Applications of the Discriminant 1

Question 1 of 7

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

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

$5x^2-6x+4=0$

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

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

$5x^2-6x+4=0$

$a = 5$ $a=5$ $b = - 6$ $b=-6$ $c = 4$ $c=4$

$Δ$ $Delta$	$=$ $=$	${\color{#9a00c7}{b}}^2-4\color{#00880A}{a}\color{#007DDC}{c}$	Discriminant Formula
	$=$ $=$	${\color{#9a00c7}{(-6)}}^2-4\color{#00880A}{(5)}\color{#007DDC}{(4)}$	Substitute values
	$=$ $=$	$36 - 80$ $36-80$
	$=$ $=$	$- 44$ $-44$

This is a negative value, which means

Δ

$Delta$

<

$<$

0

$0$

Therefore, the function has No real roots

No real roots

Question 2 of 7

Identify which values of

k

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

x^{2} - (k - 8) x + 4 = 0

$x^2-(k-8)x+4=0$

1. $k = - 4, 2$ $k=-4,2$
2. $k = 4, 12$ $k=4,12$
3. $k = - 12, - 4$ $k=-12,-4$
4. $k = 0$ $k=0$

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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} - (k - 8) x + 4 = 0

$x^2-(k-8)x+4=0$

$a = 1$ $a=1$ $b = k - 8$ $b=k-8$ $c = 4$ $c=4$

$Δ$ $Delta$	$=$ $=$	${\color{#9a00c7}{b}}^2-4\color{#00880A}{a}\color{#007DDC}{c}$	Discriminant Formula
	$=$ $=$	${\color{#9a00c7}{(k-8)}}^2-4\color{#00880A}{(1)}\color{#007DDC}{(4)}$	Substitute values
	$=$ $=$	$k^{2} - 16 k + 64 - 16$ $k^2-16k+64-16$
	$=$ $=$	$k^{2} - 16 k + 48$ $k^2-16k+48$

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

$k$

$Δ$ $Delta$	$=$ $=$	$0$ $0$
$k^{2} - 16 k + 48$ $k^2-16k+48$	$=$ $=$	$0$ $0$

[insert cross method with two

k

$k$ ’s on the right and

- 4

$-4$ &

- 12

$-12$ on the left]

$(k - 4) (k - 12)$ $(k-4)(k-12)$	$=$ $=$	$0$ $0$
$k = 4, 12$ $k=4,12$

k = 4, 12

$k=4,12$

Question 3 of 7

Identify which values of

k

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

x^{2} - 3 k x + 9 = 0

$x^2-3kx+9=0$

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

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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} - 3 k x + 9 = 0

$x^2-3kx+9=0$

$a = 1$ $a=1$ $b = - 3 k$ $b=-3k$ $c = 9$ $c=9$

$Δ$ $Delta$	$=$ $=$	${\color{#9a00c7}{b}}^2-4\color{#00880A}{a}\color{#007DDC}{c}$	Discriminant Formula
	$=$ $=$	${\color{#9a00c7}{(-3k)}}^2-4\color{#00880A}{(1)}\color{#007DDC}{(9)}$	Substitute values
	$=$ $=$	$9 k^{2} - 36$ $9k^2-36$

Remember that for a function to have two real roots,

Δ

$Delta$

>

$>$

0

$0$

Substitute the

Δ

$Delta$ computed previously, and then solve for

k

$k$

$Δ$ $Delta$	$>$ $>$	$0$ $0$
$9 k^{2} - 36$ $9k^2-36$	$>$ $>$	$0$ $0$
$9 (k^{2} - 4)$ $9(k^2-4)$	$>$ $>$	$0$ $0$
$9 (k - 2) (k + 2)$ $9(k-2)(k+2)$	$>$ $>$	$0$ $0$
$k = 2$ $k=2$		$k = - 2$ $k=-2$

To determine which region around

k = - 2

$k=-2$ and

k = 2

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

9 k^{2} - 36

$9k^2-36$

Replace the

x

$x$ axis with

k

$k$ axis and draw an upward parabola since

9

$9$ is positive

Remember that

Δ

$Delta$ must be positive

Therefore,

k

$k$

<

$<$

- 2

$-2$ and

k

$k$

>

$>$

2

$2$

k

$k$

<

$<$

- 2

$-2$ and

k

$k$

>

$>$

2

$2$

Question 4 of 7

Identify which values of

k

$k$ will make the function below have real roots

x^{2} + (k + 2) x + 4 = 0

$x^2+(k+2)x+4=0$

1. $k$ $k$ $\leq$ $≤$ $- 6$ $-6$ and $k$ $k$ $\geq$ $≥$ $2$ $2$
2. $k$ $k$ $\leq$ $≤$ $- 2$ $-2$ and $k$ $k$ $\geq$ $≥$ $6$ $6$
3. $- 6$ $-6$ $\leq$ $≤$ $k$ $k$ $\leq$ $≤$ $2$ $2$
4. $- 2$ $-2$ $\leq$ $≤$ $k$ $k$ $\leq$ $≤$ $6$ $6$

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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} + (k + 2) x + 4 = 0

$x^2+(k+2)x+4=0$

$a = 1$ $a=1$ $b = k + 2$ $b=k+2$ $c = 4$ $c=4$

$Δ$ $Delta$	$=$ $=$	${\color{#9a00c7}{b}}^2-4\color{#00880A}{a}\color{#007DDC}{c}$	Discriminant Formula
	$=$ $=$	${\color{#9a00c7}{(k+2)}}^2-4\color{#00880A}{(1)}\color{#007DDC}{(4)}$	Substitute values
	$=$ $=$	$k^{2} + 4 k + 4 - 16$ $k^2+4k+4-16$
	$=$ $=$	$k^{2} + 4 k - 12$ $k^2+4k-12$

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

k

$k$

$Δ$ $Delta$	$\geq$ $≥$	$0$ $0$
$k^{2} + 4 k - 12$ $k^2+4k-12$	$\geq$ $≥$	$0$ $0$

$(k + 6) (k - 2)$ $(k+6)(k-2)$	$\geq$ $≥$	$0$ $0$
$k = - 6$ $k=-6$		$k = 2$ $k=2$

To determine which region around

k = - 6

$k=-6$ and

k = 2

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

k^{2} + 4 k - 12

$k^2+4k-12$

Replace the

x

$x$ axis with

k

$k$ axis and draw an upward parabola since

1

$1$ is positive

Remember that

Δ

$Delta$ must be positive

Therefore,

k

$k$

\leq

$≤$

- 6

$-6$ and

k

$k$

\geq

$≥$

2

$2$

k

$k$

\leq

$≤$

- 6

$-6$ and

k

$k$

\geq

$≥$

2

$2$

Question 5 of 7

Identify which values of

k

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

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

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

1. $k = - 2, 2$ $k=-2,2$
2. $k = - 2$ $k=-2$
3. $k = 2$ $k=2$
4. $k = 0, 2$ $k=0,2$

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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 + k = 0

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

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

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

Remember that the function must have one real root, hence

Δ = 0

$Delta=0$

Substitute the

Δ

$Delta$ computed previously, and then solve for

k

$k$

$Δ$ $Delta$	$=$ $=$	$0$ $0$
$16 - 4 k^{2}$ $16-4k^2$	$=$ $=$	$0$ $0$
$4 (4 - k^{2})$ $4(4-k^2)$	$=$ $=$	$0$ $0$
$4 (2 - k) (2 + k)$ $4(2-k)(2+k)$	$=$ $=$	$0$ $0$
$k = - 2$ $k=-2$		$k = 2$ $k=2$

k = - 2, 2

$k=-2,2$

Question 6 of 7

Identify which values of

m

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

2 x^{2} + m x + 8 = 0

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

1. $m = 0$ $m=0$
2. $m = - 8$ $m=-8$
3. $m = 0, 8$ $m=0,8$
4. $m = - 8, 8$ $m=-8,8$

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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 + 8 = 0

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

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

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

Remember that the function must have one real root, hence

Δ = 0

$Delta=0$

Substitute the

Δ

$Delta$ computed previously, and then solve for

m

$m$

$Δ$ $Delta$	$=$ $=$	$0$ $0$
$m^{2} - 64$ $m^2-64$	$=$ $=$	$0$ $0$
$(m + 8) (m - 8)$ $(m+8)(m-8)$	$=$ $=$	$0$ $0$
$m = - 8, 8$ $m=-8,8$

m = - 8, 8

$m=-8,8$

Question 7 of 7

For what value of

k

$k$ will the line

y = 3 x - k

$y=3x-k$ be tangent to the parabola of

y = 2 x^{2} - x + 3

$y=2x^2-x+3$

$k =$ $k=$ (-1)

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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

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

Equate the two functions and solve for

k

$k$ in such a way that the discriminant (

Δ

$Delta$ ) will be

0

$0$

First, equate the two functions to create a single equation

y = 2 x^{2} - x + 3

$y=2x^2-x+3$

y = 3 x - k

$y=3x-k$

$2 x^{2} - x + 3$ $2x^2-x+3$	$=$ $=$	$3 x - k$ $3x-k$
$2 x^{2} - x + 3 - 3 x + k$ $2x^2-x+3-3x+k$	$=$ $=$	$0$ $0$	Move all values to the left
$2 x^{2} - 4 x + 3 + k$ $2x^2-4x+3+k$	$=$ $=$	$0$ $0$

Next, compute for the discriminant

2 x^{2} - 4 x + 3 + k

$2x^2-4x+3+k$

$a = 2$ $a=2$ $b = - 4$ $b=-4$ $c = 3 + k$ $c=3+k$

$Δ$ $Delta$	$=$ $=$	${\color{#9a00c7}{b}}^2-4\color{#00880A}{a}\color{#007DDC}{c}$	Discriminant Formula
	$=$ $=$	${\color{#9a00c7}{(-4)}}^2-4\color{#00880A}{(2)}\color{#007DDC}{(3+k)}$	Substitute values
	$=$ $=$	$16 - 24 - 8 k$ $16-24-8k$
	$=$ $=$	$- 8 - 8 k$ $-8-8k$

Remember that the line must be a tangent to the parabola which means they meet at one point and that the previous equation must have one root, hence

Δ = 0

$Delta=0$

Substitute the

Δ

$Delta$ computed previously, and then solve for

$k$

$Delta$	$=$	$0$
$-8-8k$	$=$	$0$
$-8-8k$ $+8$	$=$	$0$ $+8$	Add $8$ to both sides
$-8k$	$=$	$8$
$-8k$ $divide(-8)$	$=$	$8$ $divide(-8)$	Divide both sides by $-8$
$k$	$=$	$-1$

$k=-1$

Applications of the Discriminant 1

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Quiz summary

Information

1. Question

Hint

Well Done!

Discriminant Formula

2. Question

Hint

Keep Going!

Discriminant Formula

3. Question

Hint

Nice Job!

Discriminant Formula

4. Question

Hint

Correct!

Discriminant Formula

5. Question

Hint

Nice Job!

Discriminant Formula

6. Question

Hint

Fantastic!

Discriminant Formula

7. Question

Hint

Exceptional!

Discriminant Formula

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