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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 $3^{2m+1} = 35 - 8(3^m)$, whe... 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 equation f(x)=3x2+8x35=0f(x) = 3x^2 + 8x - 35 = 0, we can use 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. Plugging these values into the formula gives:

x=8±8243(35)23x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 3 \cdot (-35)}}{2 \cdot 3}

Calculating the discriminant:

b24ac=64+420=484b^2 - 4ac = 64 + 420 = 484

Thus,

x=8±4846=8±226x = \frac{-8 \pm \sqrt{484}}{6} = \frac{-8 \pm 22}{6}

This provides us with two potential solutions:

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

So, 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)

Using the property 32m=(3m)23^{2m} = (3^m)^2, we can rewrite:

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

Letting x=3mx = 3^m, we rewrite the equation as:

3x2+8x35=03x^2 + 8x - 35 = 0

Factoring this results in:

(3x5)(x+7)=0(3x - 5)(x + 7) = 0

This gives the solutions:

  1. x=53x = \frac{5}{3} (thus, 3m=533^m = \frac{5}{3})
  2. x=7x = -7 (not applicable since 3m>03^m > 0)

Taking the valid solution:3m=53 3^m = \frac{5}{3} Taking logarithms:

m=log3(53)m = \log_3(\frac{5}{3})

To express this in the required form m=logbpqm = \log_b p - q, we can rewrite:

m=log35log33=log351m = \log_3 5 - \log_3 3 = \log_3 5 - 1

Thus, we have p=5p = 5, q=1q = 1, and b=3b = 3.

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