TRẦN HƯNG ĐẠO invites teachers to refer to the document Solving exercises in Math Textbook Grade 9, Volume 2, pages 49, 50 to see suggestions for solving exercises in Lesson 5: Condensed root formula in chapter 4 of Algebra 9.
The document is compiled with content that closely follows the textbook program on pages 49 and 50 of Grade 9 Math, volume 2. Thereby, students will know how to solve all the exercises of lesson 5, Chapter 4 in the Grade 9 Math textbook. Episode 2. Good luck with your studies.
Theory Compact solution formula
1. Compact solution formula
For Eq
+ If 0″ data-latex=”Delta ‘ >0″ class=”lazy” data-src=”https://external-content.duckduckgo.com/iu/?u=https://pgdphurieng.edu.vn/wp-content/uploads/2023/04/holder.png” title=”tex-vdoc_-111-8″/> then the equation has two distinct solutions:
+ If then the equation has a double solution
+ If then the equation has no solution.
2. Pay attention
– When a > 0 and Eq If there is no solution, then the expression 0″ data-latex=”a{x^2} + bx + c > 0″ class=”lazy” data-src=”https://external-content.duckduckgo.com/iu/?u=https://pgdphurieng.edu.vn/wp-content/uploads/2023/04/holder.png” title=”tex-vdoc_-117-6″/> for all values of x.
– If Eq If a < 0, you should change the signs of both sides of the equation to have a > 0, then it will be easier to solve.
– For missing quadratic equations It is faster to use direct solution.
Solve math exercises 9, page 49, volume 2Lesson 17 (page 49 Math Textbook Grade 9 Part 2)
Determine a, b’, c then use the condensed solution formula to solve the equations:
a) 4×2 + 4x + 1 = 0 ;
b) 13852×2 – 14x + 1 = 0;
c) 5×2 – 6x + 1 = 0;
d) -3×2 + 4√6.x + 4 = 0.
See suggested answers
a) Quadratic equation 4×2 + 4x + 1 = 0
Yes a = 4; b’ = 2; c = 1; Δ’ = (b’)2 – ac = 22 – 4.1 = 0
The equation with double solution is:
b) Equation 13852×2 – 14x + 1 = 0
Yes a = 13852; b’ = -7; c = 1; Δ’ = (b’)2 – ac = (-7)2 – 13852.1 = -13803 < 0
So the equation has no solution.
c) Quadratic equation 5×2 – 6x + 1 = 0
Yes: a = 5; b’ = -3; c = 1.; Δ’ = (b’)2 – ac = (-3)2 – 5.1 = 4 > 0
The equation has two distinct solutions:
We have:
Inferred 0″ data-latex=”Delta ‘ = {(2sqrt 6 )^2} – ( – 3).4 = 36 > 0″ class=”lazy” data-src=”https://external-content.duckduckgo.com/iu/?u=https://pgdphurieng.edu.vn/wp-content/uploads/2023/04/holder.png” title=”tex-vdoc_-125-5″/>
Therefore the equation has two distinct solutions:
Lesson 18 (page 49 Math Textbook Grade 9 Part 2)
Put the following equations into the form ax2 + 2b’x + c = 0 and solve them. Then, use a number table or calculator to write an approximate solution (round the result to the second decimal place):
a) 3×2 – 2x = x2 + 3;
b) (2x – √2)2 – 1 = (x + 1)(x – 1);
c) 3×2 + 3 = 2(x + 1);
d) 0.5x(x + 1) = (x – 1)2.
See suggested answers
a) 3×2 – 2x = x2 + 3
⇔ 3×2 – 2x – x2 – 3 = 0
⇔ 2×2 – 2x – 3 = 0
Yes a = 2; b’ = -1; c = -3; Δ’ = b’2 – ac = (-1)2 – 2.(-3) = 7 > 0
{x_2} = dfrac{1 – sqrt 7 }{2} approx – 0.82
tex-vdoc_-129-5
b) (2x – √2)2 – 1 = (x + 1)(x – 1);
⇔ 4×2 – 2.2x.√2 + 2 – 1 = x2 – 1
⇔ 4×2 – 2.2√2.x + 2 – 1 – x2 + 1 = 0
⇔ 3×2 – 2.2√2.x + 2 = 0
0″ data-latex=”{x_1} = dfrac{2sqrt 2 + sqrt 2 }{3} = sqrt 2 approx 1,41″ class=”lazy” data-src=”https://external-content.duckduckgo.com/iu/?u=https://pgdphurieng.edu.vn/wp-content/uploads/2023/04/holder.png” title=”Because Δ’ > 0, the equation has two distinct solutions:”/>
{x_2} = dfrac{2sqrt 2 – sqrt 2 }{3} = dfrac{sqrt 2 }{3} approx 0.47
tex-vdoc_-131-4
c) 3×2 + 3 = 2(x + 1)
⇔ 3×2 + 3 = 2x + 2
⇔ 3×2 + 3 – 2x – 2 = 0
⇔ 3×2 – 2x + 1 = 0
The equation has a = 3; b’ = -1; c = 1; Δ’ = b’2 – ac = (-1)2 – 3.1 = -2 < 0
So the equation has no solution.
d) 0.5x(x + 1) = (x – 1)2
⇔ 0.5×2 + 0.5x = x2 – 2x + 1
⇔ x2 – 2x + 1 – 0.5×2 – 0.5x = 0
⇔ 0.5×2 – 2.5x + 1 = 0
Rightarrow Delta ‘ = {( – 2.5)^2} – 1.2 = 4.25 > 0
{x_1} = 2.5 + sqrt {4.25} approx 4.56
tex-vdoc_-133-3
x2 ∼ 0.44
Lesson 19 (page 49 Math Textbook Grade 9 Part 2)
Quiz. Do you know why, when a > 0 and the equation ax2 + bx + c = 0 has no solution, then ax2 + bx + c > 0 for all values of x?
-dfrac{b^{2}-4ac}{4a} > 0
=aleft ( x + dfrac{b}{2a} right )^{2}+ {left(-dfrac{b^{2}-4ac}{4a}right)}
tex-vdoc_-139-3
tex-vdoc_-135-3
tex-vdoc_-140-3
tex-vdoc_-141-2for every x.
Solve math exercises 9, page 49, volume 2: Practice
Lesson 20 (page 49 Math Textbook Grade 9 Part 2)
Solve the equations:
a) 25×2 – 16 = 0;
b) 2×2 + 3 = 0;
c) 4.2×2 + 5.46x = 0;
d) 4×2 – 2√3.x = 1 – √3.
⇔ x = ±sqrt{dfrac{16}{25}} = ±dfrac{4}{5}
2{x^2} + {rm{ }}3{rm{ }} = {rm{ }}0
tex-vdoc_-146-2
c) 4.2{x^2} + {rm{ }}5.46x{rm{ }} = {rm{ }}0
Leftrightarrow left[matrix{x=0hfillcr21x+273=0hfillcr}rightLeftrightarrowleft[matrix{x=0hfillcrx=-13hfillcr}right
d) 4{x^2} – {rm{ }}2sqrt 3 x{rm{ }} = {rm{ }}1{rm{ }} – {rm{ }}sqrt 3
Leftrightarrow {rm{ }}4{x^2} – {rm{ }}2sqrt 3 x{rm{ }}-{rm{ }}1{rm{ }} + {rm{ }}sqrt 3 {rm{ } } = {rm{ }}0
a = 4, b’ = -sqrt{3}, c = -1 + sqrt{3}
Rightarrow sqrt {Delta ‘} {rm{ }} = {rm{ }}2{rm{ }} – {rm{ }}sqrt 3
{x_2} = dfrac{{ – b’ + sqrt {Delta ‘} }}{a} =dfrac{sqrt{3} +2 – sqrt{3}}{4} =dfrac{1}{2}
tex-vdoc_-158-2
Lesson 21 (page 49 Math Textbook Grade 9 Part 2)
b) dfrac{1}{12}{x^2} + dfrac{7 }{12}x = 19
tex-vdoc_-159-2
See suggested answers
a) x2 = 12x + 288
⇔ x2 – 12x – 288 = 0
{x_2} =dfrac{6+sqrt{324}}{1}=6+18=24.
b) dfrac{1}{12}{x^2} + dfrac{7 }{12}x = 19
tex-vdoc_-159-2
⇔ x2 + 7x = 228
⇔ x2 + 7x – 228 = 0
{x_1} =dfrac{ – 7 – 31}{2} = 12,
tex-vdoc_-163-2
So the equation has two solutions x1 = 12 and x2 = -19.
b) displaystyle – {{19} over 5}{x^2} – sqrt 7 x + 1890 = 0
tex-vdoc_-165-2
Rightarrow ac=15.(-2005) <0.
b) displaystyle – {{19} over 5}{x^2} – sqrt 7 x + 1890 = 0
Rightarrow ac=-dfrac{19}{5}.1890 <0.
tex-vdoc_-168-2
⇒ the given equation has two distinct solutions.
Lesson 23 (page 49 Math Textbook Grade 9 Part 2)
A helicopter’s radar tracks the car’s movement for 10 minutes, detecting that the car’s velocity v changes depending on time by the formula:
v = 3t2 -30t + 135
(t is in minutes, v is in km/h)
a) Calculate the speed of the car when t = 5 minutes.
b) Calculate the value of t when the car speed is 120km/h (round the result to the second decimal place).
See suggested answers
a) At t = 5, we have: v = 3.52 – 30.5 + 135 = 60 (km/h)
b) When v = 120 km/h
⇔ 3t2 – 30t + 135 = 120
⇔ 3t2 – 30t + 15 = 0
Rightarrow {t_1} = {rm{ }}5{rm{ }} + {rm{ }}2sqrt 5 {rm{ }} approx {rm{ }}9.47; , , {rm{ }}{t_2} = {rm{ }}5{rm{ }} – {rm{ }}2sqrt 5 {rm{ }} approx {rm{ }}0.53.
tex-vdoc_-169-2
Since the radar observes the car’s movement for 10 minutes, t1 and t2 are both satisfied.
So at t = 9.47 minutes or t = 0.53 minutes, the car’s speed is 120km/h.
Lesson 24 (page 49 Math Textbook Grade 9 Part 2)
Given the equation (unknown x) x2 – 2(m – 1)x + m2 = 0.
a) Calculate Δ’.
b) For what value of m does the equation have two distinct solutions? Is there a double solution? No solution.
See suggested answers
a) Equation x2 – 2(m – 1)x + m2 = 0 (1)
Yes a = 1; b’ = -(m – 1); c = m2
⇒ Δ’ = b’2 – ac = (1 – m)2 – 1.m2 = 1 – 2m + m2 – m2 = 1 – 2m.
frac{1}{2}
frac{1}{2}
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