If $3x^{2}-18x-15=0$ , what is the value of $x^{2}-6x$ ?

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Algebra 2

Evaluate expression when a complex numbers and a variable term is given

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Q. If

3x^{2}-18x-15=0

, what is the value of

x^{2}-6x

Factor the quadratic equation: Given the equation $3x^2 - 18x - 15 = 0$ , we want to find the value of $x^2 - 6x$ . To do this, we can first solve for $x$ by factoring or using the quadratic formula.
Factor by grouping: Let's try to factor the quadratic equation. We look for two numbers that multiply to $(3)(-15) = -45$ and add up to $-18$ . The numbers that satisfy these conditions are $-15$ and $-3$ .
Find possible solutions for x: We can write the quadratic equation as $3x^2 - 15x - 3x - 15 = 0$ . Now, we can factor by grouping.
Substitute x values: Grouping the terms, we get $(3x^2 - 15x) - (3x + 15) = 0$ . Factoring out the common factors, we have $3x(x - 5) - 3(x + 5) = 0$ .
Determine single value for $x^2 - 6x$ : Now, we can factor out a $3$ from both groups, getting $3(x - 5)(x + 5) = 0$ .
Determine single value for $x^2 - 6x$ : Now, we can factor out a $3$ from both groups, getting $3(x - 5)(x + 5) = 0$ .Setting each factor equal to zero gives us the possible solutions for $x$ : $x - 5 = 0$ or $x + 5 = 0$ . Therefore, $x = 5$ or $x = -5$ .
Determine single value for $x^2 - 6x$ : Now, we can factor out a $3$ from both groups, getting $3(x - 5)(x + 5) = 0$ .Setting each factor equal to zero gives us the possible solutions for $x$ : $x - 5 = 0$ or $x + 5 = 0$ . Therefore, $x = 5$ or $x = -5$ .Now that we have the values of $x$ , we can substitute them into the expression $x^2 - 6x$ to find the corresponding values.
Determine single value for $x^2 - 6x$ : Now, we can factor out a $3$ from both groups, getting $3(x - 5)(x + 5) = 0$ .Setting each factor equal to zero gives us the possible solutions for $x$ : $x - 5 = 0$ or $x + 5 = 0$ . Therefore, $x = 5$ or $x = -5$ .Now that we have the values of $x$ , we can substitute them into the expression $x^2 - 6x$ to find the corresponding values.First, let's substitute $x = 5$ into $x^2 - 6x$ : $3$ $2$ .
Determine single value for $x^2 - 6x$ : Now, we can factor out a $3$ from both groups, getting $3(x - 5)(x + 5) = 0$ .Setting each factor equal to zero gives us the possible solutions for $x$ : $x - 5 = 0$ or $x + 5 = 0$ . Therefore, $x = 5$ or $x = -5$ .Now that we have the values of $x$ , we can substitute them into the expression $x^2 - 6x$ to find the corresponding values.First, let's substitute $x = 5$ into $x^2 - 6x$ : $3$ $2$ .Next, let's substitute $x = -5$ into $x^2 - 6x$ : $3$ $5$ .
Determine single value for $x^2 - 6x$ : Now, we can factor out a $3$ from both groups, getting $3(x - 5)(x + 5) = 0$ .Setting each factor equal to zero gives us the possible solutions for $x$ : $x - 5 = 0$ or $x + 5 = 0$ . Therefore, $x = 5$ or $x = -5$ .Now that we have the values of $x$ , we can substitute them into the expression $x^2 - 6x$ to find the corresponding values.First, let's substitute $x = 5$ into $x^2 - 6x$ : $3$ $2$ .Next, let's substitute $x = -5$ into $x^2 - 6x$ : $3$ $5$ .We have two possible values for $x^2 - 6x$ , which are $3$ $7$ and $3$ $8$ . However, since the original equation is a quadratic, it should have a single value for $x^2 - 6x$ that satisfies the equation for both roots.
Determine single value for $x^2 - 6x$ : Now, we can factor out a $3$ from both groups, getting $3(x - 5)(x + 5) = 0$ .Setting each factor equal to zero gives us the possible solutions for $x$ : $x - 5 = 0$ or $x + 5 = 0$ . Therefore, $x = 5$ or $x = -5$ .Now that we have the values of $x$ , we can substitute them into the expression $x^2 - 6x$ to find the corresponding values.First, let's substitute $x = 5$ into $x^2 - 6x$ : $3$ $2$ .Next, let's substitute $x = -5$ into $x^2 - 6x$ : $3$ $5$ .We have two possible values for $x^2 - 6x$ , which are $3$ $7$ and $3$ $8$ . However, since the original equation is a quadratic, it should have a single value for $x^2 - 6x$ that satisfies the equation for both roots.To find the single value that works for both roots, we can use the original equation $3(x - 5)(x + 5) = 0$ $0$ and divide through by $3$ to simplify it to $3(x - 5)(x + 5) = 0$ $2$ .
Determine single value for $x^2 - 6x$ : Now, we can factor out a $3$ from both groups, getting $3(x - 5)(x + 5) = 0$ .Setting each factor equal to zero gives us the possible solutions for $x$ : $x - 5 = 0$ or $x + 5 = 0$ . Therefore, $x = 5$ or $x = -5$ .Now that we have the values of $x$ , we can substitute them into the expression $x^2 - 6x$ to find the corresponding values.First, let's substitute $x = 5$ into $x^2 - 6x$ : $3$ $2$ .Next, let's substitute $x = -5$ into $x^2 - 6x$ : $3$ $5$ .We have two possible values for $x^2 - 6x$ , which are $3$ $7$ and $3$ $8$ . However, since the original equation is a quadratic, it should have a single value for $x^2 - 6x$ that satisfies the equation for both roots.To find the single value that works for both roots, we can use the original equation $3(x - 5)(x + 5) = 0$ $0$ and divide through by $3$ to simplify it to $3(x - 5)(x + 5) = 0$ $2$ .Now, we can see that $x^2 - 6x$ is equal to $3(x - 5)(x + 5) = 0$ $4$ because $3(x - 5)(x + 5) = 0$ $2$ implies $3(x - 5)(x + 5) = 0$ $6$ .