Free Basic Algebra - 02 Practice Test - CAT 

$D=(ab{)}^2+4a^2{b}^2=5a^2{b}^2$

$a^2{b}^2$has to be greater than or equal to 0 (i.e.,)$D>0.$So,roots are always real.

 

Let the roots be a,b.

$ab=k,a+b=8$

$a^2+b^2=30$

$(a+b{)}^2=64,a^2+b^2+2ab=64$

$30+2k=64,k=17$

Substitute x by (x-2)
The equation becomes:$(x-2{)}^2-3(x-2)+2=0,x^2-7x+12=0.$

Substitute x by$\left(\frac{x}{2}\right)$

The equation becomes:${\left(\frac{x}{2}\right)}^2-3\left(\frac{x}{2}\right)+2=0,x^2-6x+8=0.$ 

Soln:

Interchange ‘a’ and ‘c’ in the equation$a{x}^2+bx+c=0$

The equation becomes:$2x^2-3x+1=0.$ Hence option (a)

Sum of the roots:$p+q=-5$

Product of the roots:$pq=2$

$\frac{p}{q}+\frac{q}{p}=\frac{(p^2+q^2)}{pq}=\frac{\left[\right(p+q{)}^2-2pq]}{pq}=\frac{\left[\right(p+q{)}^2]}{pq}-2=\frac{25}{2}-2=\frac{21}{2}$

For a quadratic equation$a{x}^2+bx+c=0$, if the roots are in ratio$m:n$then$mn{b}^2=(m+n{)}^{2}ac.$

In the equation given:

$4\times 3\times b^2=(4+3{)}^2(3\times 3)$

$12b^2=49\times 9$

$b=\frac{7\times 3}{\left(2\sqrt{3}\right)}=\frac{7\sqrt{3}}{2}$

The equations$x^2-kx-21=0$and$x^2-3kx+35=0$have a common root.

So, equating the two equations:

$x^2-kx-21=x^2-3kx+35$

$2kx=56$

$k=\frac{28}{x}$

Putting in the equation:

$x^2-28-21=0$

$x^2=49$

$x=+7$

$ork=+4,As,k>0,k=4.$ 

Both the roots are greater than 1 and a$>$0. So, the graph of the equation will be:

$f\left(x\right)=a{x}^2+bx+c$

$f\left(1\right)>a+b+c>0$

For the equation given
$D=(k+5{)}^2+4(1\left)\right(k+5)=k^2+10k+25+4k+20=k^2+14k+45$

For the equation to have two distinct roots:
$D>0$

$k^2+14k+45>0(k+9)(k+5)>0$

Value of k will be$(-\infty ,-9)\cup (-5,\infty )$

Soln:

$a+\frac{1}{a}={(\sqrt{a}-\frac{1}{\sqrt{a}})}^2+2$

For minimum value of$a+\frac{1}{a}$,we want minimum value of${(\sqrt{a}-\frac{1}{\sqrt{a}})}^2$that will be zero.

So, minimum value of$a+\frac{1}{a}$will be 2. Hence option (b)

2nd method:- By, $A.M\ge G.M\frac{a+\frac{1}{a}}{2}\ge \sqrt{(a\times \frac{1}{a}}thusa+\frac{1}{a}>2(i.e.,)(a+\frac{1}{a})$min will be 2 .  option (b).

Given equation:$x^2-3x+2=0$
with roots p, q

When the roots become 2p and 2q,
the equation becomes:
${\left(\frac{x}{2}\right)}^2-3\left(\frac{x}{2}\right)+2=0,x^2-6x+8=0$ 

When the roots become
$(2p+1)\&(2q+1)$
the equation becomes:

$(x-1{)}^2-6(x-1)+8=0,x^2-8x+15=0$.
Hence option (D)

$f\left(x\right)=x^8+6x^7-5x^4-3x^2+2x-5$

Check the number of positive roots= no. of sign changes in f(x) = 3

Check the number of negative roots = no. of sign changes in f(-x)

$f(-x)=x^8-6x^7-5x^4-3x^2-2x-5$. No. of sign changes = 1

Now check for zero as a root.$f\left(0\right)=-5\ne 0$

So, maximum number of real roots = 3 + 1 = 4. Hence option (a)

a and b are the roots of the equation

$a+b=(\alpha -4),ab=\alpha$

$5ab-a^2-b^2$

$5ab-(a^2+b^2)$

$5ab-\left[\right(a+b{)}^2-2ab]$

$5\alpha -\left[\right(\alpha -4{)}^2-2\alpha ]$

$5\alpha -[{\alpha }^2-10\alpha +16]$

$-{\alpha }^2+15\alpha -16=-({\alpha }^2+15\alpha -16)=-({\alpha }^2+16\alpha -\alpha -16)=-\left[\alpha \right(\alpha +16)-1(\alpha +16\left)\right]=-(\alpha +16)(\alpha -1)$

So,the maximum value$=\frac{D}{4a}=\frac{[{15}^2-4(-1\left)\right(-16\left)\right]}{4\times 1}=\frac{161}{4}.$

Let the two digit number be ab.

$a\times b=12$

When 9 is added to the number

$10a+b+9=10b+a$

$9a-9b+9=0$

$a-b+1=0$

$a-\frac{12}{a}+1=0$

$a^2+a-12=0$

$(a+4)(a-3)=0$

a = -4 is not accepted. So, a = 3, b = 4

The number is 34 .