Answer :
To solve the quadratic equation [tex]\(2x^2 + x = 15\)[/tex], follow these steps:
1. Rewrite the equation in standard form:
[tex]\[ 2x^2 + x - 15 = 0 \][/tex]
2. Factor the quadratic equation:
We need to find two numbers that multiply to [tex]\(2 \cdot (-15) = -30\)[/tex] and add to [tex]\(1\)[/tex] (the coefficient of [tex]\(x\)[/tex]).
These two numbers are [tex]\(6\)[/tex] and [tex]\(-5\)[/tex], because:
[tex]\[ 6 \times (-5) = -30 \quad \text{and} \quad 6 + (-5) = 1 \][/tex]
Rewrite the middle term using these two numbers:
[tex]\[ 2x^2 + 6x - 5x - 15 = 0 \][/tex]
3. Group the terms and factor by grouping:
Group the first two terms and the last two terms:
[tex]\[ (2x^2 + 6x) + (-5x - 15) = 0 \][/tex]
Factor out the greatest common factor from each group:
[tex]\[ 2x(x + 3) - 5(x + 3) = 0 \][/tex]
Factor out the common binomial factor [tex]\((x + 3)\)[/tex]:
[tex]\[ (2x - 5)(x + 3) = 0 \][/tex]
4. Set each factor equal to zero and solve for [tex]\(x\)[/tex]:
[tex]\[ 2x - 5 = 0 \quad \text{or} \quad x + 3 = 0 \][/tex]
Solving each equation:
[tex]\[ 2x - 5 = 0 \implies 2x = 5 \implies x = \frac{5}{2} \][/tex]
[tex]\[ x + 3 = 0 \implies x = -3 \][/tex]
Thus, the solutions to the equation [tex]\(2x^2 + x = 15\)[/tex] are:
[tex]\[ x = \frac{5}{2} \quad \text{and} \quad x = -3 \][/tex]
Therefore, the correct option is:
[tex]\( x = \frac{5}{2} \)[/tex] and [tex]\( x = -3 \)[/tex]
1. Rewrite the equation in standard form:
[tex]\[ 2x^2 + x - 15 = 0 \][/tex]
2. Factor the quadratic equation:
We need to find two numbers that multiply to [tex]\(2 \cdot (-15) = -30\)[/tex] and add to [tex]\(1\)[/tex] (the coefficient of [tex]\(x\)[/tex]).
These two numbers are [tex]\(6\)[/tex] and [tex]\(-5\)[/tex], because:
[tex]\[ 6 \times (-5) = -30 \quad \text{and} \quad 6 + (-5) = 1 \][/tex]
Rewrite the middle term using these two numbers:
[tex]\[ 2x^2 + 6x - 5x - 15 = 0 \][/tex]
3. Group the terms and factor by grouping:
Group the first two terms and the last two terms:
[tex]\[ (2x^2 + 6x) + (-5x - 15) = 0 \][/tex]
Factor out the greatest common factor from each group:
[tex]\[ 2x(x + 3) - 5(x + 3) = 0 \][/tex]
Factor out the common binomial factor [tex]\((x + 3)\)[/tex]:
[tex]\[ (2x - 5)(x + 3) = 0 \][/tex]
4. Set each factor equal to zero and solve for [tex]\(x\)[/tex]:
[tex]\[ 2x - 5 = 0 \quad \text{or} \quad x + 3 = 0 \][/tex]
Solving each equation:
[tex]\[ 2x - 5 = 0 \implies 2x = 5 \implies x = \frac{5}{2} \][/tex]
[tex]\[ x + 3 = 0 \implies x = -3 \][/tex]
Thus, the solutions to the equation [tex]\(2x^2 + x = 15\)[/tex] are:
[tex]\[ x = \frac{5}{2} \quad \text{and} \quad x = -3 \][/tex]
Therefore, the correct option is:
[tex]\( x = \frac{5}{2} \)[/tex] and [tex]\( x = -3 \)[/tex]
