Math / Algebra

Questionc. $(2-5 \sqrt{6})(6+\sqrt{8})$ $3 \sqrt{3}+6 \sqrt{3}-5 \sqrt{3}-\sqrt[3]{3}$ $2 \sqrt{3}$ $\sqrt{3}$ .
23 $\qquad$
3. Determine the value of $k$ such that $f(x)=3 x^{2}-12 x+7+k$ , has only one zel [3]

Studdy Solution

STEP 1

1. The function $f(x) = 3x^2 - 12x + 7 + k$ is a quadratic function.
2. A quadratic function has only one zero when its discriminant is zero.
3. The discriminant of a quadratic function $ax^2 + bx + c$ is given by $b^2 - 4ac$ .

STEP 2

1. Identify the coefficients of the quadratic function.
2. Write the expression for the discriminant.
3. Set the discriminant equal to zero for the function to have only one zero.
4. Solve for $k$ .

STEP 3

Identify the coefficients of the quadratic function $f(x) = 3x^2 - 12x + 7 + k$ .
The coefficients are: - $a = 3$ - $b = -12$ - $c = 7 + k$

STEP 4

Write the expression for the discriminant of the quadratic function.
The discriminant $\Delta$ is given by:
$\Delta = b^2 - 4ac$
Substitute the coefficients:
$\Delta = (-12)^2 - 4 \cdot 3 \cdot (7 + k)$

STEP 5

Set the discriminant equal to zero for the function to have only one zero.
$(-12)^2 - 4 \cdot 3 \cdot (7 + k) = 0$
Simplify:
$144 - 12(7 + k) = 0$

STEP 6

Continue simplifying the equation:
$144 - 84 - 12k = 0$
$60 = 12k$

STEP 7

Solve for $k$ :
$k = \frac{60}{12}$
$k = 5$
The value of $k$ such that the function has only one zero is $\boxed{5}$ .

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Questionc. $(2-5 \sqrt{6})(6+\sqrt{8})$ $3 \sqrt{3}+6 \sqrt{3}-5 \sqrt{3}-\sqrt[3]{3}$ $2 \sqrt{3}$ $\sqrt{3}$ .
23 $\qquad$
3. Determine the value of $k$ such that $f(x)=3 x^{2}-12 x+7+k$ , has only one zel [3]

Studdy Solution

STEP 1

1. The function $f(x) = 3x^2 - 12x + 7 + k$ is a quadratic function.
2. A quadratic function has only one zero when its discriminant is zero.
3. The discriminant of a quadratic function $ax^2 + bx + c$ is given by $b^2 - 4ac$ .

STEP 2

1. Identify the coefficients of the quadratic function.
2. Write the expression for the discriminant.
3. Set the discriminant equal to zero for the function to have only one zero.
4. Solve for $k$ .

STEP 3

Identify the coefficients of the quadratic function $f(x) = 3x^2 - 12x + 7 + k$ .
The coefficients are: - $a = 3$ - $b = -12$ - $c = 7 + k$

STEP 4

Write the expression for the discriminant of the quadratic function.
The discriminant $\Delta$ is given by:
$\Delta = b^2 - 4ac$
Substitute the coefficients:
$\Delta = (-12)^2 - 4 \cdot 3 \cdot (7 + k)$

STEP 5

Set the discriminant equal to zero for the function to have only one zero.
$(-12)^2 - 4 \cdot 3 \cdot (7 + k) = 0$
Simplify:
$144 - 12(7 + k) = 0$

STEP 6

Continue simplifying the equation:
$144 - 84 - 12k = 0$
$60 = 12k$

STEP 7

Solve for $k$ :
$k = \frac{60}{12}$
$k = 5$
The value of $k$ such that the function has only one zero is $\boxed{5}$ .

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Studdy Solution

STEP 1

1. The function f(x)=3x2−12x+7+k f(x) = 3x^2 - 12x + 7 + k f(x)=3x2−12x+7+k is a quadratic function.2. A quadratic function has only one zero when its discriminant is zero.3. The discriminant of a quadratic function ax2+bx+c ax^2 + bx + c ax2+bx+c is given by b2−4ac b^2 - 4ac b2−4ac.

STEP 2

1. Identify the coefficients of the quadratic function.2. Write the expression for the discriminant.3. Set the discriminant equal to zero for the function to have only one zero.4. Solve for k k k.

STEP 3

Identify the coefficients of the quadratic function f(x)=3x2−12x+7+k f(x) = 3x^2 - 12x + 7 + k f(x)=3x2−12x+7+k.The coefficients are: - a=3 a = 3 a=3 - b=−12 b = -12 b=−12 - c=7+k c = 7 + k c=7+k

STEP 4

Write the expression for the discriminant of the quadratic function.The discriminant Δ \Delta Δ is given by:Δ=b2−4ac \Delta = b^2 - 4ac Δ=b2−4acSubstitute the coefficients:Δ=(−12)2−4⋅3⋅(7+k) \Delta = (-12)^2 - 4 \cdot 3 \cdot (7 + k) Δ=(−12)2−4⋅3⋅(7+k)

STEP 5

Set the discriminant equal to zero for the function to have only one zero.(−12)2−4⋅3⋅(7+k)=0 (-12)^2 - 4 \cdot 3 \cdot (7 + k) = 0 (−12)2−4⋅3⋅(7+k)=0Simplify:144−12(7+k)=0 144 - 12(7 + k) = 0 144−12(7+k)=0

STEP 6

Continue simplifying the equation:144−84−12k=0 144 - 84 - 12k = 0 144−84−12k=060=12k 60 = 12k 60=12k

STEP 7

Solve for k k k:k=6012 k = \frac{60}{12} k=1260​k=5 k = 5 k=5 The value of k k k such that the function has only one zero is 5 \boxed{5} 5​.

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1. The function $f(x) = 3x^2 - 12x + 7 + k$ is a quadratic function.
2. A quadratic function has only one zero when its discriminant is zero.
3. The discriminant of a quadratic function $ax^2 + bx + c$ is given by $b^2 - 4ac$ .

1. Identify the coefficients of the quadratic function.
2. Write the expression for the discriminant.
3. Set the discriminant equal to zero for the function to have only one zero.
4. Solve for $k$ .

Identify the coefficients of the quadratic function $f(x) = 3x^2 - 12x + 7 + k$ .
The coefficients are: - $a = 3$ - $b = -12$ - $c = 7 + k$

Write the expression for the discriminant of the quadratic function.
The discriminant $\Delta$ is given by:
$\Delta = b^2 - 4ac$
Substitute the coefficients:
$\Delta = (-12)^2 - 4 \cdot 3 \cdot (7 + k)$

Set the discriminant equal to zero for the function to have only one zero.
$(-12)^2 - 4 \cdot 3 \cdot (7 + k) = 0$
Simplify:
$144 - 12(7 + k) = 0$

Continue simplifying the equation:
$144 - 84 - 12k = 0$
$60 = 12k$

Solve for $k$ :
$k = \frac{60}{12}$
$k = 5$
The value of $k$ such that the function has only one zero is $\boxed{5}$ .