Find the real roots of the equation. X²-2x+5=0
1. Find the real roots of the equation. X²-2x+5=0
x² - 2x + 5 = 0
a = 1; b = -2; c = 5
.............................................
D = b² - 4ac
D = ( - 2)² - 4 . 1 . 5
D = 4 - 20
D = -16
Karena D < 0, maka akar-akarnya tidak real,
HP = { }
***************************************
Kelas 10
Pelajaran Matematika
Bab 2 Persamaan dan Fungsi Kuadrat
Kata kunci : -
Kode kategorisasi : 10.2.2
2. Find the real roots of the equation. X²-2x+5=0
There is no real roots in this equationd
Why?
Because D=b^2-4ac=4-20=-16<0tidak ada akar akar yang real karena Diskriminan <0
b2-4ac = (-2)2 - 4(1)(5)=-16 (-)
3. If the equation 4x2 - 12x + c = 0, where c is a constant, has two equal roots, find the value of c and the roots of the equation.
Jawaban:
(2x-3)(2x-3)
4x²-6x-6x+9
4x²-12x+9
c=9
4. Given the equation x2 - 3x + m = 0. The sum of cubeof its roots is 189. Find the value of m! Butuh respon cepat dan jawaban yang lengkap!!!
x^2-3x+m=0
A=1; B=-3; C=m
x1+x2=-B/A=3
x1*x2=C/A=m
x1^3+x2^3=(x1+x2)^3-3x1*x2(x1+x2)
=(3)^3-3(m)(3)
189=27-9m
9m=-162
m=-18
5. If X1 and x2 is quadratic equation roots x2 - x+2 = 0. the newquadratic equation that has roots 2x1-2 and 2x2 - 2 is .....
Penjelasan dengan langkah-langkah:
Itu ya de jawabannya. Mohon maaf kalau kurang jelas
6. Find the solution roots of the form 2x² + 7x + 3
Jawaban:
QUADRATICEQUATIONFind the solution roots of the form 2x² + 7x + 3
→{-1/2, -3}————————————————
SOLUTION![tex]2x + 7x + 3 = 0[/tex]
[tex]2x + 7x = - 3[/tex]
[tex] {x}^{2} + \frac{7}{2}x = - \frac{3}{2} [/tex]
[tex] {x}^{2} + \frac{7}{2}x + {( \frac{7}{4}) }^{2} = - \frac{3}{2} + \frac{49}{16} [/tex]
[tex] {(x + \frac{7}{4} })^{2} = \frac{ - 24 + 49}{16} [/tex]
[tex] {(x + \frac{7}{4} })^{2} = \frac{25}{16} [/tex]
[tex](x + \frac{7}{4}) = (tambah \: kurang) \sqrt{ \frac{25}{16} } [/tex]
[tex]x + \frac{7}{4} = (tambah \: kurang) \frac{5}{4} [/tex]
[tex] x_{1} = - \frac{7}{4} + \frac{5}{4} = - \frac{1}{2} [/tex]
[tex] x_{2} = - \frac{7}{4} - \frac{5}{4} = - 3 [/tex]
[tex] x_{1} = - \frac{1}{2} \: or \: x_{2} = - 3[/tex]
So,thesolutionrootsis{-1/2,-3}NOTE:Tambah kurang means '±'
————————————————
EXPLANATIONBefore you study this material, you should also have studied the root form material in chapter 1, because this QUESTION uses a little root form.
••
A one-variable quadratic equation is an equation whose highest power is two. In general, the form of a quadratic equation is ax² + bx + c = 0 where a ≠ 0, a, b, c, € R. The constants a,b,c in this equation, are called coefficients. Some examples of quadratic equations are:
3x - 7x + 5 = 0, x² - x + 12 = 0, x² - 9 = 0, 2x(x - 7) = 0,and others.The root of the quadratic equation of ax² + bx + c = 0 is the value of x that satisfies the equation. There are 3 ways to determine the roots of a quadratic equation, namely:
FactoringComplete the perfect squareQuadratic formula (Abc formula)The abc formula, like this:
x1,2 = -b ± √b² - 4ac / 2aCharacteristics of a quadratic equation based on the coefficients of its quadratic equation:
If x1 and x2 are the roots of the quadratic equation ax² + bx + c = 0, then:
x1 + x2 = -b / a and x1 x2 = c/aFor example a quadratic equation ax² + bx + c = 0 with its discriminant value is D = b² - 4ac then for D < 0 the quadratic equation has no roots, D = 0 the quadratic equation has twin roots, D > 0 the quadratic equation has two different roots .
————————————————
CONCLUSIONBASED ON THE EXPLANATION ABOVE THEN,:
the solution roots of the form 2x² + 7x + 3is{-1/2, -3}————————————————
LEARNMOREABOUT QUADRATIC FUNCTION
diketahui fungsi f(x)=2x²+4x-10. nilai dari f(5) adalah :
https://brainly.co.id/tugas/40948283HERE ARE THE MATERIAL ABOUT ROOT FORM, CLICK THIS LINK
https://brainly.co.id/tugas/46756815https://brainly.co.id/tugas/42856863https://brainly.co.id/tugas/42657285————————————————
ANSWER DETAILSSubject : Mathematics (English)Class / Grade: 9Category : JHSMaterial / Sub-Category : Chapter 2 - Quadratic Equation.Subject Code : 2Categorization : 9.2.2Keyword : Quadratic Equation.
#OptiTimCompetitionJawaban:
x = - ½ , - 3
Penjelasan:
Find the solution roots of the form 2x² + 7x + 3
======================
2x² + 7x + 3
( 2x + 1 ) ( x + 3 )
x = - ½ , - 3
7. Find the value of m which the equation x^2+ (m – 3)x + 4 = 0 has two real roots.
• Persamaan Kuadrat
-
Nilai m agar x² + (m - 3)x + 4 = 0 memiliki 2 akar - akar real adalah m ≤ -1 atau m ≥ 7
• Pembahasan •
D ≥ 0 → Memiliki dua akar - akar reaL
x² + (m - 3)x + 4 = 0
a = 1
b = m - 3
c = 4
0 ≤ b² - 4ac
0 ≤ (m - 3)² - 4(1)(4)
0 ≤ m² - 6m + 9 - 16
0 ≤ m² - 6m - 7
0 ≤ (m + 1)(m - 7)
HP = { m ≤ -1 atau m ≥ 7 }
•••
-AL
Jawab:
m ≥ 7 atau m ≤ - 1
Penjelasan dengan langkah-langkah:
akan memiliki akar akar bilangan real jika D ≥ 0
bentuk umum persamaan kuadrat adalah ax² + bx + c = 0, maka dari soal diperoleh
a = 1
b = m - 3
c = 4
karena D ≥ 0, maka
D ≥ 0
b² - 4ac ≥ 0
(m-3)² - 4.1.4 ≥ 0
m² - 6m + 9 - 16 ≥ 0
m² - 6m - 7 ≥ 0
(m - 7)(m + 1) ≥ 0
maka didapat m ≥ 7 atau m ≤ -1
jadi, HP = {m | m ≥ 7 atau m ≤ - 1}
8. for what value of 'a' the roots of the equation 2x²+6x+a=0 ,satisfy the condition (alpha/beta)+beta/alpha)<2 {where alpha and beta are the roots of the equation}
A/B + B/A < 2
( A^2 + B^2 ) / AB < 2
{ ( A + B )^2 - 2AB } / AB < 2
{ (-6/2)^2 -2(a/2) } / a/2 < 2
( 9 - a ) / (a/2) < 2
2(9-a) < 2a
18- 2a < 2a
18 - 2a - 2a < 0
18 -4a < 0
-4a < -18
a > 9/2
9. What is the equation of axis of symmetry of the given quadratic equation? * y = x2 – 8x - 9
Jawaban:
x = 4
Penjelasan dengan langkah-langkah:
x = -b/2a
= -(-8)/2(1)
= 4
Jawaban:
x = 4
Penjelasan dengan langkah-langkah:
semoga membantu:)
10. The equation mx^2+nx+2n=8x+4 has the roots n and 1/m, m≠0. (a) Find the value of m and of n. (b) Hence, using the value of m and of n in (a),find the quadratic equation that has the roots 2m and - 1/n. .
Penjelasan dengan langkah-langkah:
It's known that:
f(x) → mx² + nx + 2n = 8x + 4
f(x) → mx² + nx - 8x + 2n - 4 = 0
f(x) → mx² + x(n-8) + (2n - 4), has roots:
x1 = n x2 = 1/m
it's also known that there is a formula for a quadratic equation f(x) → ax² + bx + c = 0 which has roots x1 and x2, so it can be applies:
x1 + x2 = -b/a
x1 × x2 = c/a
in the question:
a = m b = (n-8) c = (2n-4)
a. x1 + x2 = -b/a
n + 1/m = -((n-8)/m)
(nm+1)/m = -((n-8)/m)
nm+1 = -n+8
nm+n = 7 ..................... (1) 4m + 4 = 7 → m = ¾
x1 × x2 = c/a
n×1/m = (2n-4)/m
n = 2n-4
4 = n
b. x1 = 2m = 2.¾ = 3/2
x2 = -1/4
x1 + x2 = -b/a x1 × x2 = c/a
3/2 - ¼ = -b/a 3/2 × (-¼) = c/a
5/4 = -b/a -⅜ = c/a
10/8 = -b/a
b= -10
a = 8
c = -3
g(x) = 8x² - 10x - 3
11. If p = 2 is a solution of the equation 2p - hp +6=0, find the value of h. Hence, find the other solution for the equation.
remember p = 2
Then,
2p² - hp + 6 = 0
2(2)² - h(2) + 6 = 0
2(4) - 2h + 6 = 0
8 + 6 - 2h = 0
14 - 2h = 0
2h = 14
h = 7
so, we have solution h is a 7
12. The equation 2x2 - 13x + m = 0 has roots of p and q. If p : q = 5 : 8, find the value of m! Butuh respon yang cepat dan jawaban yang lengkap!!!!
2x^2 - 13 x + m = 0
akar-akarnya p dan q
p+q = 13/2
pq = m/2
dan p : q = 5 ; 8
misalkan p = 5n dan q = 8n
p+ q = 13/2 ---> 13n = 13/2 --> n = 1/2
pq = 5n . 8n = 40 n^2
m/2 = 40(1/2)^2
m = 2 (10)
m = 20
...
atau cara lain
p= 5n dan q = 8n
y1 = y2
2x^2 -13 x + m = a(x -5n)(x-8n)
2x^2 - 13x + m = a (x^2 - 13 n x + 40 n^2)
2( x^2 -13/2 x + m/2) = a (x^2 - 13 n x + 40 n^2)
13n = 13/2 --> n = 1/2
m/2 = 40 b^2 --> m = 80 n^2 = 80(1/2)^2 = 20
13. the roots of the quadratic equation 2x^2+px+1=0 are α and β.Given α^2+β^2= 8 find the possible values of p.
2x² + px + 1 = 0
a = 2
b = p
c = 1
α + β
= -b/a
= -p/2
α.β
= c/a
= 1/2
α² + β² = 8
(α + β)² - 2.α.β = 8
(-p/2)² - 2(1/2) = 8
p²/4 - 1 = 8
p²/4 = 8 + 1
p²/4 = 9
p² = 9 x 4
p² = 36
p = √36
p = ± 6
p = 6
or
p = -6
14. One of the roots of the equation 9x^2 - 9(k+2)x +9k^2 + 18k+11=0 is two times the other root. Find thevalues of k.
Jawab:
[tex]k_1=-\frac{3}{7}\\k_2=-1[/tex]
cara terlampir di gambar
semoga membantu,
jangan lupa jadikan jawaban terbaik :)
15. Given one of the roots of the quadratic equation px2- 4x + 3p -8 =0 is 1 . Calculate the value of p
Jawab:
p = 3
Penjelasan dengan langkah-langkah:
salah satu akar = 1
persamaan nya= p[tex]x^{2}[/tex]-4x+3p-8 = 0
masukkan angka 1 ke persamaan:
p[tex](1^{2})[/tex] -4(1) +3p -8 = 0
p-4+3p-8 = 0
4p = 12
p=3
16. Given one of the roots of the quadratic equation x 2 + kx – 3 = 0 is 3. Find the value of k.
Jawaban:
x²+kx-3=0
3²+3k-3=0
9+3k-3=0
3k+6=0
3k=-6
k=-2
barangkali ada yg ditanyalan dm dan follow ig @alwi_dj
17. Solve ! If x₁ and x₂ be the roots of x₂ - 3x + 3 = 0, find the quadratic equation whose roots are (x₁– 3) and (x₂– 3).
Jawaban:
x²+3x+3=0
Penjelasan dengan langkah-langkah:
[tex]pers \: kuadrat \: {x}^{2} - 3x + 3 = 0 \\ akar \: akar \: x1 \: dan \: x2 \\ \\ x1 + x2 = \frac{ - b}{a} = \frac{ - ( - 3)}{1} = 3 \\ \\ x1 \times x2 = \frac{c}{a} = \frac{3}{1} = 3 \\ \\ maka \\ pers \: .kuadrat \: yang \: akar \\ akarnya \: (x1 - 3) \: dan \: (x2 - 3) \\ \\ (x1 - 3) + (x2 - 3) \\ = x1 + x2 - 6 \\ = 3 - 6 \\ = - 3 \\ \\ (x1 - 3)(x2 - 3) \\ \\ = (x1.x2) - 3(x1 + x2) + 9 \\ \\ = (3)- 3(3) + 9 \\ \\ = 3 \\ \\ pers.kuadrat \: baru bentuk \: umum \: \\ {x}^{2} - (x1 + x2)x + (x1.x2) = 0 \\ \\ {x}^{2} - ( - 3)x + 3 = 0 \\ \\ {x}^{2} + 3x + 3 = 0[/tex]
Terimakasih...
18. The quadratics equation 2x^2 - 2(p - 4)x+p=0 has two distinct real roots, find the value of p.
Jawaban terlampir
Terimakasih
• Persamaan Kuadrat
-
Nilai p agar persamaan 2x² - 2(p - 4)x + p = 0 memiliki dua akar - akar real berbeda adalah {p | p < 2 atau p > 8 }
• Pembahasan •
2x² - 2(p - 4)x + p = 0
a = 2
b = -2(p - 4)
c = p
D > 0 → Dua akar real berbeda
0 < b² - 4ac
0 < (-2p + 8)² - 4(2)(p)
0 < 4p² - 32p + 64 - 8p
0 < 4p² - 40p + 64
0 < p² - 10p + 16
0 < (p - 2)(p - 8)
HP = {p | p < 2 atau p > 8 }
•••
-AL
19. Find the value of x of the following equation. x2+(x+2)2
Jawab:
(x + 2)² = (x + 2) (x + 2)
=x.x + x.2 + x.2 + 2.2
=x² + 2x + 2x + 4
=x² + 4x + 4
x² + x² + 4x + 4=2x² + 4x + 4
Penjelasan dengan langkah-langkah:
cari nlai x sendiri ya will aku juga bingung
20. The roots of the quadratic equation 3x²-17+45=25x-54 are......
• 3x² -17x + 45 = 25x - 54
• 3x² - 17x + 45 - 25x + 54 = 0
• 3x² - 17x - 25x + 45 + 54 = 0
• 3x² - 42x + 99 = 0
------------------------- : 3
• x² - 14x + 33 = 0
• (x-3)(x-11) = 0
~ x¹ → x-3 = 0
~ x¹ → x = 3
~ x¹ = 3
~ x² → x-11 = 0
~ x² → x = 11
~ x² = 11
The roots :
x1 = 3 & x2 = 11
→ x = { 3 , 11 }3x² - 17 + 45 = 25x - 54
3x² + 28 = 25x - 54
3x² - 25x + 82 = 0
Solve it by quadratic formula!
a = 3, b = -25, c = 82
x₁₂ = [-b +- √(b² - 4ac)] / (2a)
= {-(-25) +- √[(-25)² - 4.3.82]} / (2.3)
= (25 +- i√359) / 6
x₁ = 25/6 + (i√359) / 6 or
x₂ = 25/6 - (i√359) / 6
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