Question
(a) Solve $3m^{2}+5m-2=0.$
(b) Hence or otherwise, solve $3 \times 9^{x}+5 \times 3^{x}-2=0.$
▶️Answer/Explanation
Detailed solution
Let’s tackle this problem step-by-step, solving part (a) first and then using that result to address part (b).
Part (a): Solving \(3m^2 + 5m – 2 = 0\)
This is a quadratic equation in \(m\). We can solve it using the quadratic formula, which for an equation of the form \(am^2 + bm + c = 0\) is:
\[
m = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}.
\]
Here, identify the coefficients:
\(a = 3\),
\(b = 5\),
\(c = -2\).
First, compute the discriminant:
\[
\Delta = b^2 – 4ac = 5^2 – 4 \cdot 3 \cdot (-2) = 25 – 4 \cdot 3 \cdot (-2) = 25 + 24 = 49.
\]
Since the discriminant is positive (\(\Delta = 49\)), there are two real roots. Now, apply the quadratic formula:
\[
m = \frac{-5 \pm \sqrt{49}}{2 \cdot 3} = \frac{-5 \pm 7}{6}.
\]
Calculate the two solutions:
\(m_1 = \frac{-5 + 7}{6} = \frac{2}{6} = \frac{1}{3}\),
\(m_2 = \frac{-5 – 7}{6} = \frac{-12}{6} = -2\).
Part (b): Solving \(3 \times 9^x + 5 \times 3^x – 2 = 0\)
We know, \(9^x = (3^2)^x = 3^{2x}\), so rewrite the equation:
\[
3 \cdot 3^{2x} + 5 \cdot 3^x – 2 = 0.
\]
This resembles the quadratic from part (a), but with \(3^{2x}\) and \(3^x\). Let’s make a substitution to align it. Set \(y = 3^x\), where \(y > 0\) since the exponential function is always positive. Then:
\(3^{2x} = (3^x)^2 = y^2\),
\(3^x = y\).
Substitute into the equation:
\[
3y^2 + 5y – 2 = 0.
\]
This is identical to the equation in part (a) with \(m = y\). Using the solutions from part (a):
\(y = \frac{1}{3}\),
\(y = -2\).
Since \(y = 3^x > 0\), discard \(y = -2\) (as an exponential cannot be negative). Thus:
\[
y = 3^x = \frac{1}{3}.
\]
Since \(\frac{1}{3} = 3^{-1}\), we have:
\[
3^x = 3^{-1}.
\]
Equate the exponents (since the bases are equal):
\[
x = -1.
\]
————Markscheme—————–
Solution: –
(a) $(3m-1)(m+2)=0$ OR $m=\frac{-5\pm\sqrt{25+24}}{6}$ (or equivalent)
$m=\frac{1}{3}, m=-2$
(b) Setting their m-value(s) $3^x$ OR recognizing a quadratic in $3^x$
$3^x=\frac{1}{3}$
(or $3^x=-2$) OR $3 \times (3^x)^2 + 5 \times 3^x – 2 = 0 $
$ x=-1 $