Thứ Sáu, Tháng Hai 7, 2025
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How to solve quadratic equations

Solve quadratic equations

A. What is a quadratic equation?

A quadratic equation is an equation of the form ax^2+bx+c=0 (a≠0) (1).

With x being the unknown unknown and because there is only 1 unknown, it is also called a 'single variable' equation. The numbers a, b, and c are known numbers, called coefficients of the equation; can be distinguished by calling them quadratic coefficients, first-order coefficients, and free or constant coefficients.

A quadratic equation is a type of polynomial equation, it only contains powers of x that are natural numbers.

Solving a quadratic equation is to find values ​​of x such that when substituting x into equation (1), ax is satisfied.2+bx+c=0. There are 4 common ways to solve quadratic equations: factoring; square complement method; use the solution formula; graph.

B. Solve quadratic equations

Step 1: Calculate Δ=b2-4ac

Step 2: Compare Δ with 0

  • Δ equation (1) has no solution
  • Δ = 0 => equation (1) has a double solution x_{1} =x_{2} = - \frac{b}{2a}
  • Δ > 0 => equation (1) has 2 distinct solutions, we use the following solution formula:

x_{1} =\frac{-b+\sqrt{\triangle } }{2a} and x_{2} =\frac{-b-\sqrt{\triangle } }{2a}

C. Mentally evaluate quadratic equations

  • If Eq ax^{2} + bx + c = 0;(a \neq 0) Have a + b + c = 0 then the equation has a solution \left\{ \begin{matrix}x_{1} = 1 \\x_{2} = \dfrac{c}{a} \\end{matrix} \right.
  • If Eq ax^{2} + bx + c = 0;(a \neq 0)Have a – b + c = 0 then the equation with solution is: \left\{ \begin{matrix}x_{1} = - 1 \\x_{2} = \dfrac{- c}{a} \\end{matrix} \right.
How to solve quadratic equations
How to solve quadratic equations

D. Using Vi – et

Viet's theorem

If x_{1};x_{2}is the solution of the equation ax^{2} + bx + c = 0;(a \neq 0)then \left\{ \begin{matrix}S = x_{1} + x_{2} = \dfrac{- b}{a} \\P = x_{1}.x_{2} = \dfrac{c}{ a} \\end{matrix} \right.

Inverse Vi – et theorem

If two numbers x_{1};x_{2}Have \left\{ \begin{matrix} S = x_{1} + x_{2} \\ P = x_{1}.x_{2} \\ \end{matrix} \right.then x_{1};x_{2}is the solution of the equation x^{2} - Sx + P = 0(x_{1};x_{2} exists when S^{2} - 4P \geq 0)

E. Example of solving a quadratic equation

Example 1: Solve the following quadratic equation: x2 – 49x – 50 = 0

Solution instructions

Method 1: Use the solution formula (a = 1; b = -49; c = -50)

\begin{matrix} \Delta = ( - 49)^{2} - 4.1.( - 50) = 2601 \\ \Rightarrow \sqrt{\Delta} = 51 \\ \end{matrix}

Because ∆ > 0, the equation has two distinct solutions \left\{ \begin{matrix}x_{1} = \dfrac{- ( - 49) - 51}{2} = - 1 \\x_{2} = \dfrac{- ( - 49) + 51}{ 2} = 50 \\end{matrix} \right.

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Method 2: Contemplation

Because a – b + c = -1 – (-49) + (-50) = 0

So the equation has two solutions \left\{ \begin{matrix} x_{1} = - 1 \\ x_{2} = 50 \\ \end{matrix} \right.

Method 3: \Delta = ( - 49)^{2} - 4.1.( - 50) = 2601 > 0″ width=”256″ height=”18″ data-type=”2″ data-latex=”\Delta = ( -
49)^{2} – 4.1.( – 50) = 2601 > 0″ class=”lazy” data-src=”https://tex.vdoc.vn?tex=%5CDelta%20%3D%20(%20-%0A49)%5E%7B2%7D%20-%204.1.(%20-%2050)%20%3D%202601%20%3E%200″/></span></p>
<p>According to the Vi-et theorem we have:</p>
<p><span class=\left\{ \begin{matrix} S = x_{1} + x_{2} = 49 = ( - 1) + 50 \\ P = x_{1}.x_{2} = - 50 = ( - 1) .50 \\ \end{matrix} \right.

So the equation has two solutions: \left\{\begin{matrix}x_{1} = - 1 \\x_{2} = - \dfrac{- 50}{1} = 50 \\\end{matrix} \right.

Example 2: Solve the 4x equation2 – 2x – 6 = 0 (2)

Δ=(-2)2 – 4.4.(-6) = 4 + 96 = 100 > 0 => given equation (2) has 2 distinct solutions.

x_{1} =\frac{-(-2)+\sqrt{100} }{2.4} =\tfrac{3}{2} and x_{2} = \frac{-(-2)-\sqrt{100} }{2.4} =-1

You can also do quick mental math, because you realize that 4-(-2)+6=0, so x1 = -1, x2 = -c/a = -(-6)/4=3/2. The solution is still the same as above.

Example 3: Solve the equation 2x2 – 7x + 3 = 0 (3)

Calculate Δ = (-7)2 – 4.2.3 = 49 – 24= 25 > 0 => (3) has 2 distinct solutions:

x_{1} =\frac{-(-7)+\sqrt{25} }{2.2} = 3 and x_{1} =\frac{-(-7)-\sqrt{25} }{2.2} = \frac{1}{2}

To check if you have calculated the correct solution is very easy, just replace x1x2 Enter equation 3, if the result is 0, it is standard. For example, replace x12.32-7.3+3=0.

Example 4: Solving the 3x equation2 + 2x + 5 = 0 (4)

Calculate Δ = 22 – 4.3.5 = -56 equation (4) has no solution.

Example 5: Solve the x equation2 – 4x +4 = 0 (5)

Calculate Δ = (-4)2 – 4.4.1 = 0 => equation (5) has a double solution:

x_{1} =x_{2} =\frac{-(-4)}{2.1} =2

In fact, if you think quickly, you can see that this is the memorable equation (ab)2 = a2 – 2ab + b2 so it's easy to rewrite (5) as (x – 2)2 = 0 x=2.

F. Factoring polynomials

If equation (1) has 2 distinct solutions x1x2you can always write it in the following form: ax2 + bx + c = a(xx1)(xx2) = 0.

Returning to equation (2), after finding 2 solutions x1, x2 you can write it in the form: 4(x-3/2)(x+1)=0.

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G. Solve quadratic equations containing parameters

1. The equation has a solution \Leftrightarrow \Delta \geq 0

2. Equation without solution \Leftrightarrow \Delta < 0

3. Equation with unique solution (Double solution or two equal solutions) \Leftrightarrow \Delta = 0

4. The equation has two distinct (different) solutions. \Leftrightarrow \Delta > 0″ width=”64″ height=”12″ data-type=”2″ data-latex=”\Leftrightarrow \Delta > 0″ class=”lazy” data-src=”https://tex.vdoc.vn?tex=%5CLeftrightarrow%20%5CDelta%20%3E%200″/></span></p>
<p>5. The equation has two solutions with the same sign <span class=\Leftrightarrow \left\{ \begin{matrix} \Delta \geq 0 \\ P > 0 \\ \end{matrix} \right.” width=”76″ height=”39″ data-type=”2″ data-latex=”\Leftrightarrow \left\{ \begin{matrix}
\Delta \geq 0 \\
P > 0 \\
\end{matrix} \right.” class=”lazy” data-src=”https://tex.vdoc.vn?tex=%5CLeftrightarrow%20%5Cleft%5C%7B%20%5Cbegin%7Bmatrix%7D%0A%5CDelta%20%5Cgeq%200%20%5C%5C%0AP%20%3E%200%20%5C%5C%0A%5Cend%7Bmatrix%7D%20%5Cright.”/></span></p>
<p>6. The equation has two solutions of opposite sign <span class=\Leftrightarrow \left\{ \begin{matrix} \Delta > 0 \\ P < 0 \\ \end{matrix} \right.\ \Leftrightarrow ac < 0

7. The equation has two positive roots (Two roots greater than 0) \Leftrightarrow \left\{ \begin{matrix} \Delta \geq 0 \\ \begin{matrix} S > 0 \\ P > 0 \\ \end{matrix} \\ \end{matrix} \right.” width=”78″ height=”62″ data-type=”2″ data-latex=”\Leftrightarrow \left\{ \begin{matrix}
\Delta \geq 0 \\
\begin{matrix}
S > 0 \\
P > 0 \\
\end{matrix} \\
\end{matrix} \right.” class=”lazy” data-src=”https://tex.vdoc.vn?tex=%5CLeftrightarrow%20%5Cleft%5C%7B%20%5Cbegin%7Bmatrix%7D%0A%5CDelta%20%5Cgeq%200%20%5C%5C%0A%5Cbegin%7Bmatrix%7D%0AS%20%3E%200%20%5C%5C%0AP%20%3E%200%20%5C%5C%0A%5Cend%7Bmatrix%7D%20%5C%5C%0A%5Cend%7Bmatrix%7D%20%5Cright.”/></span></p>
<p>8. The equation has two negative roots (Two roots are less than 0) <span class=\Leftrightarrow \left\{ \begin{matrix} \Delta \geq 0 \\ \begin{matrix} S < 0 \\ P > 0 \\ \end{matrix} \\ \end{matrix} \right.” width=”78″ height=”62″ data-type=”2″ data-latex=”\Leftrightarrow \left\{ \begin{matrix}
\Delta \geq 0 \\
\begin{matrix}
S < 0 \\
P > 0 \\
\end{matrix} \\
\end{matrix} \right.” class=”lazy” data-src=”https://tex.vdoc.vn?tex=%5CLeftrightarrow%20%5Cleft%5C%7B%20%5Cbegin%7Bmatrix%7D%0A%5CDelta%20%5Cgeq%200%20%5C%5C%0A%5Cbegin%7Bmatrix%7D%0AS%20%3C%200%20%5C%5C%0AP%20%3E%200%20%5C%5C%0A%5Cend%7Bmatrix%7D%20%5C%5C%0A%5Cend%7Bmatrix%7D%20%5Cright.”/></span></p>
<p>9. The equation has two opposite solutions <span class=\Leftrightarrow \left\{ \begin{matrix} \Delta \geq 0 \\ S = 0 \\ \end{matrix} \right.

10. The two solutions are inverse \Leftrightarrow \left\{ \begin{matrix} \Delta \geq 0 \\ P = 1 \\ \end{matrix} \right.

Things to remember:\left\{ \begin{matrix}S = x_{1} + x_{2} = \dfrac{- b}{a} \\P = x_{1}.x_{2} = \dfrac{c}{ a} \\end{matrix} \right.

Along with quadratic equations, there is also Vi-et's theorem with many applications such as mentally calculating the roots of the quadratic equation mentioned above, finding 2 numbers when knowing the sum and product, determining the signs of the roots, or analyzing accumulate into factors. This is all necessary knowledge that will stick with you in the process of learning algebra, or later in solving and arguing for quadratic equations, so you need to remember it carefully and practice it fluently.

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If you intend to study programming, you also need basic math knowledge, or even in-depth math knowledge, depending on the project you will do.

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