Solve the equation by completing the square and then solving for x. 1. x^2-6x=122. 2x^2-24x=-20
1 answer:
Answer: (x - 3)² = 21
<u>Step-by-step explanation:</u>
x² - 6x + ____ = 12 + ____
↓ ↑ ↑
x² - 6x + <u> 9 </u> = 12 + <u> 9 </u>
(x - 3)² = 21
*******************************************************
Answer: (x - 6)² = 26
<u>Step-by-step explanation:</u>
2x² - 24x = -20
<u>÷2 </u> <u> ÷2 </u> <u> ÷2 </u>
x² - 12x = -10
x² - 12x + ____ = -10 + ____
↓ ↑ ↑
x² - 12x + <u> 36 </u> = -10 + <u> 36 </u>
(x - 6)² = 26
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Solution<span><span>
1. x =(12-√228)/2=6-√ 57 = -1.550</span><span>
2. x =(12+√228)/2=6+√ 57 = 13.550
H3LL0 this is How to get this^ is down there i suggest you take some notesL0L
</span></span>The first term is, <span> <span>x2</span> </span><span> its coefficient is </span><span> 1 </span>.
<span>The middle term is, </span><span> -12x </span><span> its coefficient is </span><span> -12 </span>.
<span>The last term, "the constant", is </span><span> -21
</span>
Multiply the coefficient of the first term by the constant <span> <span> 1</span> • -21 = -21
</span>
-7 + 3 = -4</span><span>
-3 + 7 = 4</span><span>
-1 + 21 = 20
</span></span>Find the Vertex of <span>y = x2-12x-21
</span>For any parabola,<span>Ax2+Bx+C,</span><span>the </span> x <span>-coordinate of the vertex is given by </span> -B/(2A) is 6.0000 <span>Plugging into the parabola formula </span> 6.0000. <span> for </span> x <span> we can calculate the </span> y -coordinate :<span>
</span><span> y = 1.0 * 6.00 * 6.00 - 12.0 * 6.00 - 21.0
</span><span>or </span> y = -57.000 <span>Root plot for : </span><span> y = x2-12x-21</span>
<span>Axis of Symmetry (dashed) </span> {x}={ 6.00}
<span>Vertex at </span> {x,y} = { 6.00,-57.00}
x -Intercepts (Roots) :
Root 1 at {x,y} = {-1.55, 0.00}
Root 2 at<span> {x,y} = {13.55, 0.00}
</span>
read the squar <span> 2.2 </span> Solving <span> x2-12x-21 = 0</span><span> by Completing The Square</span><span> .
</span><span>Add </span><span> 21 </span><span> to both side of the equation : </span>
<span> x2-12x = 21 </span>Now the clever bit:<span> Take the coefficient of </span> x <span>, which is </span><span> 12</span><span> , divide by two, giving </span><span> 6</span><span> , and finally square it giving </span><span> 36</span><span>
</span>Adding <span> 36</span> has completed the left hand side into a perfect square :
<span> <span>x2-12x+36</span> </span> =
(x-6) • (x-6) =
<span>(x-6)2
</span>Since
<span> x2-12x+36 = 57</span><span> and</span>
<span> x2-12x+36 = (x-6)2
</span>then, according to the law of transitivity,
<span> (x-6)2 = 57
</span>We'll refer to this Equation as <span> Eq. #2.2.1
</span>The <span>Square Root Principle </span><span>says that When two things are equal, their square roots are equal.
</span>the square root of <span><span>(x-6)2 </span> </span><span> is</span>
<span> <span>(x-6)2/2</span> =
<span>(x-6)1</span> =
<span>x-6
</span></span>Now, applying the Square Root Principle to Eq. #2.2.1 we get:
<span> x-6 = <span>√<span> 57
</span></span></span>Add <span> 6 </span><span> to both sides to obtain:</span>
<span> x = 6 + √<span> 57
</span></span>Since a square root has two values, one positive and the other negative
<span> x2 - 12x - 21 = 0</span>
has two solutions:
<span>x = 6 + √<span> 57 </span></span>
or
<span>x = 6 - √<span> 57
</span></span>using the Quadratic Formula
</span> <span>The prime factorization of </span> 228 is
<span> 2•2•3•19</span><span>
</span>To be able to remove something from under the radical, there have to be <span> 2 </span><span> instances of it because we are taking a </span>square<span> i.e.</span><span> second </span>root
<span>√<span> 228 </span> = √<span> 2•2•3•19 </span> =
± <span>2 </span>• √<span> 57 </span></span>
<span> √<span> 57 </span></span><span> , rounded to 4 decimal digits, is </span><span> <span> 7.5498
</span></span> So now we are looking at:
x = ( 12 ± 2 • 7.550 ) / 2
Two real solutions:
<span> x =(12+√228)/2=6+√<span> 57 </span>= 13.550 </span>
or:
<span> x =(12-√228)/2=6-√<span> 57 </span>= -1.550 </span>
1. x =(12-√228)/2=6-√ 57 = -1.550</span><span>
2. x =(12+√228)/2=6+√ 57 = 13.550
H3LL0 this is How to get this^ is down there i suggest you take some notesL0L
</span></span>The first term is, <span> <span>x2</span> </span><span> its coefficient is </span><span> 1 </span>.
<span>The middle term is, </span><span> -12x </span><span> its coefficient is </span><span> -12 </span>.
<span>The last term, "the constant", is </span><span> -21
</span>
Multiply the coefficient of the first term by the constant <span> <span> 1</span> • -21 = -21
</span>
Find two factors of -21 whose sum equals the coefficient of the middle term, which is <span> -12 </span>.
-7 + 3 = -4</span><span>
-3 + 7 = 4</span><span>
-1 + 21 = 20
</span></span>Find the Vertex of <span>y = x2-12x-21
</span>For any parabola,<span>Ax2+Bx+C,</span><span>the </span> x <span>-coordinate of the vertex is given by </span> -B/(2A) is 6.0000 <span>Plugging into the parabola formula </span> 6.0000. <span> for </span> x <span> we can calculate the </span> y -coordinate :<span>
</span><span> y = 1.0 * 6.00 * 6.00 - 12.0 * 6.00 - 21.0
</span><span>or </span> y = -57.000 <span>Root plot for : </span><span> y = x2-12x-21</span>
<span>Axis of Symmetry (dashed) </span> {x}={ 6.00}
<span>Vertex at </span> {x,y} = { 6.00,-57.00}
x -Intercepts (Roots) :
Root 1 at {x,y} = {-1.55, 0.00}
Root 2 at<span> {x,y} = {13.55, 0.00}
</span>
read the squar <span> 2.2 </span> Solving <span> x2-12x-21 = 0</span><span> by Completing The Square</span><span> .
</span><span>Add </span><span> 21 </span><span> to both side of the equation : </span>
<span> x2-12x = 21 </span>Now the clever bit:<span> Take the coefficient of </span> x <span>, which is </span><span> 12</span><span> , divide by two, giving </span><span> 6</span><span> , and finally square it giving </span><span> 36</span><span>
</span>Adding <span> 36</span> has completed the left hand side into a perfect square :
<span> <span>x2-12x+36</span> </span> =
(x-6) • (x-6) =
<span>(x-6)2
</span>Since
<span> x2-12x+36 = 57</span><span> and</span>
<span> x2-12x+36 = (x-6)2
</span>then, according to the law of transitivity,
<span> (x-6)2 = 57
</span>We'll refer to this Equation as <span> Eq. #2.2.1
</span>The <span>Square Root Principle </span><span>says that When two things are equal, their square roots are equal.
</span>the square root of <span><span>(x-6)2 </span> </span><span> is</span>
<span> <span>(x-6)2/2</span> =
<span>(x-6)1</span> =
<span>x-6
</span></span>Now, applying the Square Root Principle to Eq. #2.2.1 we get:
<span> x-6 = <span>√<span> 57
</span></span></span>Add <span> 6 </span><span> to both sides to obtain:</span>
<span> x = 6 + √<span> 57
</span></span>Since a square root has two values, one positive and the other negative
<span> x2 - 12x - 21 = 0</span>
has two solutions:
<span>x = 6 + √<span> 57 </span></span>
or
<span>x = 6 - √<span> 57
</span></span>using the Quadratic Formula
<span> 2.3 </span> Solving <span> x2-12x-21 = 0</span> by the Quadratic Formula<span> .
</span>According to the Quadratic Formula,<span> x ,</span> the solution for <span><span> Ax2+Bx+C </span> = <span>0 </span> </span>, where A, B and C are numbers, often called coefficients, is given by :
<span>
- B ± √ B2-4AC
x = ————————
2A
</span> In our case, <span>A = 1
B = -12
C = -21
</span>Accordingly, <span>B2 - 4AC =
144 - (-84) =
228
</span>Applying the quadratic formula :
<span>12 ± √<span> 228 </span>
x = ——————
</span>
</span> <span>The prime factorization of </span> 228 is
<span> 2•2•3•19</span><span>
</span>To be able to remove something from under the radical, there have to be <span> 2 </span><span> instances of it because we are taking a </span>square<span> i.e.</span><span> second </span>root
<span>√<span> 228 </span> = √<span> 2•2•3•19 </span> =
± <span>2 </span>• √<span> 57 </span></span>
<span> √<span> 57 </span></span><span> , rounded to 4 decimal digits, is </span><span> <span> 7.5498
</span></span> So now we are looking at:
x = ( 12 ± 2 • 7.550 ) / 2
Two real solutions:
<span> x =(12+√228)/2=6+√<span> 57 </span>= 13.550 </span>
or:
<span> x =(12-√228)/2=6-√<span> 57 </span>= -1.550 </span>