Solved on Jan 22, 2024

Solve $3(t^2 - 9)^2 + 16(t^2 - 9) = -5$ using substitution. Select the correct solution set.

STEP 1

Assumptions
1. We are given a quadratic equation in the form of $3(t^2-9)^2 + 16(t^2-9) = -5$ .
2. We will use substitution to solve the equation, by letting $u = t^2 - 9$ .
3. We will solve the resulting quadratic equation in $u$ .
4. We will then find the values of $t$ by reversing the substitution.

STEP 2

Let's perform the substitution by setting $u = t^2 - 9$ . The equation then becomes a quadratic equation in terms of $u$ .
$3u^2 + 16u = -5$

STEP 3

To solve the quadratic equation, we first move all terms to one side to set the equation to zero.
$3u^2 + 16u + 5 = 0$

STEP 4

Now we attempt to factor the quadratic equation, or if that is not possible, use the quadratic formula to find the values of $u$ . The quadratic formula is given by:
$u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
where $a$ , $b$ , and $c$ are the coefficients of $u^2$ , $u$ , and the constant term respectively.

STEP 5

In our equation $3u^2 + 16u + 5 = 0$ , we identify $a = 3$ , $b = 16$ , and $c = 5$ .

STEP 6

We calculate the discriminant $b^2 - 4ac$ to determine the nature of the roots.
$\Delta = b^2 - 4ac = 16^2 - 4 \cdot 3 \cdot 5$

STEP 7

Calculate the discriminant.
$\Delta = 256 - 60 = 196$

STEP 8

Since the discriminant is positive, we have two real and distinct roots. We use the quadratic formula to find the values of $u$ .
$u = \frac{-16 \pm \sqrt{196}}{2 \cdot 3}$

STEP 9

Simplify the square root of the discriminant and the expression.
$u = \frac{-16 \pm 14}{6}$

STEP 10

Calculate the two values for $u$ .
$u_1 = \frac{-16 + 14}{6} = \frac{-2}{6} = -\frac{1}{3}$
$u_2 = \frac{-16 - 14}{6} = \frac{-30}{6} = -5$

STEP 11

Now that we have the values for $u$ , we reverse the substitution to find the corresponding values of $t$ . Recall that $u = t^2 - 9$ .

STEP 12

For $u_1 = -\frac{1}{3}$ , we have:
$t^2 - 9 = -\frac{1}{3}$

STEP 13

Solve for $t^2$ .
$t^2 = 9 - \frac{1}{3}$

STEP 14

Combine the terms on the right-hand side.
$t^2 = \frac{27}{3} - \frac{1}{3} = \frac{26}{3}$

STEP 15

Take the square root of both sides to find $t$ .
$t = \pm \sqrt{\frac{26}{3}} = \pm \frac{\sqrt{78}}{3}$

STEP 16

For $u_2 = -5$ , we have:
$t^2 - 9 = -5$

STEP 17

Solve for $t^2$ .
$t^2 = 9 - 5 = 4$

STEP 18

Take the square root of both sides to find $t$ .
$t = \pm \sqrt{4} = \pm 2$

STEP 19

Combine the values of $t$ from both $u_1$ and $u_2$ to get the solution set for $t$ .
$t = \left\{ \pm \frac{\sqrt{78}}{3}, \pm 2 \right\}$

STEP 20

Match the solution set for $t$ with the given options.
The correct answer is: d. $\left\{ \pm \frac{\sqrt{78}}{3}, \pm 2\right\}$

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Solve 3(t2−9)2+16(t2−9)=−53(t^2 - 9)^2 + 16(t^2 - 9) = -53(t2−9)2+16(t2−9)=−5 using substitution. Select the correct solution set.

STEP 1

STEP 2

Let's perform the substitution by setting u=t2−9u = t^2 - 9u=t2−9. The equation then becomes a quadratic equation in terms of uuu.3u2+16u=−53u^2 + 16u = -53u2+16u=−5

STEP 3

To solve the quadratic equation, we first move all terms to one side to set the equation to zero.3u2+16u+5=03u^2 + 16u + 5 = 03u2+16u+5=0

STEP 4

STEP 5

In our equation 3u2+16u+5=03u^2 + 16u + 5 = 03u2+16u+5=0, we identify a=3a = 3a=3, b=16b = 16b=16, and c=5c = 5c=5.

STEP 6

We calculate the discriminant b2−4acb^2 - 4acb2−4ac to determine the nature of the roots.Δ=b2−4ac=162−4⋅3⋅5\Delta = b^2 - 4ac = 16^2 - 4 \cdot 3 \cdot 5Δ=b2−4ac=162−4⋅3⋅5

STEP 7

Calculate the discriminant.Δ=256−60=196\Delta = 256 - 60 = 196Δ=256−60=196

STEP 8

Since the discriminant is positive, we have two real and distinct roots. We use the quadratic formula to find the values of uuu.u=−16±1962⋅3u = \frac{-16 \pm \sqrt{196}}{2 \cdot 3}u=2⋅3−16±196​​

STEP 9

Simplify the square root of the discriminant and the expression.u=−16±146u = \frac{-16 \pm 14}{6}u=6−16±14​

STEP 10

Calculate the two values for uuu.u1=−16+146=−26=−13u_1 = \frac{-16 + 14}{6} = \frac{-2}{6} = -\frac{1}{3}u1​=6−16+14​=6−2​=−31​u2=−16−146=−306=−5u_2 = \frac{-16 - 14}{6} = \frac{-30}{6} = -5u2​=6−16−14​=6−30​=−5

STEP 11

Now that we have the values for uuu, we reverse the substitution to find the corresponding values of ttt. Recall that u=t2−9u = t^2 - 9u=t2−9.

STEP 12

For u1=−13u_1 = -\frac{1}{3}u1​=−31​, we have:t2−9=−13t^2 - 9 = -\frac{1}{3}t2−9=−31​

STEP 13

Solve for t2t^2t2.t2=9−13t^2 = 9 - \frac{1}{3}t2=9−31​

STEP 14

Combine the terms on the right-hand side.t2=273−13=263t^2 = \frac{27}{3} - \frac{1}{3} = \frac{26}{3}t2=327​−31​=326​

STEP 15

Take the square root of both sides to find ttt.t=±263=±783t = \pm \sqrt{\frac{26}{3}} = \pm \frac{\sqrt{78}}{3}t=±326​​=±378​​

STEP 16

For u2=−5u_2 = -5u2​=−5, we have:t2−9=−5t^2 - 9 = -5t2−9=−5

STEP 17

Solve for t2t^2t2.t2=9−5=4t^2 = 9 - 5 = 4t2=9−5=4

STEP 18

Take the square root of both sides to find ttt.t=±4=±2t = \pm \sqrt{4} = \pm 2t=±4​=±2

STEP 19

Combine the values of ttt from both u1u_1u1​ and u2u_2u2​ to get the solution set for ttt.t={±783,±2}t = \left\{ \pm \frac{\sqrt{78}}{3}, \pm 2 \right\}t={±378​​,±2}

STEP 20

Match the solution set for ttt with the given options.The correct answer is: d. {±783,±2}\left\{ \pm \frac{\sqrt{78}}{3}, \pm 2\right\}{±378​​,±2}

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Solve $3(t^2 - 9)^2 + 16(t^2 - 9) = -5$ using substitution. Select the correct solution set.

Let's perform the substitution by setting $u = t^2 - 9$ . The equation then becomes a quadratic equation in terms of $u$ .
$3u^2 + 16u = -5$

To solve the quadratic equation, we first move all terms to one side to set the equation to zero.
$3u^2 + 16u + 5 = 0$

In our equation $3u^2 + 16u + 5 = 0$ , we identify $a = 3$ , $b = 16$ , and $c = 5$ .

We calculate the discriminant $b^2 - 4ac$ to determine the nature of the roots.
$\Delta = b^2 - 4ac = 16^2 - 4 \cdot 3 \cdot 5$

Calculate the discriminant.
$\Delta = 256 - 60 = 196$

Since the discriminant is positive, we have two real and distinct roots. We use the quadratic formula to find the values of $u$ .
$u = \frac{-16 \pm \sqrt{196}}{2 \cdot 3}$

Simplify the square root of the discriminant and the expression.
$u = \frac{-16 \pm 14}{6}$

Calculate the two values for $u$ .
$u_1 = \frac{-16 + 14}{6} = \frac{-2}{6} = -\frac{1}{3}$
$u_2 = \frac{-16 - 14}{6} = \frac{-30}{6} = -5$

Now that we have the values for $u$ , we reverse the substitution to find the corresponding values of $t$ . Recall that $u = t^2 - 9$ .

For $u_1 = -\frac{1}{3}$ , we have:
$t^2 - 9 = -\frac{1}{3}$

Solve for $t^2$ .
$t^2 = 9 - \frac{1}{3}$

Combine the terms on the right-hand side.
$t^2 = \frac{27}{3} - \frac{1}{3} = \frac{26}{3}$

Take the square root of both sides to find $t$ .
$t = \pm \sqrt{\frac{26}{3}} = \pm \frac{\sqrt{78}}{3}$

For $u_2 = -5$ , we have:
$t^2 - 9 = -5$

Solve for $t^2$ .
$t^2 = 9 - 5 = 4$

Take the square root of both sides to find $t$ .
$t = \pm \sqrt{4} = \pm 2$

Combine the values of $t$ from both $u_1$ and $u_2$ to get the solution set for $t$ .
$t = \left\{ \pm \frac{\sqrt{78}}{3}, \pm 2 \right\}$

Match the solution set for $t$ with the given options.
The correct answer is: d. $\left\{ \pm \frac{\sqrt{78}}{3}, \pm 2\right\}$