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Sagot :
Answer:
x = 4, or x = 3
Step-by-step explanation:
given equation is:
x² - 7x + 12 = 0
steps to solve the equation by completing the square method are as follows:
subtract the equation by constant: 12
x² - 7x = -12
make the half of the xth term: 7/2
add the half of the xth term to the equation:
x² - 7x + (7/2)² = -12 + (7/2)²
(x - 7/2)² = -12 + 49/4
(x - 7/2)² = -48 + 49 / 4
(x - 7/2)² = 1/4
taking square root of both sides:
√(x - 7/2)² = √1/4
x - 7/2 = ± 1/2
x = ±1/2 + 7/2
x = 1/2 + 7/2 or x = -1/2 + 7/2
x = 1 + 7 / 2 or x = -1 + 7 / 2
x = 8/2 or x = 6/2
x = 4, or x = 3
Answer:
Either [tex]x = 3[/tex] or [tex]x = 4[/tex].
Step-by-step explanation:
In the square completion method for solving quadratic equations, the key step is to rewrite the equation in the form [tex](x + h)^{2} = k[/tex], so that taking the square root of both sides of the equation would eliminate the [tex]x^{2}[/tex] term. To do so, note that after applying binomial expansion to [tex](x + h)^{2}[/tex], the equation becomes:
[tex]x^{2} + (2\, h)\, x + h^{2} = k[/tex].
In this question, the quadratic equation is:
[tex]x^{2} - 7\, x + 12 = 0[/tex].
The first step is to ensure that the coefficient of [tex]x^{2}[/tex] is [tex]1[/tex], which is already the case in this equation. (Otherwise, divide both sides of the equation by this coefficient so that the coefficient of [tex]x^{2}[/tex] becomes [tex]1[/tex].)
Separate the constant [tex]12[/tex] from the variable terms:
[tex]x^{2} - 7\, x = (-12)[/tex]
Next, match the coefficient of the "[tex]x[/tex]" term in the two equations. In the given equation, this coefficient is [tex](-7)[/tex], whereas this coefficient is [tex]2\, h[/tex] in the reference equation. Hence:
[tex]2\, h = (-7)[/tex].
[tex]\displaystyle h = \left(-\frac{7}{2}\right)[/tex].
Rewrite the equation [tex]x^{2} - 7\, x + 12 = 0[/tex] accordingly:
[tex]x^{2} + 2\, h = (-12)[/tex].
[tex]\displaystyle x^{2} + 2\, \left(-\frac{7}{2}\right)\, x= (-12)[/tex].
This equation is currently missing the term [tex]h^{2}[/tex], which is required to complete a square. Add this term to both sides of the equation:
[tex]x^{2} + 2\, h + h^{2} = (-12) + h^{2}[/tex].
[tex]\displaystyle x^{2} + 2\, \left(-\frac{7}{2}\right)\, x + \left(-\frac{7}{2}\right)^{2}= (-12) + \left(-\frac{7}{2}\right)^{2}[/tex].
The left side of the equation is now a perfect square. Simplify this equation to obtain:
[tex](x + h)^{2} = (-12) + h^{2}[/tex].
[tex]\displaystyle \left(x + \left(- \frac{7}{2}\right)\right)^{2} = (-12) + \left(-\frac{7}{2}\right)^{2}[/tex].
Take the square root of both sides of this equation to eliminate the [tex]x^{2}[/tex] term. When taking the square root of the right side of the equation, the plus-minus ([tex]\pm[/tex]) sign is required to account for both roots of the equation.
[tex]\displaystyle \sqrt{\left(x + \left(- \frac{7}{2}\right)\right)^{2}} = \pm \sqrt{(-12) + \left(-\frac{7}{2}\right)^{2}}[/tex]
[tex]\displaystyle x + \left(- \frac{7}{2}\right) = \pm \sqrt{(-12) + \left(-\frac{7}{2}\right)^{2}}[/tex].
[tex]\displaystyle x + \left(- \frac{7}{2}\right) = \pm \frac{1}{2}[/tex].
[tex]\displaystyle x = \frac{7}{2} \pm \frac{1}{2}[/tex].
Hence, [tex]x = 4[/tex] and [tex]x = 3[/tex] are the two solutions to this quadratic equation.
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