# CIE A level -Pure Mathematics 1 : Topic : 1.1 Quadratics: solve quadratic equations : Exam Questions Paper 1

Question

(i)Express in the form $$4x^{2}-12x$$ in the form $$\left ( 2x+a \right )^{2}+b$$

(ii)Hence ,or otherwise,find the set of values of x satisfying $$4x^{2}-12x> 7$$

(i)To express $$4x^{2}-12x$$ in the form $$\left(2x+a\right)^{2}+b$$, we need to complete the square.
We start with $$4x^{2}-12x$$. First, we factor out the common factor of 4:
$$4x^{2}-12x = 4(x^{2}-3x)$$
Next, we take half of the coefficient of the $$x$$ term, which is $$-3/2$$, square it, and add it inside the parentheses:
$$4\left [ \left ( x-\frac{3}{2} \right )^{2}-\left ( \frac{3}{2} \right )^{2} \right ]$$
$$4\left [ \left ( x-\frac{3}{2} \right )^{2}-\frac{9}{4} \right ]$$
$$4 \left ( x-\frac{3}{2} \right )^{2}-4\frac{9}{4}$$
$$\left [ 2\left ( x-\frac{3}{2} \right ) \right ]^{2}-9$$
$$\left [ 2\left ( x-\frac{3}{2} \right ) \right ]^{2}-9$$
$$\left ( 2x-3 \right )^{2}-9$$
So, $$4x^{2}-12x$$ can be expressed in the form $$\left ( 2x-3 \right )^{2}-9$$
(ii) To solve the inequality $$4x^{2}-12x > 7$$, we can first rewrite it as $$4x^{2}-12x-7 > 0$$.
$$x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$$
In this case, $$a = 4$$, $$b = -12$$, and $$c = -7$$.
$$x = \frac{-(-12) \pm \sqrt{(-12)^{2}-4(4)(-7)}}{2(4)}$$
$$x = \frac{12 \pm \sqrt{144+112}}{8}$$
$$x = \frac{12 \pm \sqrt{256}}{8}$$
$$x = \frac{12 \pm 16}{8}$$
$$x_1 = \frac{12+16}{8} = \frac{28}{8} = \frac{7}{2}$$
$$x_2 = \frac{12-16}{8} = \frac{-4}{8} = -\frac{1}{2}$$
Therefore, the set of values of $$x$$ satisfying $$4x^{2}-12x > 7$$ is $$x > \frac{7}{2}$$ or $$x < -\frac{1}{2}$$.

### Question

(i) Express $$-x^{2}+6x-5$$ in the form $$a(x+b)^{2}+c$$, where a, b and c are constants.
The function $$f:x\rightarrow -x^{2}+6x-5$$ defined for $$x\geq m$$ ,where m is constant.
(ii) State the smallest value of m for which f is one-one.
(iii) For the case where m = 5, find an expression for $$f^{-1}(x)$$ and state the domain of $$f^{-1}$$.

(i) To express the quadratic function$$-x^{2}+6x-5$$ in the form $$a(x+b)^{2}+c$$ we need to complete the square.
First, let’s rewrite the quadratic function as follows:
$$-x^{2}+6x-5=-(x^{2}-6x+5)$$
Now, we want to complete the square inside the parentheses:
$$x^{2}-6x+5=(x^{2}-6x+9)-9+5=(x-3)^{2}-4$$
Substituting this back into the original expression:
$$-x^{2}+6x-5=-\left [ (x-3)^{2}-4 \right ]=-1\left ( x-3 \right )^{2}+4$$
So, the quadratic function can be expressed in the form $$a(x+b)^{2}+c$$ as
$$-x^{2}+6x-5=-1(x-3)^{2}+4$$
Therefore, we have a=-1,b=3 and c=4.
(ii) For a quadratic function to be one-to-one (i.e., injective), it should pass the horizontal line test, which means each horizontal line intersects the graph at most once.
In this case, the quadratic function is f:x\rightarrow $$-x^{2}+6x-5$$ for $$x\geq m$$ , where m is a constant .
For the function to be one-to-one, the vertex of the parabola (which occurs at$$x=-\frac{b}{2a})$$ should lie on or to the right of the line x=m.
Using the coefficients from the quadratic function, we have:
$$x=\frac{-6}{2(-1)}=3$$
Since the vertex is at x=3, we need m\leq 3 for the function to be one-to-one.
The smallest value of m that satisfies this condition is m=3.
(iii)$$(x-3)^{2}=4-y$$
Correct order of operations
$$f^{-1}(x)=3+\sqrt{4-x}$$
Domain is $$x\leq 0$$

Question

(i) Express $$x^{2}+6x+2$$ in the form $$\left ( x+a \right )^{2}+b$$,where a and b are constants.

(ii)Hence,or otherwise, find the set of values of x for which $$x^{2}+6x+2> 9$$.

(i) To express $$x^{2}+6x+2$$ in the form $$(x+a)^{2}+b$$, we complete the square.
We start with $$x^{2}+6x+2$$. To determine the value of $$a$$, we take half of the coefficient of the $$x$$ term, which is 6, and square it:
$$\left(\frac{6}{2}\right)^{2} = 9$$
$$x^{2}+6x+9+2-9 = (x+3)^{2}+(-7)$$
Therefore, $$x^{2}+6x+2$$ can be expressed as $$(x+3)^{2}-7$$.
(ii) We have the inequality $$x^{2}+6x+2>9$$. Let’s use the expression we derived in part (i) to simplify the inequality:
$$(x+3)^{2}-7>9$$
$$(x+3)^{2}>16$$
Taking the square root of both sides (considering both positive and negative roots):
$$x+3>4$$ or $$x+3<-4$$
For $$x+3>4$$: $$x>1$$
For $$x+3<-4$$: $$x<-7$$
Therefore, the set of values of $$x$$ that satisfies $$x^{2}+6x+2>9$$ is $$x<-7$$ or $$x>1$$.

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