/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 40 Solve. $$x^{2}+4 x=3$$... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve. $$x^{2}+4 x=3$$

Short Answer

Expert verified
The solutions are \( x = -2 + \sqrt{7} \) and \( x = -2 - \sqrt{7} \).

Step by step solution

01

Move All Terms to One Side

Start by moving all terms to one side of the equation to set it to zero. Subtract 3 from both sides:\[ x^2 + 4x - 3 = 0 \]
02

Factor the Quadratic Equation

Next, factor the quadratic equation. We need to find two numbers that multiply to -3 and add to 4. These numbers are 6 and -1. So, the equation can be factored as:\[ (x + 6)(x - 1) = 0 \]
03

Set Each Factor to Zero

Set each factor equal to zero and solve for x:\[ x + 6 = 0 \] \[ x - 1 = 0 \]Solving these equations gives:\[ x = -6 \]\[ x = 1 \]
04

Verify the Solutions

Plug each solution back into the original equation to verify:For \( x = -6 \):\[ (-6)^2 + 4(-6) = 36 - 24 = 12 eq 3 \]For \( x = 1 \):\[ 1^2 + 4(1) = 1 + 4 = 5 eq 3 \]Both solutions do not satisfy the original equation, indicating a mistake in our factorization.
05

Use the Quadratic Formula

Since the factorization was incorrect, apply the quadratic formula:\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \]For our equation \( x^2 + 4x - 3 \), \(a = 1\), \(b = 4\), and \(c = -3\) gives:\[ x = \frac{{-4 \pm \sqrt{{4^2 - 4(1)(-3)}}}}{2(1)} \]\[ x = \frac{{-4 \pm \sqrt{{16 + 12}}}}{2} \]\[ x = \frac{{-4 \pm \sqrt{28}}}{2} \]\[ x = \frac{{-4 \pm 2\sqrt{7}}}{2} \]\[ x = -2 \pm \sqrt{7} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

factoring quadratics
Quadratic equations often can be solved by factoring. Factoring is the process of breaking down an equation into simpler components called factors. These factors, when multiplied, give back the original equation.
For the quadratic equation \(x^{2}+4 x=3\), to start factoring, you first need to set the equation to zero. This means moving all terms to one side: \[ x^{2} + 4x - 3 = 0 \]
Next, you need to find two numbers that multiply to the constant term (-3) and add to the linear term (4). However, sometimes factoring directly may not yield correct solutions, as seen in the mistake in the steps.
That's where it becomes crucial to apply trial and error or use alternative methods like the quadratic formula to ensure correctness.
quadratic formula
When factoring doesn't work or seems complicated, the quadratic formula is a reliable method to find solutions to any quadratic equation. The quadratic formula is given by: \[ x = \frac{{-b \ pm \ \sqrt{{b^2 - 4ac}}}}{2a} \]
For our equation \( x^2 + 4x - 3 = 0 \), the coefficients are:
  • \( a = 1 \)
  • \( b = 4 \)
  • \( c = -3 \)
Substituting these into the quadratic formula:
  • \[ x = \frac{{-4 \ pm \ \sqrt{{4^2 - 4(1)(-3)}}}}{2(1)} \]
  • \[ x = \frac{{-4 \ pm \ \sqrt{{16 + 12}}}}{2} \]
  • \[ x = \frac{{-4 \ \ pm \ \sqrt{28}}}{2} \]
  • \[ x = \frac{{-4 \ pm \ 2\ \sqrt{7}}}{2} \]
  • \[ x = -2 \ \ pm \ \sqrt{7} \]
This means the solutions are \( x = -2 + \sqrt{7} \) and \( x = -2 - \sqrt{7} \).
checking solutions
Always verify your solutions to ensure accuracy. Plug the solutions back into the original equation to see if they hold true.
For instance, let's check the solutions \( x = -2 + \sqrt{7} \) and \( x = -2 - \sqrt{7} \) in the original equation \( x^2 + 4x = 3 \).
Plugging \( x = -2 + \sqrt{7} \) into the left-hand side:
  • \[ \left( -2 + \sqrt{7} \right)^2 + 4\left( -2 + \sqrt{7} \right) \]
  • \[ = (-2 + \sqrt{7})^2 + 4(-2 + \sqrt{7}) \]
  • \[ = 4 - 4\sqrt{7} + 7 - 8 + 4\sqrt{7} \]
  • \[ = 3 \]
Similarly, checking \( x = -2 - \sqrt{7} \):
  • \[ \left( -2 - \sqrt{7} \right)^2 + 4\left( -2 - \sqrt{7} \right) \]
  • \[ = (-2 - \sqrt{7})^2 + 4(-2 - \sqrt{7}) \]
  • \[ = 4 + 4\sqrt{7} + 7 - 8 - 4\sqrt{7} \]
  • \[ = 3 \]
Both solutions satisfy the equation, ensuring they are correct.

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