Jawaban:
x² + 11x + 28 = 0
[tex]\begin{aligned}\sf x_{1,2} &= \sf \frac{ - b± \sqrt{ {b}^{2} - 4ac } }{2a} \\&= \sf \frac{ - ( 11)± \sqrt{ { 11}^{2} - 4( 1)( 28)} }{2( 1)} \\&= \sf \frac{ - 11± \sqrt{121 - 4( 28)} }{ 2} \\ &= \sf \frac{ - 11± \sqrt{121 -112} }{ 2} \\ &= \sf \frac{ - 11± \sqrt{9} }{ 2} \\ &= \sf \frac{ - 11±3}{ 2} \\ \\ \sf x_{1} &= \sf \frac{ - 11 + 3}{ 2} \\ &= \sf - \frac{8}{ 4} \\ &= \sf \red{ -2} \\ \\ \sf x_{2} &= \sf \frac{ - 11 - 3}{ 2} \\ &= \sf -\frac{ 14}{ 2} \\ \sf &= \sf \red{ -7} \end{aligned}[/tex]
[tex] \sf HP = \{ \red{-4,-7} \}[/tex]
'조슈아' (Svt)
1. x^2 + 11x + 28 = 0
a = 1 | b = 11 | c = 28
-11 ± √11^2 - 4.1.28/2.1
= -11 ± √121 - 112/2
= -11 ± √9/2
= -11 ± 3/2
x1 = -11 + 3/2
= -8/2
= -4
x2 = -11 - 3/2
= -14/2
= -7
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Jawaban:
Penyelesaian :
x² + 11x + 28 = 0
Rumus ABC
[tex]\begin{aligned}\sf x_{1,2} &= \sf \frac{ - b± \sqrt{ {b}^{2} - 4ac } }{2a} \\&= \sf \frac{ - ( 11)± \sqrt{ { 11}^{2} - 4( 1)( 28)} }{2( 1)} \\&= \sf \frac{ - 11± \sqrt{121 - 4( 28)} }{ 2} \\ &= \sf \frac{ - 11± \sqrt{121 -112} }{ 2} \\ &= \sf \frac{ - 11± \sqrt{9} }{ 2} \\ &= \sf \frac{ - 11±3}{ 2} \\ \\ \sf x_{1} &= \sf \frac{ - 11 + 3}{ 2} \\ &= \sf - \frac{8}{ 4} \\ &= \sf \red{ -2} \\ \\ \sf x_{2} &= \sf \frac{ - 11 - 3}{ 2} \\ &= \sf -\frac{ 14}{ 2} \\ \sf &= \sf \red{ -7} \end{aligned}[/tex]
[tex] \sf HP = \{ \red{-4,-7} \}[/tex]
'조슈아' (Svt)
1. x^2 + 11x + 28 = 0
a = 1 | b = 11 | c = 28
-11 ± √11^2 - 4.1.28/2.1
= -11 ± √121 - 112/2
= -11 ± √9/2
= -11 ± 3/2
x1 = -11 + 3/2
= -8/2
= -4
x2 = -11 - 3/2
= -14/2
= -7