Question
Question: How do you factor the expression \({x^2} - 20x + 48?\)...
How do you factor the expression x2−20x+48?
Solution
Here we use the standard quadratic equation and will find the roots of the equation comparing the given equation with the standard quadratic equations-ax2+bx+c=0 where roots will be defined as x=2a−b±Δ
Complete step by step answer:
Take the given equation –
x2−20x+48
Arrange in the form of the standard equation –
x2−20x+48=0
Compare the above equation with the standard equation : ax2+bx+c=0
⇒a=1 ⇒b=−20 ⇒c=48
Also, Δ=b2−4ac
Place the values from the given comparison
⇒Δ=(−20)2−4(1)(48)
Simplify the above equation –
⇒Δ=400−192
Do subtraction –
⇒Δ=208
Take square root on both the sides of the equation –
⇒Δ=208
The above equation can be re-written as –
⇒Δ=16×13
Simplify the above equation applying the square of the known number.
⇒Δ=413
Now, roots of the given equation can be expressed as –
x=2a−b±Δ
Place values in the above equation –
x=2−(−20)±413
Product of minus and minus is plus. Simplify the above equation –
x=220±1613
Take out common from the numerator from both the terms
x=22(10±213)
Common multiple from the numerator and the denominator cancel each other. Therefore remove from the numerator and the denominator.
⇒x=10±213
Therefore x=10+213 or x=10−213
This is the required solution.
Note: Be careful regarding the sign convention. Always remember that the square of negative number or the positive number is always positive. Also, product of two negative numbers is always positive whereas, product of one positive and one negative number gives us the negative number.
Every quadratic polynomial has almost two roots. Remember, the quadratic equations having coefficients as the rational numbers has the irrational roots. Also, the quadratic equations whose coefficients are all the distinct irrationals but both the roots are the rational.