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CBSE Questions for Class 9 Maths Polynomials Quiz 11 - MCQExams.com
CBSE
Class 9 Maths
Polynomials
Quiz 11
Write whether the following statement is True or False ? Justify your answer.
Zero of polynomial is always $$0$$
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True
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False
Explanation
False, because zero of a polynomial means the value of the variable for which the polynomial value becomes zero. It need not be zero and depends on the polynomial.
If one zero of the polynomial $$p(x)={x}^{3}-6{x}^{2}+11x-6$$ is $$3$$, find the other two zeroes.
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$$0$$ and $$2$$
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$$2$$ and $$-2$$
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$$1$$ and $$2$$
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$$2$$ and $$-3$$
The number of rational roots of $$(2x+3)(2x+5)(x-1)(x-2)=30$$ is
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$$4$$
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$$3$$
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$$1$$
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$$5$$
Using identities, evaluate:
$$71^2$$.
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$$5041$$
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$$3041$$
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$$6041$$
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$$2041$$
Explanation
Given, $$71^2$$,
i.e.
$$71^2$$
$$=(70+1)^2$$
.
We know,
$$(x+y)^{ 2 }=x^{ 2 }+y^{ 2 }+2xy$$.
Then,
$$71^{ 2 }=(70+1)^{ 2 }\\ =70^{ 2 }+(2\times 70\times 1) + 1^2\\ =4900+140+1\\ =5040+1 \\=5041.$$
Hence, $$71^2=5041$$.
Therefore, option $$A$$ is correct.
Find the zeros of the polynomial $$3\sqrt2 x^2+ 13x + 6 \sqrt2 $$ and verify the relationship between the zeros.
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$$x=\dfrac{\sqrt2}{3}\text{ and } \dfrac{-3}{\sqrt{2}}$$
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$$x={2\sqrt2}\text{ and } \dfrac{-3}{\sqrt{2}}$$
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$$x=\dfrac{-2\sqrt2}{3}\text{ and } {-3}{\sqrt{2}}$$
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$$x=\dfrac{2\sqrt2}{3}\text{ and } \dfrac{-3}{\sqrt{2}}$$
Explanation
Let us first factorize the given equation $$3\sqrt 2x^2+13x+6\sqrt2$$ as shown below:
$$\Rightarrow 3\sqrt { 2 } x^{ 2 }+13x+6\sqrt { 2 } \\ =3\sqrt { 2 } x^{ 2 }+9x+4x+6\sqrt { 2 } \\ =3x(\sqrt { 2 } x+3)+2\sqrt { 2 } (\sqrt { 2 } x+3)\\ =(3x+2\sqrt { 2 } )(\sqrt { 2 } x+3)$$
Therefore, the zeroes of the given polynomial are:
$$(3x+2\sqrt { 2 } )=0\\ \Rightarrow x=-\dfrac { 2\sqrt { 2 } }{ 3 }=\alpha\ \\ (\sqrt { 2 } x+3)=0\\ \Rightarrow x=-\dfrac { 3 }{ \sqrt { 2 } }=\beta$$
Now, the sum and product of the roots is as follows:
$$\alpha +\beta =-\dfrac { 2\sqrt { 2 } }{ 3 } -\dfrac { 3 }{ \sqrt { 2 } } =-\left( \dfrac { 2\sqrt { 2 } }{ 3 } +\dfrac { 3 }{ \sqrt { 2 } } \right) =\dfrac { -4-9 }{ 3\sqrt { 2 } } =-\dfrac { 13 }{ 3\sqrt { 2 } } =-\dfrac { b }{ a } \\ \alpha \beta =\left( -\dfrac { 2\sqrt { 2 } }{ 3 } \right) \left( -\dfrac { 3 }{ \sqrt { 2 } } \right) =2=\dfrac { 6\sqrt { 2 } }{ 3\sqrt { 2 } } =\dfrac { c }{ a }$$
Hence verified.
The value of $${ \left( 1.02 \right) }^{ 2 }+{ \left( 0.98 \right) }^{ 2 }$$, corrected to three decimal places is:
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$$2.001$$
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$$2.004$$
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$$2.006$$
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$$1.995$$
Explanation
Given,
$$(1.02)^2+(0.98)^2$$.
We know,
$$(a+b)^2= a^2+2ab+b^2$$.
Then,
$$(1.02)^2+(0.98)^2$$
$$=(1.02)^2+(0.98)^2+(2\times 1.02\times 0.98)-(2\times 1.02\times 0.98$$)
$$=(1.02+0.98)^2-1.9992$$
$$=(2)^2-1.9992$$
$$=4-1.9992$$
$$=2.0008$$.
When corrected to three decimal digit,
$$2.0008$$
$$=2.001$$.
Therefore, option $$A$$ is correct.
Which of the following is the zeroes of the polynomial $$x^2 + 4x + 4 =$$?
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$$2$$
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$$-2$$
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$$4$$
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$$-4$$
Explanation
Now,
$$x^2 + 4x + 4$$
$$=(x+2)^2$$.
If $$x^2+4x+4=0$$ then $$x=-2,-2$$.
So the the zeroes of $$x^2+4x+4 $$ is $$-2$$.
The zeroes of the polynomial $$P(x)=x(x-2)(x+3)$$ is
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$$0$$
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$$0,2,3$$
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$$0,2,-3$$
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$$None\ of\ these$$
Explanation
$$P\left(x\right)=x\left(x-2\right) \left(x+3\right)$$
$$\therefore \ $$To find zeroes,
$$P\left(x\right)=0$$
$$\therefore \left(x\right) \left(x-2\right) \left(x+3\right)=0$$
$$\therefore x=0$$ OR $$x-2=0$$
OR
$$x+3=0$$
$$\therefore x=0$$
OR
$$x=2$$
OR
$$x=-3$$
$$\therefore$$ value of $$x$$ is $$ 0, 2, -3$$
Degree of the polynomial $$13 + 11x + 12x^3 + 3x^2$$ is
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$$2$$
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$$1$$
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$$3$$
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$$4$$
Explanation
The degree of an individual term of a polynomial is the exponent of its variable
$$13+11x+12{x}^{3}+3{x}^{2}=12{x}^{3}+3{x}^{2}+11x+13$$
The highest exponent of $$x$$ is $$3$$
$$\therefore\,$$ Degree$$ =3$$
The roots of $$x^{3}-4x^{2}-8x+8$$ are :
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$$1,2\pm \sqrt{3}$$
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$$-2,3\pm \sqrt{5}$$
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$$-2,2\pm \sqrt{3}$$
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$$1,3\pm \sqrt{5}$$
Explanation
\begin{array}{l} { x^{ 3 } }-4{ x^{ 2 } }-8x+8 \\ { { suppose } }, \\ y={ x^{ 3 } }-4{ x^{ 2 } }-8x+8 \\ and\, \, \, replace\, y\, with\, \, 0. \\ { x^{ 3 } }-4{ x^{ 2 } }-8x+8=0 \\ Now,\, factorizing\, the\, equ: \\ (x+2)\, ({ x^{ 2 } }-6x+4)=0 \\ \Rightarrow x+2=0 \\ \therefore \, \, \, x=-2 \\ and,\, \\ \, \, \, { x^{ 2 } }-6x+4=0\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left[ { u\sin g\, quadratic{ { } }formula=-\frac { { b\pm \sqrt { { b^{ 2 } }-4(ac) } } }{ { 2a } } } \right. \\ \, \, \, \, substitute\, the\, values\& \, find\, the\, value\, of\, x: \\ \, \, \, \, \, \, \, a=1,\, b=-6,\, c=4 \\ \, \Rightarrow x=\, \, \frac { { -6\pm \sqrt { { { (-6) }^{ 2 } }-4(1)\, (4) } } }{ { 2\, .1 } } \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\frac { { 6\pm \sqrt { 36-16 } } }{ { 2\, } } =\frac { { 6\pm \sqrt { 20 } } }{ 2 } =\frac { { 6\pm 2\sqrt { 5 } } }{ { 2\, } } =3\pm \sqrt { 5 } \\ \, \, \, \, \, \, \, \, \, \, \, \, \therefore \, \, \, \, x=3+\sqrt { 5 } \, ,3-\sqrt { 5 } \\ so,that\, we\, can\, say\, the\, value\, of\, \, x=-2,\, 3+\sqrt { 5 } \, ,3-\sqrt { 5 } \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, or\, \, x=-2,\, 3\pm \sqrt { 5 } \, \\ And\, the\, \, correct\, option\, is\, B. \end{array}
Factorise : $${ (ax+by) }^{ 2 }+{ (2bx-2ay) }^{ 2 }-6abxy$$
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$${ a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }+2{ b }^{ 2 }{ x }^{ 2 }+2{ a }^{ 2 }{ y }^{ 2 }+6abxy$$
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$${ a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }+4{ b }^{ 2 }{ x }^{ 2 }+4{ a }^{ 2 }{ y }^{ 2 }-12abxy$$
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$${ a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }+{ b }^{ 2 }{ x }^{ 2 }+{ a }^{ 2 }{ y }^{ 2 }+4abxy$$
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$${ a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }+4{ b }^{ 2 }{ x }^{ 2 }+4{ a }^{ 2 }{ y }^{ 2 }+6abxy$$
Explanation
$$\left( ax+by\right)^2+\left( 2bx-2ay\right)^2-babxy$$
Here we will use the following identity,
$$\Rightarrow \left( m+n\right)^2=m^2+n^2+2mn$$
and $$\left(m-n\right)^2=m^2+n^2-2mn$$
Using this we have :
$$= \left(ax\right)^2+\left(by\right)^2+2axby+4\left(bx\right)^2+4\left(ay\right)^2-8axby-6abxy$$
$$=a^2x^2+b^2y^2+4b^2x^2+4a^2y^2-14axby+2axby$$
$$=a^2x^2+b^2y^2+4b^2x^2+4a^2y^2-12axby$$
Hence, the answer is $$a^2x^2+b^2y^2+4b^2x^2+4a^2y^2-12axby.$$
If one zero of a quadratic polynomial $$x^2+3x+k$$ is $$2$$. Find the value of $$k$$.
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$$2$$
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$$-10$$
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$$-5$$
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$$9$$
Explanation
$$2$$ is a zero of $$x^2+3x+k$$.
Thus, $$(2)^2+3(2)+k=0$$
$$k+10=0$$
$$k=-10$$
If $$xy=20$$ and $$(x+y)^{2}=70$$, then $$x^{2}+y^{2}$$ is equal to
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$$25$$
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$$35$$
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$$30$$
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$$50$$
Explanation
$$(x+y)^2=x^2+y^2+2xy$$
It is given that $$(x+y)^2=70$$ and $$xy=20$$
$$\Rightarrow 70=x^2+y^2+(2\times 20)$$
$$x^2+y^2=70-40$$
$$=30$$
The zeroes of the polynomial $${ p }({ x })={ x }({ x }-1)({ x }-2)$$ are
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$$0$$
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$$0 , - 1 , - 2$$
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$$0,1 , - 2$$
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$$0,1,2$$
Explanation
$$\begin{array}{l} We\, have \\ p\left( x \right) =x\left( { x-1 } \right) \left( { x-2 } \right) \\ for\, the\, \, zero\, polynomial \\ p\left( x \right) =x \\ \therefore x\left( { x-1 } \right) \left( { x-2 } \right) =0 \\ Here\, we\, get \\ x=0\, \, \, or\, \, \left( { x-1 } \right) =0\, \, and\, \left( { x-2 } \right) =0 \\ \therefore x=0\, \, ,\, \, \, x=1\, \, and\, x=2 \\ So,\, \left( { 0,1,2 } \right) \, are\, the\, zero\, polynomila\, of\, p\left( x \right) \, \\ Hence,\, option\, D\, is\, the\, required\, asnwer. \end{array}$$
if $$(a-3)x^{2}+(2a+b)x+(b-c)$$ is a zero polynomial, then
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$$b+2a=0$$
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$$ab+c=0$$
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$$a=3$$
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$$b-c=0$$
The zero of the polynomial $$P\left( x \right) =\surd 5x-5$$ is ......
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$$\surd 5$$
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$$-\surd 5$$
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$$-5$$
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$$4$$
Explanation
Given, $$p(x)=\sqrt{5}x-5$$.
Here, we find zero of a polynomial when $$p(x)=0$$.
Then,
$$\sqrt{5}x-5=0$$
$$\implies$$
$$\sqrt{5}x=5$$
$$\implies$$
$$x=\dfrac{5}{\sqrt{5}}$$
$$\implies$$
$$x=\dfrac{5}{\sqrt{5}}\times\dfrac{\sqrt{5}}{\sqrt{5}}$$
$$\implies$$
$$x=\dfrac{5\sqrt{5}}{5}=\sqrt{5}$$.
Therefore, option $$A$$ is correct.
The degree of the polynomial $$\frac { 4 }{ 5 } { x }^{ 2 }-\frac { 7 }{ 5 } x+\frac { 2 }{ 3 } { x }^{ 3 }+6$$ is :
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2
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1
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3
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0
The value of $$(a +b)^{2} + (a-b)^{2}$$ is
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$$ 2a + 2b $$
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$$ 2a - 2b$$
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$$ 2a^{2} + 2b^{2}$$
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$$ 2a^{2} - 2b^{2}$$
Explanation
$$\textbf{Step-1: Apply the relevant formula & simplify.}$$
$$\text{We have,}$$
$$(a + b)^{2} + (a - b)^{2}$$
$$\text{Now,}$$
$$(a+b)^2 = a^2+2ab+b^2$$ $$......(i)$$
$$(a-b)^2= a^2-2ab+b^2$$ $$......(ii)$$
$$\textbf{Step-2: Add the above expressions to get the result}$$
$$(a + b)^{2} + (a - b)^{2}$$
$$a^2+2ab+b^2 + a^2-2ab+b^2$$
$$=2a^2 +2b^2$$
$$\textbf{Hence, option - C is the answer}$$
Write whether the following statement is True or False. Justify your answer.
A polynomial cannot have more than one zero
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True
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False
Explanation
The given statement is False, because a polynomial can have any number of zeroes which depends on the degree of the polynomial.
Write whether the following statement is True or False. Justify your answer.
Zero of a polynomial is always $$0$$
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True
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False
Explanation
The given statement is False, because zero of polynomial can be any real number.
Write whether the following statement is True or False. Justify your answer.
The degree of the sum of two polynomials each of degree $$5$$ is always $$5$$.
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True
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False
Explanation
The given statement is False. For example, consider the two polynomial $$-x^5 + 3x^2 + 4$$ and $$x^5 + x^4 + 2x^3 + 3.$$ The degree of each of these polynomial is $$5$$. Their sum is $$x^4 + 2x^3 + 3x^2 + 7.$$ The degree of this polynomial is $$4$$ not $$5$$.
Verify whether the following is true or false.
$$\dfrac{-4}{5}$$ is a zero of $$4 - 5y.$$
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True
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False
Explanation
A zero of a polynomial $$p(x)$$ is a number $$c$$ such that $$p(c) = 0$$
Let $$p(y) = 4 - 5y$$
$$\therefore p \left ( -\dfrac{4}{5} \right ) = 4 - 5 \left ( \dfrac{-4}{5} \right ) = 4 + 4 = 8 \neq 0$$
Hence, $$-\dfrac{4}{5}$$ is not a zero of $$4 - 5y.$$
So, given statement is false.
Verify whether the following is true or false.
$$0$$ and $$2$$ are the zeroes of $$t^2 - 2t.$$
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True
0%
False
Explanation
A zero of a polynomial $$p(x)$$ is a number $$c$$ such that $$p(c) = 0$$
Let $$p(t) = t^2 - 2t$$
$$\therefore p(0) = (0)^2 - 2(0) = 0$$
And $$p(2) = (2)^2 - 2(2) = 4 - 4 = 0$$
Hence, $$0$$ and $$2$$ are zeroes of the polynomial $$p(t) = t^2 - 2t.$$
So, given statement is true.
Verify whether the following is true or false.
$$-\dfrac{1}{3}$$ is a zero of $$3x + 1.$$
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True
0%
False
Explanation
A zero of a polynomial $$p(x)$$ is a number $$c$$ such that $$p(c) = 0$$
Let $$p(x) = 3x + 1$$
$$\therefore p\left ( -\dfrac{1}{3} \right ) = 3 \left ( -\dfrac{1}{3} \right ) + 1 = -1 + 1 = 0$$
Hence, $$-\dfrac{1}{3}$$ is a zero of $$p(x) = 3x + 1.$$
So, given statement is true.
Verify whether the following is true or false.
$$-3$$ is a zero of $$x - 3$$
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0%
True
0%
False
Explanation
A zero of a polynomial $$p(x)$$ is a number $$c$$ such that $$p(c) = 0$$
Let $$p(x) = x - 3$$
$$\therefore p(-3) = -3 - 3 = -6 \neq 0$$
Hence, $$-3$$ is not a zero of $$x - 3$$.
So, given statement is false.
Verify whether the following is true or false.
$$-3$$ is a zero of $$y^2 + y - 6.$$
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0%
True
0%
False
Explanation
A zero of a polynomial $$p(x)$$ is a number $$c$$ such that $$p(c) = 0$$
Let $$p(y) = y^2 + y - 6$$
$$\therefore p(-3) = (-3)^2 + (-3) - 6 = 9 - 3 - 6 = 0$$
Hence, $$-3$$ is a zero of the polynomial $$y^2 + y - 6.$$
So, given statement is true.
Find the zeroes of the polynomial $$4-\dfrac{1}{2}x^2$$
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2
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$$2\sqrt{2}$$
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0
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4
Explanation
$$4-\dfrac{1}{2}x^2=0$$
$$\dfrac{1}{2}x^2=4$$
$$x^2=8$$
$$\therefore x=\pm 2\sqrt{2}$$
$$\therefore$$ The zeroes of the given polynomial $$2\sqrt{2}$$.
Which of the following are the zeroes of the quadratic polynomial $$9-4x^2$$?
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4
0%
9
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$$\dfrac{3}{2}$$
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$$\dfrac{2}{3}$$
Explanation
$$9-4x^2=0$$
$$4x^2=9$$
$$x^2=\dfrac{9}{4}$$
$$\therefore x=\pm \dfrac{3}{2}$$
$$\therefore$$ The zeroes of the given polynomial are $$\dfrac{3}{2}$$and $$-\dfrac{3}{2}$$.
The zeroes of the polynomials, $$t^2 15$$ are.
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$$\pm \sqrt{15}$$
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$$\pm \sqrt{5}$$
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$$\pm \sqrt{3}$$
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$$\pm 3$$
Explanation
$$t^2 – 15=0$$
$$t^2=15$$
$$t=\pm \sqrt{15}$$
The degree of a quadratic polynomial is:
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$$0$$
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$$1$$
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$$2$$
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$$3$$
Explanation
The highest power of the variable in a polynomial in one variable is called the
degree
of the polynomial.
Quadratic polynomial is a polynomial in which highest order of the variable is $$2$$.
For example, $$2x^2+5x+1=0$$ is a quadratic equation and it has a degree $$2$$.
Thus, degree of quadratic polynomial is $$2$$.
Therefore, option $$C$$ is correct.
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