic equation whose roots are $-3$ and $\frac{-2}{5}$.Solution:
Sum of roots is $-3+(\frac{-2}{5})=\frac{-17}{5}$
Product of roots is $-3\times\frac{-2}{5}=\frac{6}{5}$
$x^2-(\frac{-17}{5})x+(\frac{6}{5})=0$
$x^2+\frac{17x}{5}+\frac{6}{5}=0$
Taking the L.C.M
$\frac{5x^2+17x+6}{5}=0$
Cross multiply
$5x^2+17x+6=0times s5$
$5x^2+17x+6=0$.
Related post
Example 7
If the roots of a quadratic equation are $\frac{-1}{3}$ and $\frac{2}{3}$, find the equation.
Solution:
Sums of roots is $\frac{-1}{3}+\frac{2}{3}=\frac{1}{3}$
Product of roots is $\frac{-1}{3}\times\frac{2}{3}=\frac{-2}{9}$
$x^2-(sum of roots)x+(product of roots)=0$
$x^2-(\frac{1}{3})x+(\frac{-2}{9})=0$
$x^2-\frac{1x}{3}-(\frac{2}{9})=0$
Taking the L.C.M
$\frac{9x^2-3x-2}{9}=0$
Cross multiply
$9x^2-3x-2=0\times 9$
$9x^2-3x-2=0$.
Post a Comment
Post a Comment