In this video lesson, I review the basic concepts of percentages and percent changes that are tested on the SAT math section.
Online SAT Prep the right way
In this video lesson, I review the basic concepts of percentages and percent changes that are tested on the SAT math section.
In this video lesson, I review the basic concepts of fractional exponents that are tested on the SAT math section.
In this video lesson, I review the basics of the general equation of a parabola that shows up in different forms on the SAT math.
Key Concept: If you are given the equation of a circle in the expanded form x2–2hx+y2–2ky=f then the center of the circle is (h,k) and the radius is given by r=√f+h2+k2.
The above result saves the effort of completing the square and turning the above equation into the general equation of a circle. We will apply this concept to solve a difficult SAT math problem.
Practice Problem: The equation of a circle in the xy-plane is shown below. What is the radius of the circle?
2x2–2x+2y2+10y=5
Solution: First divide the equation by 2
2x2–2x+2y2+10y=5
and rewrite in a form where the coefficients of x2 and y2 terms are equal to 1. We obtain
x2–x+y2+5y=52
If we compare this to the general expanded form for the equation of a circle
x2–2hx+y2–2ky=f
we obtain −2h=−1 and −2k=5, therefore h=12 and k=−52, and the center of the circle is
(12,−52).
Also, f=52, and we know that r=√f+h2+k2, where f=52, h=12, and k=−52. On substitution, we obtain r=√52+(12)2+(−52)2=√52+14+254=√10+1+254=√364=√9=3
Explanation: The result stated above can be obtained by starting with the general equation of a circle with center at (h,k) and a radius of r units:
(x−h)2+(y−k)2=r2Square the algebraic termsx2–2hx+h2+y2−2ky+k2=r2Move the constant terms to the rightx2–2hx+y2−2ky=r2–h2–k2
Compare the above equation to the general expanded form
x2–2hx+y2–2ky=f
and we find that the coordinates of the center (h,k) appear as part of the coefficients of the x and y terms. Further, f=r2–h2–k2, which on rearrangment gives r=√f+h2+k2.
A quadratic equation has the form ax2+bx+c=0, where a,b, and c are real numbers, and a≠0. The ax2 term is called the quadratic term, the bx is the linear term, and c is called the constant term. Consider the quadratic equation x2–x−6=0, if we substitute x=3 in this equation we find that the equation holds true because 32−3−6=9−9=0. Similarly if we substitute x=−2 in this equation we again find that (−2)2–(−2)–6=4+2−6=0. These two values of x are the only two numbers that satisfy this quadratic equation, and are called the roots of the quadratic equation. Sometimes the roots are also referred to as zeros of the quadratic equation, in other words the value of the quadratic expression is zero at these values.
The roots of the quadratic equation x2–x−6=0 can be obtained by factoring the equation into a product of binomials. Let’s assume that the binomials are x+r1 and x+r2, our goal is to find the values of r1 and r2 such that: (x+r1)(x+r2)=x2–x−6 x2+(r1+r2)x+r1r2=x2–x−6 The two expressions above are identical and therefore the coefficient of x terms must be equal or r1+r2=−1. Similarly, the constant terms must also be equal, therefore r1r2=−6. By using trial and error we find that the values of r1 and r2 are 2 and −3. Therefore our quadratic equation can be rewritten as x2–x−6=(x+2)(x−3)=0 For the above equation to hold true, either x+2=0 or x−3=0, which tells us that x=−2 and x=3 are the solutions to this quadratic equation.
When the coefficient of the x2 term is not 1 we follow the steps described below to factor the quadratic equation. I will use the example of 2x2−5x−33=0.
Completing a square refers to the process of adding a constant term to a quadratic expression such that the result is a perfect square. Consider the expression x2+6x+4, we can rewrite this as x2+6x+9−9+4=x2+6x+9−5 We further recognize that x2+6x+9=(x+3)2 is a perfect square, and therefore the original expression is equivalent to (x+3)2−5.
In general if we are given a quadratic of the form x2+bx, we add (b2)2 to the quadratic to obtain a perfect square. x2+bx+(b2)2=x2+bx+b24=(x+b2)2
Completing the square of a quadratic expression helps us to find the least and greatest value of such expressions. In general the quadratic expression ax2+bx+c attains its maxima (a<0) or minima (a>0) when x=−b2a. The greatest and the least value of the quadratic expression can be found by substituting x=−b2a into the expression ax2+bx+c.
Solved Example: h(t)=−16t2+24t+18
The equation above gives the height above ground h, in feet, of a ball t seconds after it is launched vertically upward from a platform that is at a height of 18 feet. What is the greatest height the ball achieves?
Explanation:The easiest way to solve this problem is by recognizing that the maximum value of the expression occurs when t=−b2a=−242(−16)=34, and we can find the maximum height by substituting t=34 in the expression −16t2+24t+18=−16(34)2+24(34)+18=−9+18+18=27.
We can also do this problem by completing the square, we first factor out the term −16 from the first two terms of the quadratic expression to obtain:
h(t)=−16t2+24t+18Factor out −16=−16[t2–32]+18Complete the square=−16[t2–32+(34)2–(34)2]+18=−16[t2–32+(34)2]+16(34)2+18=−16(t−34)2+27
Note that the expression (t−34)2≥0 because it is a square of a number. This means that the term −16(t−34)2 will always be less than or equal to zero. The maximum height will have a value of 27 and this will happen when t=34.
The quadratic formula can be used to obtain the solutions to the general quadratic equation ax2+bx+c=0:
x=−b±√b2−4ac2a
Example: 2x2−5x−33=0, we have a=2, b=−5, and c=−33, replacing these values in the above quadratic formula yields x=−(−5)±√(−5)2−4(2)(−33)2(2)=5±√2894=5±174 and the two solutions are x=112 and x=−3.
Solved Example: x22=mx+n2
In the quadratic equation above, m and n are constants. What are the solutions for x ?
Explanation: x22=mx+n2Multiply both sides by 2x2=2mx+nRewrite in standard formx2–2mx–n=0Note a=1, b=−2m, c=−nx=−b±√b2−4ac2aApply quadratic formula=−(−2m)±√(−2m)2−4(1)(−n)2(1)=2m±√4m2+4n2=2m±2√m2+n2Divide throughout by 2=m±√m2+n
Let r1 and r2 be the roots of the quadratic equation ax2+bx+c=0, then the quadratic equation in the factored form can be written as a(x−r1)(x−r2)=0. If we multiply this expression and rearrange it, we obtain:
a(x−r1)(x−r2)=0a(x2–xr2–xr1+r1r2)=0a[x2−x(r1+r2)+r1r2]=0ax2−xa(r1+r2)+ar1r2=0
This resulting expanded form is equivalent to the quadratic equation ax2+bx+c=0, and the coefficient of all the terms must be identical. If we equate the coefficient of x term we obtain −a(r1+r2)=b, or in other words:
Sum of Roots=r1+r2=−ba Further if we equate the constant part of the two equations, we obtain ar1r2=c, or in other words: Product of Roots=r1r2=ca The above two formulas are helpful in finding the sum and product of the roots without having to find the roots themselves.
Solved Example: 2x2=3x+2 What is the sum of all the solutions to the quadratic equation above?
Explanation: Rewrite the equation in the standard quadratic form ax2+bx+c=0 as 2x2−3x−2=0. We have a=2, b=−3, and c=−2. The sum of the roots is equal to −ba=−(−3)2=32. There is no need to waste time in first finding the two roots, which are −12 and 2, and adding them to find their sum of 32.
The table below summarizes how the discriminant b2−4ac determines the nature of the roots of a general quadratic equation ax2+bx+c=0, and where the corresponding graph of the parabola y=ax2+bx+c=0 falls with respect to the x−axis. I will review parabolas in a different article.
ConditionNature of RootsGraph of thecorresponding parabolab2–4ac>0Roots are distinct and realParabola intersects the x−axisat two distinct pointsb2–4ac<0No real rootsParabola does not intersect the x−axisb2−4ac=0Roots are identicalParabola is tangentalso called a double rootto the x−axis
Example: If b is a constant such that the equation 2x2=bx−3 has two distinct real solutions, which of the following could be the value of b ?
Explanation: First write the quadratic equation in the standard form 2x2−bx+3=0. For the roots to be real and distinct b2−4ac>0, which is equivalent to (−b)2–4(2)(3)>0 or b2>24. The only value in the answer choices that satisfies this condition is −5.