Applications of the Discriminant 2

Question 1 of 6

Using the discriminant, find the nature of the roots of the function:

x^{2} - 6 x - 4 = 0

$x^2-6x-4=0$

1. No real roots
2. One real root
3. Two real roots

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Nature of the Roots	Discriminant ( $Δ$ $Delta$ )
Two real roots	$Δ$ $Delta$ $>$ $>$ $0$ $0$
One real root	$Δ = 0$ $Delta=0$
No real roots	$Δ$ $Delta$ $<$ $<$ $0$ $0$

Discriminant Formula

Δ = b^{2} - 4 a c

$\Delta={\color{#9a00c7}{b}}^2-4\color{#00880A}{a}\color{#007DDC}{c}$

First, compute for the discriminant

x^{2} - 6 x - 4 = 0

$x^2-6x-4=0$

$a = 1$ $a=1$ $b = - 6$ $b=-6$ $c = - 4$ $c=-4$

$Δ$ $Delta$	$=$ $=$	${\color{#9a00c7}{b}}^2-4\color{#00880A}{a}\color{#007DDC}{c}$	Discriminant Formula
	$=$ $=$	${\color{#9a00c7}{(-6)}}^2-4\color{#00880A}{(1)}\color{#007DDC}{(-4)}$	Substitute values
	$=$ $=$	$36 + 16$ $36+16$
	$=$ $=$	$52$ $52$

This is a positive value, which means

Δ

$Delta$

>

$>$

0

$0$

Therefore, the function has Two real roots

Two real roots

Question 2 of 6

Identify which values of

k

$k$ will make the function below have no real roots

k x^{2} + 4 x + 1 = 0

$kx^2+4x+1=0$

1. $- 4$ $-4$ $<$ $<$ $k$ $k$ $<$ $<$ $4$ $4$
2. $k$ $k$ $<$ $<$ $0$ $0$
3. $k$ $k$ $<$ $<$ $4$ $4$
4. $k$ $k$ $>$ $>$ $4$ $4$

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Nature of the Roots	Discriminant ( $Δ$ $Delta$ )
Two real roots	$Δ$ $Delta$ $>$ $>$ $0$ $0$
One real root	$Δ = 0$ $Delta=0$
No real roots	$Δ$ $Delta$ $<$ $<$ $0$ $0$

Discriminant Formula

Δ = b^{2} - 4 a c

$\Delta={\color{#9a00c7}{b}}^2-4\color{#00880A}{a}\color{#007DDC}{c}$

First, compute for the discriminant

k x^{2} + 4 x + 1 = 0

$kx^2+4x+1=0$

$a = k$ $a=k$ $b = 4$ $b=4$ $c = 1$ $c=1$

$Δ$ $Delta$	$=$ $=$	${\color{#9a00c7}{b}}^2-4\color{#00880A}{a}\color{#007DDC}{c}$	Discriminant Formula
	$=$ $=$	${\color{#9a00c7}{4}}^2-4\color{#00880A}{(k)}\color{#007DDC}{(1)}$	Substitute values
	$=$ $=$	$16 - 4 k$ $16-4k$

Remember that for a function to have one real root,

Δ

$Delta$

<

$<$

0

$0$

Substitute the

Δ

$Delta$ computed previously, and then solve for

k

$k$

$Δ$ $Delta$	$<$ $<$	$0$ $0$
$16 - 4 k$ $16-4k$	$<$ $<$	$0$ $0$
$16 - 4 k$ $16-4k$ $- 16$ $-16$	$<$ $<$	$0$ $0$ $- 16$ $-16$	Subtract $16$ $16$ from both sides
$- 4 k$ $-4k$	$<$ $<$	$- 16$ $-16$
$- 4 k$ $-4k$ $\div - 4$ $divide-4$	$<$ $<$	$- 16$ $-16$ $\div - 4$ $divide-4$	Divide both sides by $- 4$ $-4$
$k$ $k$	$>$ $>$	$4$ $4$	An inequality flips when divided by a negative

k

$k$

>

$>$

4

$4$

Question 3 of 6

Identify which values of

m

$m$ will make the function below have real roots

x^{2} + (m + 2) x + 1 = 0

$x^2+(m+2)x+1=0$

1. $- 4$ $-4$ $\leq$ $≤$ $m$ $m$ $\leq$ $≤$ $0$ $0$
2. $m$ $m$ $\leq$ $≤$ $0$ $0$ and $m$ $m$ $\geq$ $≥$ $4$ $4$
3. $0$ $0$ $\leq$ $≤$ $m$ $m$ $\leq$ $≤$ $4$ $4$
4. $m$ $m$ $\leq$ $≤$ $- 4$ $-4$ and $m$ $m$ $\geq$ $≥$ $0$ $0$

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Nature of the Roots	Discriminant ( $Δ$ $Delta$ )
Two real roots	$Δ$ $Delta$ $>$ $>$ $0$ $0$
One real root	$Δ = 0$ $Delta=0$
No real roots	$Δ$ $Delta$ $<$ $<$ $0$ $0$

Discriminant Formula

Δ = b^{2} - 4 a c

$\Delta={\color{#9a00c7}{b}}^2-4\color{#00880A}{a}\color{#007DDC}{c}$

First, compute for the discriminant

x^{2} + (m + 2) x + 1 = 0

$x^2+(m+2)x+1=0$

$a = 1$ $a=1$ $b = m + 2$ $b=m+2$ $c = 1$ $c=1$

$Δ$ $Delta$	$=$ $=$	${\color{#9a00c7}{b}}^2-4\color{#00880A}{a}\color{#007DDC}{c}$	Discriminant Formula
	$=$ $=$	${\color{#9a00c7}{(m+2)}}^2-4\color{#00880A}{(1)}\color{#007DDC}{(1)}$	Substitute values
	$=$ $=$	$m^{2} + 4 m + 4 - 4$ $m^2+4m+4-4$
	$=$ $=$	$m^{2} + 4 m$ $m^2+4m$

A function that has real roots can have either one or two real roots, hence

Δ

$Delta$

\geq

$≥$

0

$0$

Substitute the

Δ

$Delta$ computed previously, and then solve for

m

$m$

$Δ$ $Delta$	$\geq$ $≥$	$0$ $0$
$m^{2} + 4 m$ $m^2+4m$	$\geq$ $≥$	$0$ $0$
$m (m + 4)$ $m(m+4)$	$\geq$ $≥$	$0$ $0$
$m = 0$ $m=0$		$m = - 4$ $m=-4$

To determine which region around

m = 0

$m=0$ and

m = - 4

$m=-4$ would be included, plot these points and make a rough sketch of

m^{2} + 4 m

$m^2+4m$

Replace the

x

$x$ axis with

m

$m$ axis and draw an upward parabola since

1

$1$ is positive

Remember that

Δ

$Delta$ must be positive

Therefore,

m

$m$

\leq

$≤$

- 4

$-4$ and

m

$m$

\geq

$≥$

0

$0$

m

$m$

\leq

$≤$

- 4

$-4$ and

m

$m$

\geq

$≥$

0

$0$

Question 4 of 6

Identify which values of

m

$m$ will make the function below have one real root

2 x^{2} + m x + m - 2 = 0

$2x^2+mx+m-2=0$

$m =$ $m=$ (4)

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Nature of the Roots	Discriminant ( $Δ$ $Delta$ )
Two real roots	$Δ$ $Delta$ $>$ $>$ $0$ $0$
One real root	$Δ = 0$ $Delta=0$
No real roots	$Δ$ $Delta$ $<$ $<$ $0$ $0$

Discriminant Formula

Δ = b^{2} - 4 a c

$\Delta={\color{#9a00c7}{b}}^2-4\color{#00880A}{a}\color{#007DDC}{c}$

First, compute for the discriminant

2 x^{2} + m x + m - 2 = 0

$2x^2+mx+m-2=0$

$a = 2$ $a=2$ $b = m$ $b=m$ $c = m - 2$ $c=m-2$

$Δ$ $Delta$	$=$ $=$	${\color{#9a00c7}{b}}^2-4\color{#00880A}{a}\color{#007DDC}{c}$	Discriminant Formula
	$=$ $=$	${\color{#9a00c7}{m}}^2-4\color{#00880A}{(2)}\color{#007DDC}{(m-2)}$	Substitute values
	$=$ $=$	$m^{2} - 8 m + 16$ $m^2-8m+16$

Remember that the function must have one real root, hence

Δ = 0

$Delta=0$

Substitute the

$Delta$ computed previously, and then solve for

$m$

$Delta$	$=$	$0$
$m^2-8m+16$	$=$	$0$

[insert cross method image with two

$m$ ’s on the left and two

$-4$ ’s on the right]

$(m-4)(m-4)$	$=$	$0$
$m=4$

$m=4$

Question 5 of 6

Identify which values of

$k$ will make the function below have no real roots

$kx^2-4x+k=0$

1. $k$ $<$ $0$ and $k$ $>$ $2$
2. $-2$ $<$ $k$ $<$ $2$
3. $0$ $<$ $k$ $<$ $2$
4. $k$ $<$ $-2$ and $k$ $>$ $2$

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Nature of the Roots	Discriminant ( $Delta$ )
Two real roots	$Delta$ $>$ $0$
One real root	$Delta=0$
No real roots	$Delta$ $<$ $0$

Discriminant Formula

$\Delta={\color{#9a00c7}{b}}^2-4\color{#00880A}{a}\color{#007DDC}{c}$

First, compute for the discriminant

$kx^2-4x+k=0$

$a=k$ $b=-4$ $c=k$

$Delta$	$=$	${\color{#9a00c7}{b}}^2-4\color{#00880A}{a}\color{#007DDC}{c}$	Discriminant Formula
	$=$	${\color{#9a00c7}{(-4)}}^2-4\color{#00880A}{(k)}\color{#007DDC}{(k)}$	Substitute values
	$=$	$16-4k^2$

Remember that the function must have no real roots, hence

$Delta$

$<$

$0$

Substitute the

$Delta$ computed previously, and then solve for

$k$

$Delta$	$<$	$0$
$16-4k^2$	$<$	$0$
$4(4-k^2)$	$<$	$0$
$4(2-k)(2+k)$	$<$	$0$
$k=-2$		$k=2$

To determine which region around

$k=-2$ and

$k=2$ would be included, plot these points and make a rough sketch of

$16-4k^2$

Replace the

$x$ axis with

$k$ axis and draw a downward parabola since

$-4$ is negative

Remember that

$Delta$ must be negative

Therefore,

$k$

$<$

$-2$ and

$k$

$>$

$2$

$k$

$<$

$-2$ and

$k$

$>$

$2$

Question 6 of 6

Identify which values of

$m$ will make the function below have two real roots

$2x^2+mx+8=0$

1. $m$ $<$ $-4$ and $m$ $>$ $8$
2. $0$ $<$ $m$ $<$ $4$
3. $m$ $<$ $-8$ and $m$ $>$ $8$
4. $-8$ $<$ $m$ $<$ $8$

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Nature of the Roots	Discriminant ( $Delta$ )
Two real roots	$Delta$ $>$ $0$
One real root	$Delta=0$
No real roots	$Delta$ $<$ $0$

Discriminant Formula

$\Delta={\color{#9a00c7}{b}}^2-4\color{#00880A}{a}\color{#007DDC}{c}$

First, compute for the discriminant

$2x^2+mx+8=0$

$a=2$ $b=m$ $c=8$

$Delta$	$=$	${\color{#9a00c7}{b}}^2-4\color{#00880A}{a}\color{#007DDC}{c}$	Discriminant Formula
	$=$	${\color{#9a00c7}{m}}^2-4\color{#00880A}{(2)}\color{#007DDC}{(8)}$	Substitute values
	$=$	$m^2-64$

Remember that the function must have two real roots, hence

$Delta$

$>$

$0$

Substitute the

$Delta$ computed previously, and then solve for

$m$

$Delta$	$>$	$0$
$m^2-64$	$>$	$0$
$(m+8)(m-8)$	$>$	$0$
$m=-8$		$m=8$

To determine which region around

$m=-8$ and

$k=8$ would be included, plot these points and make a rough sketch of

$m^2-64$

Replace the

$x$ axis with

$m$ axis and draw an upward parabola since

$1$ is positive

Remember that

$Delta$ must be positive

Therefore,

$m$

$<$

$-8$ and

$m$

$>$

$8$

$m$

$<$

$-8$ and

$m$

$>$

$8$

Applications of the Discriminant 2

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Quiz summary

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1. Question

Hint

Excellent!

Discriminant Formula

2. Question

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Great Work!

Discriminant Formula

3. Question

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Discriminant Formula

4. Question

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Well Done!

Discriminant Formula

5. Question

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Excellent!

Discriminant Formula

6. Question

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Keep Going!

Discriminant Formula

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