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Showing posts with label high school math. Show all posts
Showing posts with label high school math. Show all posts

Friday, July 10, 2026

Solving Quadratic Equations by Completing the Square The Digital SAT Math Guide to Quadratic Equations (Part 2)

 

The  Digital SAT Math Guide to Quadratic Equations (Part 2)

Solving Quadratic Equations by Completing the Square

In the previous chapter, you learned how to solve quadratic equations by factoring. Factoring is often the quickest method, but many quadratic equations on the Digital SAT cannot be factored easily. Some have large coefficients, some produce fractional values, and others have no integer factors at all.

For these equations, completing the square provides a systematic method that always works. Unlike factoring, you do not have to guess factor pairs or recognize patterns. Instead, you follow the same sequence of algebraic steps every time.

Before learning the procedure, remember one important rule.

The coefficient of x² should be 1 before you begin completing the square.

A quadratic equation whose coefficient of x² is 1 is called a monic quadratic equation.

If the equation is not monic, divide every term on both sides of the equation by the coefficient of x². This makes the remaining steps much easier and reduces mistakes.


Universal Method for Completing the Square

Always follow these steps.

  1. Write the equation in standard form.

  2. If the coefficient of x² is not 1, divide every term on both sides by that coefficient.

  3. Move the constant term to the opposite side.

  4. Take half of the coefficient of x.

  5. Square that number.

  6. Add the squared value to both sides.

  7. Rewrite the left side as the square of a binomial.

  8. Take the square root of both sides.

  9. Remember both the positive and negative square roots.

  10. Solve the resulting linear equations.

  11. Check every solution in the original equation.


Example 1

Solve

2x² + 12x + 4 = 0

Step 1

The coefficient of x² is 2.

Divide every term on both sides by 2.

2x² ÷ 2 + 12x ÷ 2 + 4 ÷ 2 = 0 ÷ 2

Simplify.

x² + 6x + 2 = 0

The equation is now monic.

Step 2

Subtract 2 from both sides.

x² + 6x + 2 − 2 = 0 − 2

Simplify.

x² + 6x = −2

Step 3

Take half of 6.

6 ÷ 2 = 3

Square it.

3² = 9

Step 4

Add 9 to both sides.

x² + 6x + 9 = −2 + 9

Simplify.

x² + 6x + 9 = 7

Step 5

Rewrite the left side.

(x + 3)² = 7

Step 6

Take square roots.

√((x + 3)²) = ±√7

Simplify.

x + 3 = ±√7

Step 7

Subtract 3 from both sides.

Positive solution:

x = −3 + √7

Negative solution:

x = −3 − √7


Example 2

Now solve a question that produces fractions immediately after making the quadratic monic.

4x² + 10x − 3 = 0

Step 1

The coefficient of x² is 4.

Divide every term by 4.

4x² ÷ 4 + 10x ÷ 4 − 3 ÷ 4 = 0 ÷ 4

Simplify.

x² + ⁵⁄₂x − ³⁄₄ = 0

Notice that fractions are perfectly acceptable. Do not convert them to decimals because exact fractions make later calculations more accurate.

Step 2

Move the constant term.

Add ³⁄₄ to both sides.

x² + ⁵⁄₂x = ³⁄₄

Step 3

Take half of the coefficient of x.

The coefficient is ⁵⁄₂.

Half of ⁵⁄₂ is

⁵⁄₂ ÷ 2 = ⁵⁄₄

Now square the result.

(⁵⁄₄)² = ²⁵⁄₁₆

Step 4

Add ²⁵⁄₁₆ to both sides.

x² + ⁵⁄₂x + ²⁵⁄₁₆ = ³⁄₄ + ²⁵⁄₁₆

Convert ³⁄₄ to sixteenths.

³⁄₄ = ¹²⁄₁₆

Now add.

¹²⁄₁₆ + ²⁵⁄₁₆ = ³⁷⁄₁₆

The equation becomes

x² + ⁵⁄₂x + ²⁵⁄₁₆ = ³⁷⁄₁₆

Step 5

Rewrite the left side.

(x + ⁵⁄₄)² = ³⁷⁄₁₆

Step 6

Take square roots.

√((x + ⁵⁄₄)²) = ±√(³⁷⁄₁₆)

Simplify.

x + ⁵⁄₄ = ±√37⁄4

Step 7

Subtract ⁵⁄₄ from both sides.

x = −⁵⁄₄ ± √37⁄4

These are the exact solutions.


Notice that completing the square works just as well with fractions as it does with whole numbers. On the Digital SAT, leaving answers in exact fractional or radical form is often the correct approach unless the question specifically asks for a decimal approximation.


Friday, December 19, 2008

trigonometry problem

show that 1/(1-cosx) =(cscx+cotx)cscx

start from the LHS, introduce the conjugate
simplify the denominator using identity then divide term by term

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index of math problems

trigonometry formulae

for some other examples
prove that (cotA+cosecA-1)/(cotA-cosecA+1)=cosecA+cotA -----> (problem on trigonometry )

Prove that (tanxsinx) / (tanx+sinx) = (tanx-sinx) / (tanx sinx) -----> (problem on trigonometry )

find exact value of tan[22½ °] or tan (pi / 8 ) -----> (problem on trigonometry )

find exact value of tan15° without using calculator --------value of tan15° without using calculator

conditional identity : If A+B+C=π,
prove that: sinA+sinB+sinC = cot(A/2).cot(B/2) [sinA+sinB-sinC] -----> (problem on conditional identity (trigonometry) )


graphs of trigonometric functions-----> (graphs of trigonometric functions )

prove that
{ tan((π/4)+x) - tan((π/4)-x)} / { tan((π/4)+x) + tan((π/4)-x) } = sin2x ---------trigonometry problem

prove that (cos²x - sin²x) / (cos²x + sinxcosx) = 1 - tanx -------------trigonometry problem

show that
tan(x)/[1-cot(x)] + cot(x)/[1-tan(x)]= 1+sec(x)csc(x) , x≠ n π /4 ---------trigonometry



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Friday, November 14, 2008

integration by substitution

evaluate integral of e^(2x) / sqrt [1 + e^(2x)]

some other examples
integral of ( ( 4 - x^2 )^( 3/2) ) with limits 0 to 1--------------integration by substitution

integral of e^{x^(1/3)}-------------- e^{x^(1/3)}

integral of ( ( 4 - x^2 )^( 3/2) ) with limits 0 to 1--------------integration by substitution

more problems on integration -----------problems on integration

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work





Find p If Two Lines Are Perpendicular | 3D Geometry Solution

 If the lines (x - 3)/1 = (1 - y)/1 = (z + 2)/p and (2 - x)/3 = (y + 1)/5 = (z + 56)/2p are perpendicular to each other, then find the value...