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**Problem Solving Lesson 17: Sign of a quadratic triangle (Connection)**

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### Solve lesson 6.15, page 24, Math textbook 10, Connecting knowledge, volume 2

Consider the signs of the following quadratic triangles:

a) \(3x^{2}-4x+1\)

b) \(x^{2}+2x+1\)

c) \(-x^{2}+3x-2\)

d) \(-x^{2}+x-1\)

Solve lesson 6.15, page 24, Math textbook 10, Connecting knowledge, volume 2

Consider the signs of the following quadratic triangles:

a) \(3x^{2}-4x+1\)

b) \(x^{2}+2x+1\)

c) \(-x^{2}+3x-2\)

d) \(-x^{2}+x-1\)

**Solution method**

For a quadratic polynomial: \(f\left( x \right) = a{x^2} + bx + c\;\;\left( {a \ne 0} \right),\;\;\) \(\Delta = {b^2} – 4ac.\)

+) If \(\Delta < 0\) then \(f(x)\) is always the same sign as the coefficient \(a,\) for every \(x \in R.\)

+) If \(\Delta = 0\) then \(f(x)\) is always the same sign as the coefficient \(a,\) unless \(x=-\frac{b}{2a}.\)

+) If \(\Delta > 0\) then \(f(x)\) always has the same sign as the coefficient \(a\) when \(x < x_1\) or \(x > x_2,\) has opposite signs with the coefficient \(a\) when \(x_1 < x < x_2\) where \(x_1, \, \, x_2 \, \, (x_1 < x_2)\) are two solutions of \(f(x) .\)

**Detailed explanation**

a) \(f(x) = 3x^{2}-4x+1\), \(\Delta >0, a>0\), has 2 distinct solutions of 1 and \(\frac{1 respectively. }{3}\)

Marking table:

So f(x) > 0 for all \(x\in \left ( -\infty ;\frac{1}{3} \right )\cup \left ( 1;+\infty \right )\) and f( x) < 0 for all \(\left ( \frac{1}{3};1 \right )\)

b) \(f(x)=x^{2}+2x+1\), \(\Delta =0, a>0\), has a double root x = -1.

So f(x) > 0 for all \(x \neq -1\).

c) \(f(x)=-x^{2}+3x-2\), \(\Delta >0, a<0\), has 2 distinct solutions 1 and 2 respectively.

Marking table:

So f(x) < 0 for all \(x\in \left ( -\infty ;1 \right )\cup \left ( 2;+\infty \right )\) and f(x) > 0 for all \ (\left ( 1,2 \right )\)

d) \(f(x)=-x^{2}+x-1\), \(\Delta <0, a<0\). It follows that f(x) is always negative for all real numbers x.

### Solve lesson 6.16 page 24 Math textbook 10 Connecting knowledge volume 2

Solve the quadratic inequalities:

a) \(x^{2}-1\geq 0\)

b) \(x^{2}-2x-1<0\)

c) \(-3x^{2}+12x+10\leq 0\)

d) \(5x^{2}+x+1\geq 0\)

Solve lesson 6.16 page 24 Math textbook 10 Connecting knowledge volume 2

Solve the quadratic inequalities:

a) \(x^{2}-1\geq 0\)

b) \(x^{2}-2x-1<0\)

c) \(-3x^{2}+12x+10\leq 0\)

d) \(5x^{2}+x+1\geq 0\)

**Solution method**

Use the sign of first-order binomial and quadratic trinomial to solve inequalities.

For a quadratic polynomial: \(f\left( x \right) = a{x^2} + bx + c\;\;\left( {a \ne 0} \right),\;\;\) \(\Delta = {b^2} – 4ac.\)

+) If \(\Delta < 0\) then \(f(x)\) is always the same sign as the coefficient \(a,\) for every \(x \in R.\)

+) If \(\Delta = 0\) then \(f(x)\) is always the same sign as the coefficient \(a,\) unless \(x=-\frac{b}{2a}.\)

+) If \(\Delta > 0\) then \(f(x)\) always has the same sign as the coefficient \(a\) when \(x < x_1\) or \(x > x_2,\) has opposite signs with the coefficient \(a\) when \(x_1 < x < x_2\) where \(x_1, \, \, x_2 \, \, (x_1 < x_2)\) are two solutions of \(f(x) .\)

**Detailed explanation**

a) \(x^{2}-1\) has \(\Delta >0, a>0\), 2 distinct solutions of -1 and 1, respectively.

\(x^{2}-1\geq 0\) \(\Leftrightarrow x\in \left ( -\infty;-1 \right )\cup \left ( 1;+\infty \right )\)

So the solution set is S = \(\left ( -\infty;-1 \right )\cup \left ( 1;+\infty \right )\)

b) \(x^{2}-2x-1\) has \(\Delta =0, a>0\), the double root is x = -1, has \(x^{2}-2x-1> 0\) for all \(x \neq -1\)

So the equation \(x^{2}-2x-1<0\) has no solution.

So the inequality has no solution.

c) \(-3x^{2}+12x+10\) has \(\Delta >0, a<0\) 2 distinct solutions of \(\sqrt{\frac{13}{3}} +2\) and \(-\sqrt{\frac{13}{3}}+2\)

\(-3x^{2}+12x+10\leq 0\) \(\Leftrightarrow x\in \left ( -\infty; \sqrt{\frac{13}{3}}+2 \right ]\cup \left [\sqrt{\frac{13}{3}}+2 ;+\infty \right )\)

Vậy tập nghiệm là S = \(\left ( -\infty; \sqrt{\frac{13}{3}}+2 \right ]\cup \left[\sqrt{\frac{13}{3}}+2;+\infty\right)\)[\sqrt{\frac{13}{3}}+2;+\infty \right)\)

d) \(5x^{2}+x+1\) has \(\Delta <0, a>0\) so \(5x^{2}+x+1 >0\) for all real numbers x.

### Solve lesson 6.17, page 24, Math textbook 10, Connecting knowledge, volume 2

Find the values of the parameter m so that the following quadratic triangle is positive for all \(x\in \mathbb{R}\).

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\(x^{2}+(m+1)x+2m+3\)

Solve lesson 6.17, page 24, Math textbook 10, Connecting knowledge, volume 2

Find the values of the parameter m so that the following quadratic triangle is positive for all \(x\in \mathbb{R}\).

\(x^{2}+(m+1)x+2m+3\)

**Solution method**

For the quadratic triangle \(a{x^2} + bx + c\) to be positive for all \(x\in \mathbb{R}\) then \(\left\{ {\begin{array}{*{ 20}{l}}

{a > 0}\\

{\Delta < 0}

\end{array}} \right.\)

**Detailed explanation**

\(x^{2}+(m+1)x+2m+3>0\) for all \(x\in \mathbb{R}\)

\(\Leftrightarrow \left\{\begin{matrix}\Delta =(m+1)^{2}-4.(2m+3)<0\\ a=1>0\end{matrix}\right. \Leftrightarrow m^{2}-6m-11<0\)

\(\Leftrightarrow -2\sqrt{5}+3

### Solve lesson 6.17, page 24, Math textbook 10, Connecting knowledge, volume 2

Find the values of the parameter m so that the following quadratic triangle is positive for all \(x\in \mathbb{R}\).

\(x^{2}+(m+1)x+2m+3\)

Solve lesson 6.17, page 24, Math textbook 10, Connecting knowledge, volume 2

\(x^{2}+(m+1)x+2m+3\)

**Solution method**

For the quadratic triangle \(a{x^2} + bx + c\) to be positive for all \(x\in \mathbb{R}\) then \(\left\{ {\begin{array}{*{ 20}{l}}

{a > 0}\\

{\Delta < 0}

\end{array}} \right.\)

**Detailed explanation**

\(x^{2}+(m+1)x+2m+3>0\) for all \(x\in \mathbb{R}\)

\(\Leftrightarrow \left\{\begin{matrix}\Delta =(m+1)^{2}-4.(2m+3)<0\\ a=1>0\end{matrix}\right. \Leftrightarrow m^{2}-6m-11<0\)

\(\Leftrightarrow -2\sqrt{5}+3

### Solution 6.18 page 24 Math textbook 10 Connecting knowledge volume 2

An object is thrown vertically downward from a height of 320 m with an initial velocity v ._{Q }= 20m/s. After at least how many seconds, the object is not more than 100m from the ground? Assume that air resistance is negligible.

Solution 6.18 page 24 Math textbook 10 Connecting knowledge volume 2

An object is thrown vertically downward from a height of 320 m with an initial velocity v ._{Q }= 20m/s. After at least how many seconds, the object is not more than 100m from the ground? Assume that air resistance is negligible.

**Solution method**

We have the vertical Oy axis, the positive direction downward, the origin O at the throw point and the time origin at the time of the throw.

Infer y = \(v_{o}tg\frac{t^{2}}{2}=20t+5t^{2}\), calculate y when the object is 100m above the ground

Solve the inequality: \(20t+5t^{2}>220\) find the number of seconds for the object to be no more than 100m above the ground

**Detailed explanation**

Choose the vertical Oy axis, the positive direction downward, the origin O at the point of the throw and the time origin at the time of the throw.

there is y= \(v_{o}tg\frac{t^{2}}{2}=20t+5t^{2}\) , where g is the free acceleration, take g = 10

If the object is 100m from the ground, the distance traveled by the object is y = 320 – 100 = 220 m.

In order for the object to be no more than 100m from the ground, the distance y traveled by the object must be greater than 220.

We have the inequality: \(20t+5t^{2}>220\)

<=> \(5t^{2}+20t-220>0\)

<=> \(t>-2+4\sqrt{3}\approx 4.93\) or \(t<-2-4\sqrt{3}\approx -8.93\) (type)

So after at least 4.93 seconds, the object is not more than 100m from the ground.

### Solve lesson 6.19 page 24 Math textbook 10 Connecting knowledge volume 2

Considering a circle with diameter AB = 4 and a point M moving on segment AB, let AM = x. Consider two circles of diameter AM and MB. The symbol S(x) is the area of the plane part inside the larger circle and outside the two smaller circles. Determine the values of x so that the area S(x) does not exceed half the sum of the areas of the two small circles.

Solve lesson 6.19 page 24 Math textbook 10 Connecting knowledge volume 2

Considering a circle with diameter AB = 4 and a point M moving on segment AB, let AM = x. Consider two circles of diameter AM and MB. The symbol S(x) is the area of the plane part inside the larger circle and outside the two smaller circles. Determine the values of x so that the area S(x) does not exceed half the sum of the areas of the two small circles.

**Solution method**

Given that the radius of a circle with diameter AM is \(\frac{x}{2}\), the radius of a circle with diameter MB is \(\frac{4-x}{2}\).

Find the area of a circle with diameter AM(S_{first}), area of circle with diameter MB(S_{2}), the area of a circle with diameter AB(S).

Then calculate the area S(x) = S – S_{first} – WILL_{2}

**Detailed explanation**

+ AM = x, AB = 4 => MB = 4 -x, so radius of circle with diameter AM is \(\frac{x}{2}\), radius of circle diameter MB is \(\frac {4-x}{2}\).

+ The area of a circle with diameter AM is: \(S_{1}=\pi \frac{x^{2}}{4}\).

The area of a circle with diameter MB is: \(S_{2}=\pi \frac{(4-x)^{2}}{4}\).

The area of a circle with diameter AB is: \(S=\pi .16\).

+ Area S(x) = \(\pi .16- \pi \frac{x^{2}}{4}-\pi \frac{(4-x)^{2}}{4}\) = \(\pi \frac{-2x^{2}+8x+48}{4}\)

+ According to the topic \(S(x) \leq \frac{1}{2}(S_{1}+S_{2})\)

\(\Leftrightarrow \) \(\pi \frac{-2x^{2}+8x+48}{4}\leq \frac{1}{2}(\pi \frac{x^{2}}{ 4} +\pi \frac{(4-x)^{2}}{4})\)

\(\Leftrightarrow \) \(-2x^{2}+8x+48 \leq \frac{1}{2}(x^{2}+(4-x)^{2}\)

\(\Leftrightarrow \) \(-2x^{2}+8x+48 \leq x^{2}-x+8\)

\(\Leftrightarrow \) \(-2.45 \leq x \leq 5.45\)

Since x > 0, we have: \(0 < x \leq 5,45\)

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