Quadratics — practice questions

This quadratic does not factorise easily, so we use the quadratic formula. Here, $a=3$, $b=-7$, and $c=-5$.

Substitute these values into the formula: $x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(3)(-5)}}{2(3)}$

$x = \frac{7 \pm \sqrt{49 - (-60)}}{6}$

$x = \frac{7 \pm \sqrt{49 + 60}}{6}$

$x = \frac{7 \pm \sqrt{109}}{6}$

Now, we find the two possible values for $x$: $x_1 = \frac{7 + \sqrt{109}}{6} \approx \frac{7 + 10.440}{6} \approx \frac{17.440}{6} \approx 2.9067...$

$x_2 = \frac{7 - \sqrt{109}}{6} \approx \frac{7 - 10.440}{6} \approx \frac{-3.440}{6} \approx -0.5733...$

Rounding to 3 significant figures, the solutions are $x = 2.91$ and $x = -0.573$.

'Two distinct real roots' means we need the discriminant to be strictly positive: $b^2 - 4ac > 0$.

Identify the coefficients: $a = 1$, $b = (k+3)$, and $c = (k+6)$.

Substitute into the discriminant: $(k+3)^2 - 4(1)(k+6) > 0$

Expand and simplify: $k^2 + 6k + 9 - 4k - 24 > 0$ $k^2 + 2k - 15 > 0$

Factorise the quadratic in $k$: $(k+5)(k-3) > 0$

The critical values are $k = -5$ and $k = 3$. Because this is an upward-opening parabola in $k$, the expression is positive outside the roots.

Therefore the set of values is $k < -5$ or $k > 3$.

First, factor out the coefficient of $x^2$ from the first two terms: $f(x) = 2(x^2 + 6x) + 5$

Next, complete the square inside the bracket. We take half the coefficient of $x$ (which is 6), square it (giving 9), and then add and subtract it inside the bracket: $f(x) = 2[(x+3)^2 - 3^2] + 5$ $f(x) = 2[(x+3)^2 - 9] + 5$

Now, expand the outer bracket: $f(x) = 2(x+3)^2 - 2(9) + 5$ $f(x) = 2(x+3)^2 - 18 + 5$

Finally, simplify: $f(x) = 2(x+3)^2 - 13$

This is in the form $a(x+p)^2 + q$ with $a=2$, $p=3$ and $q=-13$.

The vertex form is $a(x-h)^2+k$, so we compare $2(x+3)^2 - 13$ with this. We have $h = -3$ and $k = -13$. Therefore, the coordinates of the vertex are $(-3, -13)$.

Since the coefficient of $x^2$ is $a=2$, which is positive, the parabola is U-shaped. This means the vertex is a minimum point.