Solve the Quadratic Equation: 10x² - 65x + 135 = 4x² + 13x - 45

Quadratic Equations with Two-Sided Variables

Solve the following equation:

10x265x+135=4x2+13x45 10x^2-65x+135=4x^2+13x-45

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Find X
00:03 Arrange the equation so the right side equals 0
00:14 Group terms
00:29 Simplify as much as possible
00:40 Identify equation components
00:48 Use the roots formula
00:55 Substitute appropriate values and solve for X
01:15 Calculate products and squares
01:26 Calculate square root of 49
01:33 Find the 2 possible solutions
01:45 This is one solution
01:48 This is the second solution and the answer to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Solve the following equation:

10x265x+135=4x2+13x45 10x^2-65x+135=4x^2+13x-45

2

Step-by-step solution

To solve this problem, we'll follow these steps:

  • Step 1: Simplify the equation by gathering all terms on one side.
  • Step 2: Combine like terms to express the equation in standard quadratic form.
  • Step 3: Apply the quadratic formula to find the values of xx.

Now, let's work through each step:

Step 1: Starting with the equation 10x265x+135=4x2+13x4510x^2 - 65x + 135 = 4x^2 + 13x - 45, subtract 4x24x^2, 13x13x, and add 4545 on both sides:

10x265x+1354x213x+45=010x^2 - 65x + 135 - 4x^2 - 13x + 45 = 0

Step 2: Combine like terms:

(10x24x2)+(65x13x)+(135+45)=0(10x^2 - 4x^2) + (-65x - 13x) + (135 + 45) = 0

This simplifies to 6x278x+180=06x^2 - 78x + 180 = 0.

Step 3: Identify the coefficients a=6a = 6, b=78b = -78, and c=180c = 180. Use the quadratic formula:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substitute the values:

x=(78)±(78)24618026x = \frac{-(-78) \pm \sqrt{(-78)^2 - 4 \cdot 6 \cdot 180}}{2 \cdot 6}

x=78±6084432012x = \frac{78 \pm \sqrt{6084 - 4320}}{12}

x=78±176412x = \frac{78 \pm \sqrt{1764}}{12}

x=78±4212x = \frac{78 \pm 42}{12}

Calculating the two possible values:

x1=78+4212=12012=10x_1 = \frac{78 + 42}{12} = \frac{120}{12} = 10

x2=784212=3612=3x_2 = \frac{78 - 42}{12} = \frac{36}{12} = 3

Therefore, the solutions to the equation are x1=10x_1 = 10 and x2=3x_2 = 3.

The correct answer according to the provided choices is x1=10x_1=10 and x2=3x_2=3, which corresponds to choice 4.

3

Final Answer

x1=10 x_1=10 x2=3 x_2=3

Key Points to Remember

Essential concepts to master this topic
  • Collection: Move all terms to one side before simplifying
  • Formula: Use x=b±b24ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} with a=6, b=-78, c=180
  • Check: Substitute x=10: 10(100)65(10)+135=4(100)+13(10)45=485 10(100) - 65(10) + 135 = 4(100) + 13(10) - 45 = 485

Common Mistakes

Avoid these frequent errors
  • Applying quadratic formula without collecting terms first
    Don't try to use the quadratic formula on 10x265x+135=4x2+13x45 10x^2 - 65x + 135 = 4x^2 + 13x - 45 directly = wrong coefficients! This gives incorrect a, b, c values and wrong solutions. Always move all terms to one side first to get standard form ax2+bx+c=0 ax^2 + bx + c = 0 .

Practice Quiz

Test your knowledge with interactive questions

a = coefficient of x²

b = coefficient of x

c = coefficient of the constant term


What is the value of \( c \) in the function \( y=-x^2+25x \)?

FAQ

Everything you need to know about this question

Why can't I just use the quadratic formula on both sides separately?

+

The quadratic formula only works on equations in the form ax2+bx+c=0 ax^2 + bx + c = 0 . You must collect all terms on one side first to identify the correct coefficients a, b, and c.

How do I know which terms to move to which side?

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It doesn't matter! You can move everything to the left side or right side. Just make sure one side equals zero when you're done. Moving terms changes their signs, so be careful!

What if I get a negative discriminant under the square root?

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A negative discriminant means no real solutions exist. In this problem, we got 1764=42 \sqrt{1764} = 42 , which is positive, so we have two real solutions.

Can I factor instead of using the quadratic formula?

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Yes! After getting 6x278x+180=0 6x^2 - 78x + 180 = 0 , you could factor out 6 first: 6(x213x+30)=0 6(x^2 - 13x + 30) = 0 , then factor x213x+30 x^2 - 13x + 30 if possible.

Why do we get two answers for quadratic equations?

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Quadratic equations represent parabolas, which can cross the x-axis at two points. Each crossing point gives us a solution. That's why we use the ± symbol in the quadratic formula!

How can I check both answers quickly?

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Substitute each value back into the original equation (not the simplified form). For x=10 and x=3, both sides should equal 485 and 0 respectively.

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