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Answer :
We start with the equation
[tex]$$
4c^2 - 17c = 15.
$$[/tex]
Step 1. Bring all terms to one side
Subtract 15 from both sides to obtain a quadratic equation in standard form:
[tex]$$
4c^2 - 17c - 15 = 0.
$$[/tex]
Step 2. Multiply the leading coefficient and the constant
Multiply the leading coefficient ([tex]$4$[/tex]) and the constant term ([tex]$-15$[/tex]) to get
[tex]$$
4 \times (-15) = -60.
$$[/tex]
Step 3. Find two numbers that multiply to [tex]$-60$[/tex] and add to [tex]$-17$[/tex]
We need two numbers whose product is [tex]$-60$[/tex] and whose sum is [tex]$-17$[/tex]. These numbers are [tex]$-20$[/tex] and [tex]$3$[/tex], since
[tex]$$
(-20) \times 3 = -60 \quad \text{and} \quad (-20) + 3 = -17.
$$[/tex]
Step 4. Rewrite the middle term
Express the middle term ([tex]$-17c$[/tex]) using [tex]$-20c$[/tex] and [tex]$3c$[/tex]:
[tex]$$
4c^2 - 20c + 3c - 15 = 0.
$$[/tex]
Step 5. Factor by grouping
Group the terms in pairs:
- First group: [tex]$4c^2 - 20c$[/tex]. Factor out [tex]$4c$[/tex]:
[tex]$$
4c^2 - 20c = 4c(c - 5).
$$[/tex]
- Second group: [tex]$3c - 15$[/tex]. Factor out [tex]$3$[/tex]:
[tex]$$
3c - 15 = 3(c - 5).
$$[/tex]
Both groups have a common factor of [tex]$(c - 5)$[/tex]. Factor it out:
[tex]$$
(4c + 3)(c - 5) = 0.
$$[/tex]
Step 6. Solve for [tex]$c$[/tex]
Set each factor equal to zero:
1. For the factor [tex]$4c + 3 = 0$[/tex]:
[tex]$$
4c = -3 \quad \Longrightarrow \quad c = -\frac{3}{4}.
$$[/tex]
2. For the factor [tex]$c - 5 = 0$[/tex]:
[tex]$$
c = 5.
$$[/tex]
Final Answer
The solutions to the equation [tex]$$4c^2 - 17c = 15$$[/tex] are
[tex]$$
c = -\frac{3}{4} \quad \text{and} \quad c = 5.
$$[/tex]
[tex]$$
4c^2 - 17c = 15.
$$[/tex]
Step 1. Bring all terms to one side
Subtract 15 from both sides to obtain a quadratic equation in standard form:
[tex]$$
4c^2 - 17c - 15 = 0.
$$[/tex]
Step 2. Multiply the leading coefficient and the constant
Multiply the leading coefficient ([tex]$4$[/tex]) and the constant term ([tex]$-15$[/tex]) to get
[tex]$$
4 \times (-15) = -60.
$$[/tex]
Step 3. Find two numbers that multiply to [tex]$-60$[/tex] and add to [tex]$-17$[/tex]
We need two numbers whose product is [tex]$-60$[/tex] and whose sum is [tex]$-17$[/tex]. These numbers are [tex]$-20$[/tex] and [tex]$3$[/tex], since
[tex]$$
(-20) \times 3 = -60 \quad \text{and} \quad (-20) + 3 = -17.
$$[/tex]
Step 4. Rewrite the middle term
Express the middle term ([tex]$-17c$[/tex]) using [tex]$-20c$[/tex] and [tex]$3c$[/tex]:
[tex]$$
4c^2 - 20c + 3c - 15 = 0.
$$[/tex]
Step 5. Factor by grouping
Group the terms in pairs:
- First group: [tex]$4c^2 - 20c$[/tex]. Factor out [tex]$4c$[/tex]:
[tex]$$
4c^2 - 20c = 4c(c - 5).
$$[/tex]
- Second group: [tex]$3c - 15$[/tex]. Factor out [tex]$3$[/tex]:
[tex]$$
3c - 15 = 3(c - 5).
$$[/tex]
Both groups have a common factor of [tex]$(c - 5)$[/tex]. Factor it out:
[tex]$$
(4c + 3)(c - 5) = 0.
$$[/tex]
Step 6. Solve for [tex]$c$[/tex]
Set each factor equal to zero:
1. For the factor [tex]$4c + 3 = 0$[/tex]:
[tex]$$
4c = -3 \quad \Longrightarrow \quad c = -\frac{3}{4}.
$$[/tex]
2. For the factor [tex]$c - 5 = 0$[/tex]:
[tex]$$
c = 5.
$$[/tex]
Final Answer
The solutions to the equation [tex]$$4c^2 - 17c = 15$$[/tex] are
[tex]$$
c = -\frac{3}{4} \quad \text{and} \quad c = 5.
$$[/tex]
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