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Given that $f(x) = 3x^2 + 8x - 35$, where $x \in \mathbb{R}$, find the two roots of $f(x) = 0$ - Leaving Cert Mathematics - Question b - 2021

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Given-that-$f(x)-=-3x^2-+-8x---35$,-where-$x-\in-\mathbb{R}$,-find-the-two-roots-of-$f(x)-=-0$-Leaving Cert Mathematics-Question b-2021.png

Given that $f(x) = 3x^2 + 8x - 35$, where $x \in \mathbb{R}$, find the two roots of $f(x) = 0$. Hence or otherwise, solve the equation $3^{2m+1} = 35 - 8(3^{m})$, ... show full transcript

Worked Solution & Example Answer:Given that $f(x) = 3x^2 + 8x - 35$, where $x \in \mathbb{R}$, find the two roots of $f(x) = 0$ - Leaving Cert Mathematics - Question b - 2021

Step 1

Find the two roots of $f(x) = 0$

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Answer

To find the roots of the quadratic equation 3x2+8x35=03x^2 + 8x - 35 = 0, we can apply the quadratic formula:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

where a=3a = 3, b=8b = 8, and c=35c = -35. First, we calculate the discriminant:

b24ac=8243(35)=64+420=484.b^2 - 4ac = 8^2 - 4 \cdot 3 \cdot (-35) = 64 + 420 = 484.

Then we plug this into the quadratic formula:

x=8±48423=8±226.x = \frac{-8 \pm \sqrt{484}}{2 \cdot 3} = \frac{-8 \pm 22}{6}.

This gives us two roots:

  1. x=146=73x = \frac{14}{6} = \frac{7}{3}
  2. x=306=5x = \frac{-30}{6} = -5

Thus, the two roots are x=73x = \frac{7}{3} and x=5x = -5.

Step 2

Solve the equation $3^{2m+1} = 35 - 8(3^{m})$

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Answer

We start with the equation:

32m+1=358(3m).3^{2m+1} = 35 - 8(3^m).

Rewriting 32m3^{2m} in terms of 3m3^m yields:

32m=(3m)2.3^{2m} = (3^m)^2.

Substituting x=3mx = 3^m, the equation becomes:

3x2+8x35=0.3x^2 + 8x - 35 = 0.

From the previous steps, we know this factors to:

(3x7)(x+5)=0.(3x - 7)(x + 5) = 0.

The solutions for xx are x=73x = \frac{7}{3} and x=5x = -5. Since x=3mx = 3^m and must be positive, we only consider x=73x = \frac{7}{3}.

Therefore, we can express 3m3^m as:

3m=73.3^m = \frac{7}{3}.

Taking logarithms gives:

m=log3(73)=log37log33=log371.m = \log_3 \left(\frac{7}{3}\right) = \log_3 7 - \log_3 3 = \log_3 7 - 1.

This is in the form m=logbpqm = \log_b p - q, where b=3b = 3, p=7p = 7, and q=1q = 1.

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