Quadratic Equations Test

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Quadratic Equations Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following is a quadratic equation?

A
$$(a-1)x^{3}+(b -2)x^{2} + c$$

B
$$(a-1)x^{2}+(b -2)x + 5c$$

C
$$(a-1)x^{4}+(b -2)x^{2} + 2c$$

D
$$(x-1)x^{4}+(b -2)x^{2} + 2c$$

Solution

A quadratic equation can be expressed in the form $$ax^{2} + bx + c = 0$$.
Here, the only equation that has a second degree variable is $$(a-1)x^{2}+(b -2)x + 5c$$ and all other equations have third and fourth degree variables.
So, the quadratic equation is $$(a-1)x^{2}+(b -2)x + 5c$$.
Question 2
1 / -0

$$h = -16t^{2} + vt + k$$
The equation above gives the height $$h$$, in feet, of a ball $$t$$ seconds after it is thrown straight up with an initial speed of $$v$$ feet per second from a height of $$k$$ feet. Which of the following gives $$v$$ in terms of $$h, t$$ and $$k$$?

A
$$v = h + k - 16t$$

B
$$v = \dfrac {h - k + 16}{t}$$

C
$$v = \dfrac {h + k}{t} - 16t$$

D
$$v = \dfrac {h - k}{t} + 16t$$

Solution

Given, $$ h = -16t^2 + vt + k $$
$$ \Rightarrow h + 16t^2 - k = vt $$
$$ \Rightarrow v = \dfrac { h + 16t^2 - k}{t} $$
$$ \Rightarrow v = \dfrac {h-k}{t} + 16t $$
Question 3
1 / -0

$$\sqrt{2k^2+17}-x=0$$
If $$k > 0$$ and $$x = 7$$ in the equation above, what is the value of $$k$$?

A
$$2$$

B
$$3$$

C
$$4$$

D
$$5$$

Solution

Given expression is $$\sqrt{2k^2 + 17} - x = 0$$
Given that $$x = 7$$, we thus get
$$\sqrt{2k^2 + 17} = 7$$
$$\therefore 2k^2 + 17 = 49$$ [squaring both the sides]
$$\therefore 2k^2 = 32$$
$$\therefore k^2 = 16$$
$$\therefore k = \pm4$$
$$\implies k=4$$ $$[\because k>0]$$
Question 4
1 / -0

If $$f(x) = x^{2} - 4x + 1, f(x)$$, the $$x$$-intercept of $$f(x)$$ is closest to which of the following?

A
$$(-0.33, 0)$$

B
$$(0.27, 0)$$

C
$$(1.73, 0)$$

D
$$(3.27, 0)$$

E
$$(4.33, 0)$$

Solution

Given $$f(x)=x^{2}-4x+1$$, $$f(x)$$
Then $$y=x^{2}-4x+1$$ and $$y=0$$
Then $$x^{2}-4x+1=0$$
$$x=\dfrac{-(-4)\pm \sqrt{(-4)^{2}-4\times 1\times 1}}{2\times 1}$$
$$\Rightarrow x=\dfrac{4\pm \sqrt{16-4}}{2}$$
$$\Rightarrow x=\dfrac{4\pm \sqrt{12}}{2}$$
$$\Rightarrow x=2\pm \sqrt{3}$$
$$x=2-\sqrt{3}=2-1.72=0.27$$
$$x=2+\sqrt{3}=2+1.73=3.73$$
Then $$x$$-intercept of $$f(x)$$ is closest is $$(0.27,0))$$
Question 5
1 / -0

Find the value of $$x$$ which satisfies $$x^{2} - 8x + 13 = 0$$

A
$$-4\pm \sqrt {3}$$

B
$$-4\pm 2\sqrt {3}$$

C
$$4\pm \sqrt {3}$$

D
$$4\pm 3\sqrt {2}$$

E
$$4\pm 2\sqrt {2}$$

Solution

Given equation is $${x}^{2} - 8x+13 =0$$
$$a=1, b=-8$$ and $$c=13$$
By using quadratic formula:
$$x=\dfrac {-b\pm \sqrt {b^2-4ac}}{2a}$$

$$\therefore x =\dfrac { (8\pm \sqrt{64-52})}{2} $$

$$= \dfrac {(8\pm \sqrt{12})}{2} $$

$$= \dfrac {(8\pm 2\sqrt3)}{2} $$

$$= 4 \pm \sqrt3$$
Question 6
1 / -0

If A and B are whole numbers such that $$9A^{2} = 12A + 96$$ and $$B^{2} = 2B + 3$$, find the value of $$5A + 7B$$.

A
$$31$$

B
$$37$$

C
$$41$$

D
$$43$$

Solution

A,B are whole numbers
$$9{ A }^{ 2 }=12A+96$$ & $${ B }^{ 2 }=2B+3$$
$$9{ A }^{ 2 }-12A-96=0$$ & $${ B }^{ 2 }-2B-3=0$$
$$3{ A }^{ 2 }-4A-32=0$$
$$3{ A }^{ 2 }-12A+8A-32=0$$ & $${ B }^{ 2 }-3B+B-3=0$$
$$(3A+8)(A-4)=0$$ & $$(B-3)(B+1)=0$$
$$\therefore A=4,B=3$$ ($$\because A,B$$ are whole numbers $$\because (A\neq \cfrac { -8 }{ 3 }) $$ &$$(B\neq -1$$)
$$5A+7B=5(4)+7(3)$$
$$=41$$
Question 7
1 / -0

What are the solutions to $$3x^{2} + 12x + 6 = 0$$?

A
$$x = -2\pm \sqrt {2}$$

B
$$x = -2 \pm \dfrac {\sqrt {30}}{3}$$

C
$$x = -6\pm \sqrt {2}$$

D
$$x = -6 \pm 6\sqrt {2}$$

Solution

Solutions of the equation $$ ax^{ 2 }+bx+c = 0 $$ are given be $$ \dfrac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } $$
Hence, solutions of the given equation are $$ \dfrac { -12 \pm \sqrt { { 12 }^{ 2 }-4(3)(6) } }{ 2(3) } $$
$$= \dfrac { -12 \pm \sqrt { 144-72 } }{ 6 } $$
$$= - 2 \pm \dfrac {\sqrt{72}}{6} $$
$$= - 2 \pm \dfrac {6\sqrt{2}}{6} $$
$$= -2 \pm \sqrt {2} $$
Question 8
1 / -0

$$\displaystyle x^2-\frac{k}{2} x= 2p$$
In the quadratic equation above, k and p are constants. What are the solutions for x?

A
$$\displaystyle x= \frac{k}{4} \pm \frac{\sqrt{k^2 + 2p}}{4}$$

B
$$\displaystyle x= \frac{k}{4} \pm \frac{\sqrt{k^2 + 32p}}{4}$$

C
$$\displaystyle x= \frac{k}{2} \pm \frac{\sqrt{k^2 + 2p}}{2}$$

D
$$\displaystyle x= \frac{k}{2} \pm \frac{\sqrt{k^2 + 32p}}{4}$$

Solution

We know that for a quadratic equation of the form $$ax^2+bx+c=0$$,
$$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$ [Quadratic formula]

The given equation is $$x^2-\dfrac{k}{2}x=2p$$
We can rewrite the equation as $$x^2-\dfrac{k}{2}x-2p=0$$
$$\therefore \ x=\dfrac{\dfrac{k}{2}\pm\sqrt{\left ( \dfrac{-k}{2} \right )^2-4\times 1\times (-2p)}}{2\times 1}$$ [Quadratic formula]

$$\Rightarrow \dfrac{\dfrac{k}{2}\pm\sqrt{ \dfrac{k^2}{4}+ 8p}}{2}$$

$$\Rightarrow \dfrac{k\pm\sqrt{k^2+ 32p}}{4}$$

$$\Rightarrow \dfrac{k}{4}\pm\dfrac{\sqrt{k^2+ 32p}}{4}$$

Hence, option B is correct.
Question 9
1 / -0

Which of the following is a possible solution for $$x$$ in terms of $$k$$ for the equation $$x = \dfrac {2k}{x + 2}$$?

A
$$\sqrt {2k}$$

B
$$\sqrt {-2k}$$

C
$$1 - \sqrt {1 + 2k}$$

D
$$\sqrt {1 + 2k} + 1$$

E
$$\sqrt {1 + 2k} - 1$$

Solution

Given $$ x = \dfrac {2k}{x+2} $$
$$ \Rightarrow x^2 + 2x = 2k $$
$$ \Rightarrow x^2 + 2x - 2k = 0 $$
Solutions of the equation $$ ax^{ 2 }+bx+c = 0 $$ are given by $$ \dfrac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } $$
Hence, solutions of the given equation are
$$ =\dfrac { -2 \pm \sqrt { { 2 }^{ 2 }-4(1)(-2k) } }{ 2(1) } \\= \dfrac { -2 \pm \sqrt { 4 +8k } }{ 2 } \\= 1 \pm \dfrac {2\sqrt{1+2k}}{2} \\= - 1 \pm \sqrt {1 + 2k} $$
Thus one of the possible solutions of the given equation is $$ \sqrt {1 + 2k} - 1 $$.
Question 10
1 / -0

Which of the following is a possible value of $$x$$, if $$5x^{2}- 10x + 4 = 0$$?

A
$$2\sqrt {5}$$

B
$$1 + 2\sqrt {5}$$

C
$$1 + \dfrac {\sqrt {5}}{5}$$

D
$$2 + \dfrac {\sqrt {5}}{10}$$

E
$$10 + \dfrac {\sqrt {5}}{10}$$

Solution

Given expression is $$5x^2-10x+4=0$$

Solutions of the equation $$ ax^{ 2 }+bx+c = 0 $$ are given be $$ \dfrac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } $$

Hence, solutions of the given equation are $$ \dfrac { 10 \pm \sqrt { { 10 }^{ 2 }-4(5)(4) } }{ 2(5) }$$

$$ = \dfrac { 10 \pm \sqrt { 100-80 } }{ 10 } $$

$$= 1 \pm \dfrac {\sqrt{5}}{5} $$

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Quadratic Equations Test - 33

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Quadratic Equations Test - 33

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Question 1

$$(a-1)x^{3}+(b -2)x^{2} + c$$

$$(a-1)x^{2}+(b -2)x + 5c$$

$$(a-1)x^{4}+(b -2)x^{2} + 2c$$

$$(x-1)x^{4}+(b -2)x^{2} + 2c$$

Solution

Question 2

$$v = h + k - 16t$$

$$v = \dfrac {h - k + 16}{t}$$

$$v = \dfrac {h + k}{t} - 16t$$

$$v = \dfrac {h - k}{t} + 16t$$

Solution

Question 3

$$2$$

$$3$$

$$4$$

$$5$$

Solution

Question 4

$$(-0.33, 0)$$

$$(0.27, 0)$$

$$(1.73, 0)$$

$$(3.27, 0)$$

$$(4.33, 0)$$

Solution

Question 5

$$-4\pm \sqrt {3}$$

$$-4\pm 2\sqrt {3}$$

$$4\pm \sqrt {3}$$

$$4\pm 3\sqrt {2}$$

$$4\pm 2\sqrt {2}$$

Solution

Question 6

$$31$$

$$37$$

$$41$$

$$43$$

Solution

Question 7

$$x = -2\pm \sqrt {2}$$

$$x = -2 \pm \dfrac {\sqrt {30}}{3}$$

$$x = -6\pm \sqrt {2}$$

$$x = -6 \pm 6\sqrt {2}$$

Solution

Question 8

$$\displaystyle x= \frac{k}{4} \pm \frac{\sqrt{k^2 + 2p}}{4}$$

$$\displaystyle x= \frac{k}{4} \pm \frac{\sqrt{k^2 + 32p}}{4}$$

$$\displaystyle x= \frac{k}{2} \pm \frac{\sqrt{k^2 + 2p}}{2}$$

$$\displaystyle x= \frac{k}{2} \pm \frac{\sqrt{k^2 + 32p}}{4}$$

Solution

Question 9

$$\sqrt {2k}$$

$$\sqrt {-2k}$$

$$1 - \sqrt {1 + 2k}$$

$$\sqrt {1 + 2k} + 1$$

$$\sqrt {1 + 2k} - 1$$

Solution

Question 10

$$2\sqrt {5}$$

$$1 + 2\sqrt {5}$$

$$1 + \dfrac {\sqrt {5}}{5}$$

$$2 + \dfrac {\sqrt {5}}{10}$$

$$10 + \dfrac {\sqrt {5}}{10}$$

Solution

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