Free Objective Test 01 Practice Test - 11th and 12th 

Let $\alpha$, ${\alpha }^n$ be two roots,

Then $\alpha +{\alpha }^n$ = -$\frac{b}{a}$, $\alpha {\alpha }^n$ = $\frac{c}{a}$

Eliminating $\alpha$, we get ($\frac{c}{a}{)}^{\frac{1}{n+1}}$ + ($\frac{c}{a}{)}^{\frac{n}{n+1}}$ = -$\frac{b}{a}$

⇒ $a.a^{-\frac{1}{n+1}}$. $c^{\frac{1}{n+1}}$ + $a.a^{-\frac{n}{n+1}}$. $c^{\frac{n}{n+1}}$ = -b

or $(a^{n}c{)}^{\frac{1}{n+1}}$ + $(a{c}^n{)}^{\frac{1}{n+1}}$ = -b

Let the roots be $\alpha$ and n $\alpha$

Sum of roots, α + nα = -$\frac{b}{a}$ ⇒ α = -$\frac{b}{a(n+1)}$    ......(i)

and product, α.n.α = $\frac{c}{a}$ ⇒ ${\alpha }^2$ = $\frac{c}{na}$     ......(ii)

From (i) and (ii), we get

⇒ $[$ -$\frac{b}{a(n+1)}$ ${]}^2$ = $\frac{c}{na}$ ⇒ $\frac{b^2}{a^2(n+1{)}^2}$ = $\frac{c}{na}$

⇒ $n{b}^2$ = ac$(n+1{)}^2$.

Note : Students should remember this question as a fact.

Let the equation (in written form) be $x^2$ + 17$x$ + q = 0.
Roots are -2, -15.
So q = 30,
And correct equation is $x^2$ + 13$x$ + 30 = 0.
Hence roots are -3, -10.

We know that the expression a$x^2$ + b$x$ + c > 0 for all $x$, if a > 0 and $b^2$ < 4ac

∴ $(a^2-1)$$x^2$ + 2(a-1)$x$ + 2 is positive for all $x$, if $a^2-1$ > 0 and 4$(a-1{)}^2-8(a^2-1)$ < 0

⇒ $a^2-1$ > 0 and -4(a - 1)(a + 3) < 0

⇒ $a^2-1$ > 0 and (a - 1)(a + 3) > 0

⇒ $a^2$ > 1 and a < -3 or a > 1

⇒ a < -3 or a > 1

The given equations are

q$x^2$ + p$x$ + q = 0    ........(i)

and $x^2$ - 4q$x$ + $p^2$ = 0    ........(ii)

Roots of (i) are complex, therefore $p^2-4q^2$ < 0

Now discriminant of (ii) is

$16q^2-4p^2=-4(p^2-4q^2)$ > 0

Hence, roots are real and unequal.

From $k=\frac{x^2-x+1}{x^2+x+1}$

We have $x^2(k-1)+x(k+1)+k-1=0$

As given, x is real  ⇒ $(k+1{)}^2-4(k-1{)}^2\ge 0$

⇒ $3k^2-10k+3\le 0$

Which is possible only when the value of k lies between the roots of the equation $3k^2-10k+3=0$

That is, when $\frac{1}{3}\le k\le 3$ {Since roots are $\frac{1}{3}$ and 3}

 Try making perfect square the given equation by adding and subtracting  $\frac{b^2}{4a^2}$
on solving it we'll find 
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Given that, α + β = 0

α + β + γ = -p ⇒ γ = -p

Substituting γ = -p in the given equation

⇒ $-p^3+p^3-pq+r=0$ ⇒ pq =r

If $\alpha ,\beta$, γ are the roots of the equation

$\alpha +\beta +\gamma =0,\alpha \beta +\beta \gamma +\gamma \alpha =4,\alpha \beta \gamma =-1$

$therefore(\alpha +\beta {)}^{-1}+(\beta +\gamma {)}^{-1}+(\gamma +\alpha {)}^{-1}$
=$\frac{1}{-\gamma }+\frac{1}{-\alpha }+\frac{1}{-\beta }$
=$-\left(\frac{\alpha \beta +\beta \gamma +\gamma \alpha }{\alpha \beta \gamma }\right)$
=$-\left(\frac{4}{-1}\right)=4$

Let $y=x^2$. then $x=\sqrt{y}$

∴ $x^3+8=0$ ⇒ $y^{\frac{3}{2}}+8=0$

⇒ $y^3=64$ ⇒ $y^3-64=0$

Thus the equation having roots ${\alpha }^2,{\beta }^2$ and ${\gamma }^{2}is{x}^3-64=0$

Given equation is $x^3-3x+2=0$

⇒ $x^2(x-1)+x(x-1)-2(x-1)=0$

⇒ $(x-1)(x^2+x-2)=0$ ⇒ $(x-1)(x-1)(x+2)=0$

Hence roots are 1, 1, -2

We can solve this problem using options given - 
On putting x = 5, we get

$2(5{)}^5-14(5{)}^4+31(5{)}^3-64(5{)}^2+19\left(5\right)+130=0$

Hence x = 5 satisfies the given equation.

Thus 5 is a root of the equation.
Other options will not satisfy the given equation so the answer is x = 5.

Let the roots be $\frac{\alpha }{\beta },\alpha ,\alpha \beta ,\beta \ne 0$
Then the product of roots is ${\alpha }^3=-\frac{-1}{8}=\frac{1}{8}$ 
⇒ $\alpha =\frac{1}{2}$ and hence β = $\frac{1}{2}$
(Since, sum of roots $=\frac{\alpha }{\beta }+\alpha +\alpha \beta =14/8)$,
so roots are $1,\frac{1}{2}$, $\frac{1}{4}$

Alternate method: By inspecting through options, we get the numbers $1,\frac{1}{2}$, $\frac{1}{4}$ satisfying the given equation.

According to given condition,

$4a^2-4(10-3a)<0$  ⇒ $a^2+3a-10<0$

⇒ $(a+5)(a-2)<0\Rightarrow -5<a<2.$

Case 1: When x + 2 ≥ 0 i.e/ x ≥ -2,

Then given inequality becomes

$x^2-(x+2)+x>0$  ⇒ $x^2-2>0$  ⇒ $|x|>\sqrt{2}$

⇒ therefore, in this case the part of the solution set is

[-2,-$\sqrt{2}$) ∪ ($\sqrt{2}$, ∞).

Case 2: When x + 2 < 0 i.e. x < -2,

Then given inequality becomes $x^2+(x+2)+x>0$

⇒ $x^2+2x+2>0$ ⇒ $(x+1{)}^2+1>0$, which is true for all real x

Hence, the part of the solution set in this case is (-∞, -2). Combining the two cases, the solution set is (-∞, -2)∪[-2, -$\sqrt{2}$)∪($\sqrt{2},\infty )=(-\infty ,-\sqrt{2}$)∪($\sqrt{2}$, ∞)