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Sia

w44225592785's Question
Math
Posted 4 months ago

Solve the equations by completing the square.
You will need to work them out on paper.
Give exact answers for these questions:
x2+16x+5=0x214x+10=0x= or x=+ or x=
\begin{array}{l}
x^{2}+16 x+5=0 \\
x^{2}-14 x+10=0 \\
x= \\
\text { or } x= \\
-\sqrt{ } \\
+\sqrt{ } \\
\text { or } x= \\
-\sqrt{ }
\end{array}


Give your answers to 1 d.p1 \mathrm{~d} . \mathrm{p}. or better for these questions:
x25x=10x=
\begin{array}{l}
x^{2}-5 x=10 \\
x=\square
\end{array}

or x=x= \square
6x254x+30=0
6 x^{2}-54 x+30=0

x=x= \square or x=x= \square
5x260x+30=02x2+16x=5
5 x^{2}-60 x+30=0 \quad 2 x^{2}+16 x=5

x=x= \square or x=x= \square
x=x= \square or x=x= \square
[12]
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Answer from Sia
Verified answer
Posted 4 months ago
Solution by Steps
step 1
To solve the equation x2+16x+5=0x^2 + 16x + 5 = 0, we can complete the square. First, we rewrite it as x2+16x=5x^2 + 16x = -5
step 2
Next, we take half of the coefficient of xx, which is 88, and square it to get 6464. We add 6464 to both sides: x2+16x+64=59x^2 + 16x + 64 = 59
step 3
Now, we can factor the left side: (x+8)2=59(x + 8)^2 = 59. Taking the square root of both sides gives us x+8=±59x + 8 = \pm \sqrt{59}
step 4
Finally, we solve for xx: x=8±59x = -8 \pm \sqrt{59}
step 5
For the second equation x214x+10=0x^2 - 14x + 10 = 0, we rewrite it as x214x=10x^2 - 14x = -10
step 6
Taking half of the coefficient of xx, which is 7-7, and squaring it gives us 4949. We add 4949 to both sides: x214x+49=39x^2 - 14x + 49 = 39
step 7
Factoring gives us (x7)2=39(x - 7)^2 = 39. Taking the square root of both sides gives us x7=±39x - 7 = \pm \sqrt{39}
step 8
Finally, we solve for xx: x=7±39x = 7 \pm \sqrt{39}
step 9
For the equation x25x=10x^2 - 5x = 10, we rewrite it as x25x10=0x^2 - 5x - 10 = 0
step 10
Completing the square gives us x25x+254=654x^2 - 5x + \frac{25}{4} = \frac{65}{4}
step 11
Factoring gives us (x52)2=654(x - \frac{5}{2})^2 = \frac{65}{4}. Taking the square root gives us x52=±652x - \frac{5}{2} = \pm \frac{\sqrt{65}}{2}
step 12
Finally, we solve for xx: x=5±652x = \frac{5 \pm \sqrt{65}}{2}
step 13
For the equation 6x254x+30=06x^2 - 54x + 30 = 0, we divide by 6 to simplify: x29x+5=0x^2 - 9x + 5 = 0
step 14
Completing the square gives us x29x+814=894x^2 - 9x + \frac{81}{4} = \frac{89}{4}
step 15
Factoring gives us (x92)2=894(x - \frac{9}{2})^2 = \frac{89}{4}. Taking the square root gives us x92=±892x - \frac{9}{2} = \pm \frac{\sqrt{89}}{2}
step 16
Finally, we solve for xx: x=9±892x = \frac{9 \pm \sqrt{89}}{2}
step 17
For the equation 5x260x+30=05x^2 - 60x + 30 = 0, we divide by 5 to simplify: x212x+6=0x^2 - 12x + 6 = 0
step 18
Completing the square gives us x212x+36=30x^2 - 12x + 36 = 30
step 19
Factoring gives us (x6)2=30(x - 6)^2 = 30. Taking the square root gives us x6=±30x - 6 = \pm \sqrt{30}
step 20
Finally, we solve for xx: x=6±30x = 6 \pm \sqrt{30}
step 21
For the equation 2x2+16x=52x^2 + 16x = 5, we rewrite it as 2x2+16x5=02x^2 + 16x - 5 = 0
step 22
Completing the square gives us 2(x2+8x+16)=362(x^2 + 8x + 16) = 36
step 23
Factoring gives us 2(x+4)2=362(x + 4)^2 = 36. Taking the square root gives us x+4=±3x + 4 = \pm 3
step 24
Finally, we solve for xx: x=4±3x = -4 \pm 3
Answer
For the equations: 1. x=8±59x = -8 \pm \sqrt{59} 2. x=7±39x = 7 \pm \sqrt{39} 3. x=5±652x = \frac{5 \pm \sqrt{65}}{2} 4. x=9±892x = \frac{9 \pm \sqrt{89}}{2} 5. x=6±30x = 6 \pm \sqrt{30} 6. x=4±3x = -4 \pm 3
Key Concept
Completing the square is a method used to solve quadratic equations by rewriting them in a perfect square form.
Explanation
The solutions are derived by transforming the quadratic equations into a form that allows us to easily extract the roots. Each step involves manipulating the equation to isolate the variable.

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