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Applications of the Discriminant>
Applications of the Discriminant 1Applications of the Discriminant 1
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Question 1 of 7
1. Question
Using the discriminant, find the nature of the roots of the function:`5x^2-6x+4=0`Hint
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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`5x^2-6x+4=0``a=5` `b=-6` `c=4``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` `=` `-44` This is a negative value, which means `Delta``<``0`Therefore, the function has No real rootsNo real roots -
Question 2 of 7
2. Question
Identify which values of `k` will make the function below have one real root`x^2-(k-8)x+4=0`Hint
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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`x^2-(k-8)x+4=0``a=1` `b=k-8` `c=4``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^2-16k+64-16` `=` `k^2-16k+48` Remember that for a function to have one real root, `Delta``=``0`Substitute the `Delta` computed previously, and then solve for `k``Delta` `=` `0` `k^2-16k+48` `=` `0` [insert cross method with two `k`’s on the right and `-4` & `-12` on the left]`(k-4)(k-12)` `=` `0` `k=4,12` `k=4,12` -
Question 3 of 7
3. Question
Identify which values of `k` will make the function below have two real roots`x^2-3kx+9=0`Hint
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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`x^2-3kx+9=0``a=1` `b=-3k` `c=9``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 `=` `9k^2-36` Remember that for a function to have two real roots, `Delta``>``0`Substitute the `Delta` computed previously, and then solve for `k``Delta` `>` `0` `9k^2-36` `>` `0` `9(k^2-4)` `>` `0` `9(k-2)(k+2)` `>` `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 `9k^2-36`Replace the `x` axis with `k` axis and draw an upward parabola since `9` is positive
Remember that `Delta` must be positiveTherefore, `k``<``-2` and `k``>``2``k``<``-2` and `k``>``2` -
Question 4 of 7
4. Question
Identify which values of `k` will make the function below have real roots`x^2+(k+2)x+4=0`Hint
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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`x^2+(k+2)x+4=0``a=1` `b=k+2` `c=4``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+4k+4-16` `=` `k^2+4k-12` A function that has real roots can have either one or two real roots, hence `Delta``≥``0`Substitute the `Delta` computed previously, and then solve for `k``Delta` `≥` `0` `k^2+4k-12` `≥` `0` 
`(k+6)(k-2)` `≥` `0` `k=-6` `k=2` To determine which region around `k=-6` and `k=2` would be included, plot these points and make a rough sketch of `k^2+4k-12`Replace the `x` axis with `k` axis and draw an upward parabola since `1` is positive
Remember that `Delta` must be positiveTherefore, `k``≤``-6` and `k``≥``2``k``≤``-6` and `k``≥``2` -
Question 5 of 7
5. Question
Identify which values of `k` will make the function below have one real root`kx^2-4x+k=0`Hint
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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 one real root, 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` `k=-2,2` -
Question 6 of 7
6. Question
Identify which values of `m` will make the function below have one real root`2x^2+mx+8=0`Hint
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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 one real root, 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=-8,8` `m=-8,8` -
Question 7 of 7
7. Question
For what value of `k` will the line `y=3x-k` be tangent to the parabola of `y=2x^2-x+3`- `k=` (-1)
Hint
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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
$$\Delta={\color{#9a00c7}{b}}^2-4\color{#00880A}{a}\color{#007DDC}{c}$$Equate the two functions and solve for `k` in such a way that the discriminant (`Delta`) will be `0`First, equate the two functions to create a single equation`y=2x^2-x+3``y=3x-k``2x^2-x+3` `=` `3x-k` `2x^2-x+3-3x+k` `=` `0` Move all values to the left `2x^2-4x+3+k` `=` `0` Next, compute for the discriminant`2x^2-4x+3+k``a=2` `b=-4` `c=3+k``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 `=` `16-24-8k` `=` `-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 `Delta=0`Substitute the `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`
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