Solve the following Question.(1 Marks)

Question 11 Mark

Show the following quadratic equation by factorization method:
$\text{x}^2+2\text{x}+5=0$

Answer

$x^2 + 2x + 5 = 0$
Now, completing the squares, we get
$(x + 1)^2 + 4 = 0$
$\Rightarrow (x + 1)^2 - 2i^2 = 0$
$\Rightarrow (x + 1 + 2i) (x + 1 - 2i) = 0$
$\Rightarrow (x + 1 + 2i) = 0$ or $(x + 1 - 2i) = 0$
$\therefore$ $x = -1 = 2i, -1 + 2i$

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Question 21 Mark

If roots $\alpha,\beta$ of the equation $x^2− px + 16 = 0$ satisfy the relation $\alpha^2+\beta^2= 9,$ then write the value p.

Answer

$\alpha,\beta$ are the roots of the equation $x^2 - px + 16 = 0$
$\Rightarrow\alpha+\beta=\frac{-\text{b}}{\text{a}}=\frac{-(-\text{p})}{1}=\text{p}$
and $\alpha,\beta=\frac{\text{c}}{\text{a}}=\frac{16}{1}=16$
Now,
$\alpha^2+\beta^2=9$
$\Rightarrow(\alpha+\beta)^2-2\alpha\beta=9$
$\Rightarrow\text{p}^2-2\times16=9$
$\Rightarrow\text{p}^2=9+32$
$\Rightarrow\text{P}=\sqrt{41}$

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Question 31 Mark

If a and b are roots of the equation $x^2− x + 1 = 0,$ then write the value of $a^2 + b^2.$

Answer

The given equation is $x^2− x + 1 = 0 ....(i)$
a and b are the roots of $(i)$
$\therefore a + b = -1 ...(ii)$
and $ab = 1$
Now $a^2 + b^2 = (a + b)^2 - 2ab$
$= (-1)^2 - 2.1$
$= 1 - 2$
$= -1$

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Question 41 Mark

Show the following quadratic equation by factorization method:
$9\text{x}^2+4=0$

Answer

$9x^2 + 4 = 0$
$\Rightarrow (3x)^2 - (2i^2) = 0$ [$\because$ $i^2 = -1$]
$\Rightarrow (3x + 2i) (3x - 2i) = 0$
$\Rightarrow\text{x}=\frac{-2}{3}\text{i}\ \text{or}\text{ x}=\frac{2}{3}\text{i}$
$\therefore\text{x}=\frac{-2}{3}\text{ i},\frac{2}{3}\text{ i}$

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Question 51 Mark

If $2+\sqrt{3}$ is root of the equation $x^2+ px + q = 0$, than write the values of $p$ and $q$.

Answer

Since,
a root of the equation
$x^2+ px + q = 0 .......(i)$
If $2+\sqrt{3}$ in one of the solution (roots) of in $2-\sqrt{3}$ will be other roots
Now, sum of roots $=\frac{-\text{b}}{\text{a}}$
$\Rightarrow(2+\sqrt{3})+(2-\sqrt{3})=-\text{p}$
$\Rightarrow4=-\text{p}$
and product of root $=\frac{\text{c}}{\text{a}}$
$\Rightarrow(2+\sqrt{3})(2-\sqrt{3})=2$
$\Rightarrow 4 - 3 = q$
$\Rightarrow q = 1$
Thus,
$p = -4, q = 1$

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Question 61 Mark

If the difference between the roots of the equation is $x^2 + ax + 8 = 0$ is $2$, write the values of a.

Answer

The given equation in $x^2 + ax + 8 = 0 ....(i)$
Let $\alpha$ and $\beta$ are the two roots of (i) then $\alpha+\beta=\text{a}\ \&\ \alpha\beta=8$
we have given $\alpha -\beta=2$
Now $(\alpha-\beta)^2+4\alpha\beta=(\alpha+\beta)^2$
$\Rightarrow2^2+4.8=(\alpha+\beta)^2$
$\Rightarrow4+32=(\alpha+\beta)^2$
$\Rightarrow\alpha+\beta=\pm6$
But $\alpha+\beta=\text{a}$
$\therefore\alpha=\pm6$

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Question 71 Mark

Write roots of the equation $(a − b) x^2+ (b − c) x + (c − a) = 0.$

Answer

The given equation in $(a − b) x^2+ (b − c) x + (c − a) = 0 ....(i)$
Let $\alpha$ and $\beta$ are the roots of (i)
then, $\alpha+\beta=\frac{-\text{b}}{\text{a}}=\frac{-(\text{b}-\text{c})}{\text{a}-\text{b}}\ ...(\text{ii})$
and $\alpha\beta=\frac{\text{c}}{\text{a}}=\frac{\text{c}-\text{a}}{\text{a}-\text{b}}$
Now, $(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta$
$\Rightarrow\Big(-\frac{\text{b}-\text{c}}{\text{a}-\text{b}}\Big)^2-4\Big(\frac{\text{c}-\text{a}}{\text{a}-\text{b}}\Big)$
$\Rightarrow\frac{(\text{b}-\text{c})^2-4(\text{c}-\text{a})(\text{a}-\text{b})}{(\text{a}-\text{b})^2}$
$\therefore\alpha-\beta=\frac{\sqrt{(\text{b}-\text{c})^2-4(\text{c}-\text{a})(\text{a}-\text{b}})}{(\text{a}-\text{b})}\ ...(\text{ii})$
Solving (i) and (ii)
$2\alpha=\frac{-(\text{b}-\text{c})}{\text{a}-\text{b}}+\frac{\sqrt{(\text{b}-\text{c})^2-4(\text{c}-\text{a})(\text{a}-\text{b}})}{\text{a}-\text{b}}$
$=\frac{-(\text{b}-\text{c})}{\text{a}-\text{b}}+\frac{\sqrt{(2\text{a}-\text{b}-\text{c})^2}}{\text{a}-\text{b}}$
$=\frac{2(\text{a}-\text{b})}{\text{a}-\text{b}}=2$
$\therefore\alpha=1$
From (ii) $\beta=\frac{\text{c}-\text{a}}{\text{a}-\text{b}}$

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Question 81 Mark

Write the number of real roots of the equation $(x - 1)^2+ (x - 2)^2+ (x - 3)^2 = 0$

Answer

We have,$(x - 1)^2+ (x - 2)^2+ (x - 3)^2 = 0$
$\Rightarrow x^2 - 2x + 1 + x^2- 4x + 4 + x^2 - 6x + 9 = 0$
$\Rightarrow 3x^2 - 12x + 14 = 0$
Now, $D = b^2 - 4ac$
$= (-12)^2 - 4.3.14$
$= -24 < 0$
$\because$ $D < 0$ so, no real roots.

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Question 91 Mark

If $\alpha,\beta$ are roots of the equation $x^2− a (x + 1) − c = 0$, then write the value of $(1+\alpha)(1+\beta)$

Answer

$\alpha,\beta$ are the roots of $x^2 - a (x + 1) - c = 0 ...(i)$
$\Rightarrow x^2 - ax - (a + c) = 0$
$\alpha+\beta=\frac{-\text{b}}{\text{a}}=\text{a}$ and $\alpha\beta=\frac{\text{c}}{\text{a}}=-(\text{a}+\text{c})\ ...(\text{ii})$
Now,
$(1+\alpha)(1+\beta)=1+(\alpha+\beta)+\alpha\beta$
$=1+\text{a}+(-\text{a}-\text{c})$
$=1-\text{c}$

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Question 101 Mark

If $\alpha,\beta$ are roots of the equation $x^2 + lm + m = 0$, write an equation whose roots are $-\frac{1}{\alpha}$ and $-\frac{1}{\beta}$

Answer

$\alpha$ and $\beta$ are the roots of $x^2 + lm + m = 0 ...(i)$
$\therefore\alpha+\beta=\frac{-\text{b}}{\text{a}}=-\text{l}$
and $\alpha\beta=\frac{\text{c}}{\text{a}}=\text{m}$
now,
The quadratic equation whose roots are $-\frac{1}{\alpha}$ and $-\frac{1}{\beta}$ is $x^2$ - (sum of roots)$x$ + (product of roots) $= 0$
$\Rightarrow\text{x}^2-\Big(-\frac{1}{\alpha}+\frac{-1}{\beta}\Big)+\Big(-\frac{1}{\alpha}.\frac{-1}{\beta}\Big)=0$
$\Rightarrow\text{x}^2+\Big(\frac{\alpha+\beta}{\alpha\beta}\Big)\text{x}+\frac{1}{\alpha\beta}=0$
$\Rightarrow\text{x}^2+\Big(\frac{-\text{I}}{\text{m}}\Big)\text{x}+\frac{1}{\text{m}}=0$
$\Rightarrow\text{mx}^2-\text{lx}+1=0$

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Question 111 Mark

Write the number of quadratic equations, with real roots, which do not change by squaring their roots.

Answer

Let a and b be the real roots of the quadratic equation.
we need to find the number of quadratic equation such that they ramain unchanged even if roots are squared.
$a^2=a \text { and } b^2=b$
$\Rightarrow a(a-1)=0 \text { and } b(b-1)=0$
$\Rightarrow a=0 \text { or } a=1 \text { and } b=0 \text { or } b=1$
so we have four pairs of roots $(0,0),(0,1),(1,0),(1,1)$
For $(0,0)$
$(x-0)(x-0)=x^2$
for $(0,1)$
$(x-0)(x-1)=x(x-1)=x^2-1$
For $(1,0)$
$(x-1)(x-0)=(x-1) x=x^2-1$
For $(1,1)$
$(x-1)(x-1)=(x-1)^2=x^2-2 x+1$
So there are $3$ quadratic equations with real roots, which do not change by squaring their roots.

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Question 121 Mark

If a and b are roots of the equation $x^2− px + q = 0$, than write the value of $\frac{1}{\text{a}}+\frac{1}{\text{b}}.$

Answer

Since,
a, b are the roots of $x^2 - px + q = 0$
$\Rightarrow\text{a}+\text{b}=-\Big(\frac{-\text{P}}{1}\Big)=\text{P}$
$\text{a.b}=\frac{\text{q}}{1}=\text{q}$
Now,
$\frac{1}{\text{a}}+\frac{1}{\text{b}}=\frac{\text{a}+\text{b}}{\text{a.b}}=\frac{\text{P}}{\text{q}}$
$\therefore\frac{1}{\text{a}}+\frac{1}{\text{b}}=\frac{\text{p}}{\text{q}}$

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Question 131 Mark

Show the following quadratic equation by factorization method:
$x^2 + 1 = 0$

Answer

$x^2 + 1 = 0$
$\Rightarrow x^2+ i^2 = 0$ [$\because$ $i^2 = -1$]
$\Rightarrow (x + i) (x - i) = 0 [a^2 - b^2 = (a + b) (a - b)]$
$\Rightarrow x = i, -i$

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Maths Solve the following Question.(1 Marks)

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