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(2x-3)(x+4)=0
Let
x=a and
x=b be unique solutions to the given equation. What is the value of
-a-b ?

$(2 x-3)(x+4)=0$ $\newline$ Let $x=a$ and $x=b$ be unique solutions to the given equation. What is the value of $-a-b$ ?

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Solve a quadratic equation using the zero product property

Full solution

Q.

(2 x-3)(x+4)=0

\newline

Let

x=a

and

x=b

be unique solutions to the given equation. What is the value of

-a-b

?

Set Equation Equal: To find the values of $a$ and $b$ , we need to set each factor of the equation $(2x-3)(x+4)=0$ equal to zero and solve for $x$ . First, let's solve $2x - 3 = 0$ . $2x - 3 = 0$ Add $3$ to both sides: $2x = 3$ Divide both sides by $2$ : $x = \frac{3}{2}$ So, one solution is $x = \frac{3}{2}$ , which means $a = \frac{3}{2}$ .
Solve $2x - 3 = 0$ : Now, let's solve the second factor, $x + 4 = 0$ . $\newline$ $x + 4 = 0$ $\newline$ Subtract $4$ from both sides: $\newline$ $x = -4$ $\newline$ So, the other solution is $x = -4$ , which means $b = -4$ .
Solve $x + 4 = 0$ : We are asked to find the value of $-a - b$ . Now that we have $a = \frac{3}{2}$ and $b = -4$ , we can substitute these values into the expression. $\newline$ $-a - b = -\left(\frac{3}{2}\right) - (-4)$
Substitute Values: Now, let's perform the subtraction. $\newline$ -\frac{3}{2} - (-4) = -\frac{3}{2} + 4\(\newlineTo add these, we need a common denominator. The common denominator for \$2\) and \(1\) (implicit in the \(4\)) is \(2\).\(\newline\)\(-\frac{3}{2} + 4 = -\frac{3}{2} + \frac{8}{2}\)
Perform Subtraction: Now, add the fractions.\(\newline\)\(-\frac{3}{2} + \frac{8}{2} = \frac{8 - 3}{2}\)\(\newline\)\(= \frac{5}{2}\)\(\newline\)So, the value of \(-a - b\) is \(\frac{5}{2}\).

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