Step-by-step explanation:
factor 4 out of the variable terms, as this helps.
but my approach is simply to define the target and then calculate "backwards".
we want to find
(ax + b)² = a²x² + 2abx + b²
and now we compare with the original equation :
a²x² = 4x²
a² = 4
a = 2
2abx = 16x
2×2×bx = 16x
4b = 16
b = 4
b² = 16, but we have only 3, so we need to subtract 16-3 = 13 from the completed square.
so, our equation is
(2x + 4)² - 13 = 0
(2x + 4)² = 13
2x + 4 = sqrt(13)
2x = sqrt(13) - 4
x = sqrt(13)/2 - 2
Answer:
Isolate the constant.
Step-by-step explanation:
Given quadratic equation:
[tex]4x^2+16x+3=0[/tex]
Completing the Square
Step 1
Isolate the constant by subtracting 3 from both sides:
[tex]\implies 4x^2+16x+3-3=0-3[/tex]
[tex]\implies 4x^2+16x=-3[/tex]
Step 2
Factor out the coefficient of x² from the left side:
[tex]\implies 4(x^2+4x)=-3[/tex]
Divide both sides by 4:
[tex]\implies x^2+4x=-\dfrac{3}{4}[/tex]
Step 3
Add the square of half the coefficient of x to both sides. This forms a perfect square trinomial on the left side:
[tex]\implies x^2+4x+\left(\dfrac{4}{2}\right)^2=-\dfrac{3}{4}+\left(\dfrac{4}{2}\right)^2[/tex]
Simplify:
[tex]\implies x^2+4x+4=-\dfrac{3}{4}+4[/tex]
[tex]\implies x^2+4x+4=\dfrac{13}{4}[/tex]
Step 4
Factor the perfect square trinomial on the left side:
[tex]\implies (x+2)^2=\dfrac{13}{4}[/tex]
We have now completed the square.
To solve, square root both sides:
[tex]\implies \sqrt{(x+2)^2}=\sqrt{\dfrac{13}{4}}[/tex]
[tex]\implies x+2=\pm\sqrt{\dfrac{13}{4}}[/tex]
[tex]\implies x+2=\pm\dfrac{\sqrt{13}}{\sqrt{4}}[/tex]
[tex]\implies x+2=\pm\dfrac{\sqrt{13}}{2}[/tex]
Subtract 2 from both sides:
[tex]\implies x+2-2=-2\pm\dfrac{\sqrt{13}}{2}[/tex]
[tex]\implies x=-2\pm\dfrac{\sqrt{13}}{2}[/tex]
Rewrite -2 as -4/2:
[tex]\implies x=-\dfrac{4}{2}\pm\dfrac{\sqrt{13}}{2}[/tex]
[tex]\textsf{Apply the fraction rule}\quad \dfrac{a}{c} \pm \dfrac{b}{c}=\dfrac{a \pm b}{c}:[/tex]
[tex]\implies x=\dfrac{-4 \pm\sqrt{13}}{2}[/tex]
Learn more about completing the square here:
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