We have two equations,$4(x^{2}+y^{2}+z^{2}) = a$ ---(1)$4(x - y - z) = 3 + a$ ---(2)Substituting the value of a from equation 1 in equation 2, we get,$4\left(x\ -\ y\ -\ z\right)\ =\ 3\ +\ 4(x^2\ +\ y^2\ +\ z^2)$$\ 3\ +\ 4(x^2\ +\ y^2\ +\ z^2)\ -4\left(x\ -\ y\ -\ z\right)\ =\ 0$$\ 3\ +\ 4x^2\ +\ 4y^2\ +\ 4z^2\ -4x\ \ +\ 4y\ +\ 4z\ =\ 0$It can be written as,$4x^2\ -4x\ +\ 1+\ 4y^2\ +\ 4y\ +\ 1\ +\ 4z^2\ +\ 4z\ +\ 1\ =\ 0$$\left(2x\ -\ 1\right)^2\ +\ \left(2y\ +\ 1\right)^2\ +\ \left(2z\ +\ 1\right)^2\ \ =0$We know that if the sum of the&nbsp;squares of terms is 0, then all the terms must be equal to 02x - 1 = 0x =&nbsp;$\dfrac{1}{2}$2y +&nbsp;1 = 0y =&nbsp;$-\dfrac{1}{2}$2z + 1 = 0z = $-\dfrac{1}{2}$Substituting the values in equation 2, we get,$4\left(\dfrac{1}{2}\ -\ \left(-\dfrac{1}{2}\right)\ -\ \left(-\dfrac{1}{2}\right)\right)\ =\ 3\ +\ a$$4\left(\dfrac{3}{2}\right)\ =\ 3\ +\ a$$6\ =\ 3\ +\ a$$\ a\ =\ 3$Therefore, the correct answer is option A.

When given more than one equations, stating the fact that there is a common root,&nbsp;We need to equate the two equations to get discernible values for $x$Here, we are given three equations with the values of $m$, $n$$x^2+mx+9=x^2+\left(m+n\right)x+35$$mx+9=mx+nx+35$$nx=-26$Similarly, we can do it for the other equation as well,&nbsp;$x^2+nx+17=x^2+\left(m+n\right)x+35$$mx=-18$Substituting the value of either $mx$ or $nx$ in the original equations, we get&nbsp;$x^2-18+9=0$$x^2=9$$x=\pm\ 3$Since we are given that the root is negative,&nbsp;$x=-3$$n=-\dfrac{26}{-3}$$m=-\dfrac{18}{-3}$$3n=26$$2m=12$$2m+3n=38$

From the sum and product of roots, we get:&nbsp;$\alpha\ +\beta\ =-\dfrac{\lambda}{3}$ and&nbsp;$\alpha\ \beta\ =-\dfrac{1}{3}$Simplifying the expression given in the question, we get:&nbsp;$\dfrac{\alpha^2+\beta^2\ }{\alpha^2\beta^2\ }=15$and substituting the denominator's value as 1/9, we get:$\alpha^2+\beta^2\ =\dfrac{15}{9}$We want the expression&nbsp;$\alpha^3+\beta^3\ $, so multiplying both sides by&nbsp;$\alpha+\beta$, we get:$\alpha^3+\beta^3+\alpha\beta\left(a+\beta\ \right)=\dfrac{15}{9}\left(\alpha\ +\beta\ \right)$$\alpha^3+\beta^3+\dfrac{\lambda}{9}\ =\dfrac{15}{9}\left(-\dfrac{\lambda}{3}\ \right)$$\alpha^3+\beta^3+\dfrac{\lambda}{9}\ =-\dfrac{5\lambda}{9}-\dfrac{\lambda}{9}=-\dfrac{2\lambda\ }{3}\ \ $We would still need to find the value of $\lambda$This we can do from the initial relation we had:$\alpha^2+\beta^2\ =\dfrac{15}{9}$$\alpha^2+\beta^2\ =\left(\alpha+\beta\right)^2-2\alpha\ \beta\ \ \ =\dfrac{15}{9}$$\dfrac{\lambda^2}{9}+\frac{2}{3}\ \ \ =\dfrac{15}{9}$$\dfrac{\lambda^2}{9}\ =\dfrac{15-6}{9}=\dfrac{9}{9}=1$This would finally give us&nbsp;$\lambda^2=9$Using this in our required expression, we get:$\left(\alpha^3+\beta^3\right)^2=\left(-\dfrac{2\lambda}{3}\ \ \right)^2=\dfrac{4\times\ 9}{9}=4$Therefore, Option C is the correct answer.&nbsp;

In such questions, we should be trying to complete the squares.&nbsp;We see a $xy$ term; we need to accommodate that in a square that has both x and y terms.&nbsp;Since there is only one other term with x, we also need to have it entirely in the square.&nbsp;$\left(2x-y\right)^{^2}=4x^2+y^2-4xy$Using this in the given equation, we are left with&nbsp;$\left(2x-y\right)^{^2}+3y^2+3-6y$This&nbsp;can be written as&nbsp;$\left(2x-y\right)^{^2}+3\left(y^2+1-2y\right)$$\left(2x-y\right)^{^2}+3\left(y-1\right)^2=0$Since both the squares add up to 0, this is only possible when the squares themselves are 0This would give us y=1 from the second term, and using that, we get x= 1/2 from the first term.&nbsp;Therefore the value of 4x+5y will be 2+5 = 7Hence, 7 is the correct answer.&nbsp;

Given a and b are the distinct roots of the equation $2x^{2} - 6x + k = 0$⇒ a + b = -(-6/2) = 3 (Sum of the roots)⇒ ab = k/2 (Product of the roots)Now, (a+b) and ab are the roots of the quadratic equation $x^{2} + px + p = 0$⇒ a + b + ab = -p ⇒ 3 + k/2 = -p ---(1)⇒ (a + b)(ab) = p ⇒ 3(k/2) = p ---(2)$3+\dfrac{k}{2}=-\dfrac{3k}{2}$ ⇒ 2k = -3 ⇒ k = $-\dfrac{3}{2}$p = $\dfrac{3k}{2}=\dfrac{3}{2}\left(-\dfrac{3}{2}\right)=-\dfrac{9}{4}$⇒ 8(k-p) = $8\left(-\frac{3}{2}+\frac{9}{4}\right)=-12+18=6$

Given that -2 is a root of the given cubic equation.=&gt; Dividing the given equation by (x + 2), using the Horner's method of synthetic division:coefficient of $x^2$ is 1, and coefficient of x is (2r+1)-2 = 2r-1 and the constant term = (4r-1)-2(2r-1) = 1.=&gt; The quadratic obtained by dividing the cubic = $x^2 + (2r - 1)x + 1 = 0$, Since, this equation has 2 real roots =&gt; Discriminant should be greater than 0=&gt; $(2r - 1)^2 \gt 4 $ =&gt; 2r-1 &gt; 2 or 2r-1 &lt; -2 =&gt; r &gt; 3/2 or r &lt; -1/2.=&gt; Minimum possible non-negative integer value of r is 2.

It is given that $ 2^{4x^2} - 2^{2x^2+x+16} + 2^{2x+30} = 0 $, which can be written as:=&gt; $ \left(2^{2x^2}\right)^2 - 2^{2x^2} \cdot 2^{x+15} \cdot 2^1 + \left(2^{x+15}\right)^2 = 0 $=&gt; $ \left(2^{2x^2} - 2^{x+15}\right)^2 = 0 $=&gt; $ 2^{2x^2} - 2^{x+15} = 0 $ (Since $ (a-b)^2 = 0 \Rightarrow a-b = 0 $)=&gt; $ 2x^2 = x + 15 $=&gt; $ 2x^2 - x - 15 = 0 $=&gt; $ 2x^2 - 6x + 5x - 15 = 0 $=&gt; $ 2x(x-3) + 5(x-3) = 0 $=&gt; $ (2x+5)(x-3) = 0 $Hence, the possible values of x are $ -\frac{5}{2} $, and 3, respectively.Therefore, the sum of the possible values is $ \left(3 - \frac{5}{2}\right) = \frac{1}{2} $The correct option is C

It is given that $ (x-1)^2 + 2kx + 11 = 0 $ has no real roots. (Where k is the largest integer)$ (x-1)^2 + 2kx + 11 = 0 $, which can be written as:=&gt; $ x^2 - 2x + 1 + 2kx + 11 = 0 $=&gt; $ x^2 - 2(k-1)x + 12 = 0 $We know that for no real roots, $ D $ &lt; $ 0 $ =&gt; $ b^2 - 4ac $ &lt; $ 0 $Hence, $ {(2(k-1))}^2 - 4 \cdot 1 \cdot 12$ &lt; $ 0 $=&gt; $ 4(k-1)^2 $ &lt; $ 48 $=&gt; $ (k-1)^2 $ &lt; $ 12 $Since k is an integer, it implies (k-1) is also an integer.Therefore, from the above inequality, we can say that the largest possible value of (k-1) = 3=&gt; The largest possible value of k is 4.Now we need to calculate the least possible value of $ \frac{k}{4y} + 9y $.$ \frac{k}{4y} + 9y $ can be written as $ \frac{4}{4y} + 9y = \frac{1}{y} + 9y $The least possible value of $ 9y + \frac{1}{y} $ can be calculated using A.M-G.M inequality.Using A.M-G.M inequality, we get:$ \frac{9y + \frac{1}{y}}{2} \geq \sqrt{9y \times \frac{1}{y}} $=&gt; $ \frac{9y + \frac{1}{y}}{2} \geq \sqrt{9} $=&gt; $ \frac{9y + \frac{1}{y}}{2} \geq 3 $=&gt; $ 9y + \frac{1}{y} \geq 6 $Hence, the least possible value is 6

It is given that $ x^2 + bx + c = 0 $ has two real roots. Let the roots of the equation be $ \alpha, \beta $. ($ \alpha$ &gt; $\beta $)Then, we can say that $ \frac{1}{\alpha} - \frac{1}{\beta} = \frac{1}{3} $ .... Eq(1)Similarly, $ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{5}{9} $ .... Eq(2)Eq(2) can be written as $ \left(\frac{1}{\alpha} - \frac{1}{\beta}\right)^2 + 2 \cdot \frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{5}{9} $=&gt; $ \left(\frac{1}{3}\right)^2 + 2 \cdot \frac{1}{\alpha \cdot \beta} = \frac{5}{9} $=&gt; $ \frac{2}{\alpha \cdot \beta} = \frac{4}{9} \Rightarrow \frac{1}{\alpha \cdot \beta} = \frac{2}{9} $=&gt; $ \alpha \cdot \beta = \frac{9}{2} $We know that the product of the roots is equal to c, which implies $ c = \frac{9}{2} $We also know that the sum of the roots is equal to -b.=&gt; $ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \left(\frac{1}{\alpha} + \frac{1}{\beta}\right)^2 - \frac{2}{\alpha \beta} = \frac{5}{9} $=&gt; $ \left(\frac{\alpha + \beta}{\alpha \beta}\right)^2 - \frac{4}{9} = \frac{5}{9} $=&gt; $ \left(\frac{\alpha + \beta}{\alpha \beta}\right)^2 = (1)^2 $=&gt; $ \alpha + \beta = \pm \alpha \beta $Hence, the maximum value of b is $ \frac{9}{2} $.Hence, the maximum value of (b+c) is 9

$b^2$ &lt; $4ac$ means that the discriminant is less than 0. Therefore, f(x)&gt;0 for all x if the coefficient of $x^2$ is positive, and f(x)&lt;0 for all x if the coefficient of $x^2$ is negative.We are given that f(m)&lt;0 and m is an integer.So the set containing values of m will either be empty if the coefficient of $x^2$ is positive, or it will be a set of all

Let the roots of the given equation $5x^3 + cx^2 - 10x + 9 = 0$ be r, −r and pr − r + p = $-\frac{c}{5}$p = $-\frac{c}{5}$ …… (1)$-r^2 - pr + pr = -2$$r^2 = 2$ …… (2)$-r^2 p = -\frac{9}{5}$$p = \frac{9}{10}$ …… (3)Substituting p in (1), we get$\frac{9}{10} = -\frac{c}{5}$$-\frac{9}{2} = c$

$2x^2 + kx + 5 = 0$ has no real roots so $D$ &lt; $0$$k^2 - 40$ &lt; $0$$(k - \sqrt{40})(k + \sqrt{40})$ &lt; $0$$k \in (-\sqrt{40}, \sqrt{40})$$x^2 + (k - 5)x + 1 = 0$ has two distinct real roots so $D$ &gt; $0$$(k - 5)^2 - 4$ &gt; $0$$k^2 - 10k + 21$ &gt; $0$$(k - 3)(k - 7)$ &gt; $0$$k \in (-\infty, 3) \cup (7, \infty)$Therefore possible value of k are -6, -5, -4, -3, -2, -1, 0, 1, 2

Let $\frac{x^2 - 6x + 10}{3 - x} = p$$x^2 - 6x + 10 = 3p - px$$x^2 - (6 - p)x + 10 - 3p = 0$Since the equation will have real roots,$(6 - p)^2 - 4 \times (10 - 3p) \ge 0$$p^2 - 12p + 12p + 36 - 40 \ge 0$$p^2 \ge 4$$p \ge 2 ,\ p \le -2$Now, when $p = -2$, $x = 4$. Since it is given that $x &lt; 3$, thus this value will be discarded.Now, $\frac{1}{2}$ and $-\frac{1}{2}$ do not come in the mentioned range.When $p = 2$, $x = 2$Thus, the minimum possible value of p will be 2.Thus, the correct option is C.Alternate explanation:Since $x &lt; 3$,$3 - x &gt; 0$Let $3 - x = y$. So, $y &gt; 0$.Now, $\frac{x^2 - 6x + 10}{3 - x} = \frac{x^2 - 6x + 9 + 1}{3 - x}$$\Rightarrow \frac{(3 - x)^2 + 1}{3 - x}$Since $3 - x = y$, the equation will transform to $\frac{y^2 + 1}{y}$ or $y + \frac{1}{y}$The minimum value of the expression $y + \frac{1}{y}$ for $y &gt; 0$ will be at $y = 1$i.e. Minimum value = $1 + 1 = 2$Thus, the correct option is C.

$a, b, c, m$ and $n$ are integers so if one root is $3 + 2\sqrt{2}$ then the other root is $3 - 2\sqrt{2}$Sum of roots $= 6 = -\dfrac{b}{a}$ or $b = -6a$Product of roots $= 1 = \dfrac{c}{a}$ or $c = a$$a, b, c, m$ and $n$ are integers so if one root is $4 + 2\sqrt{3}$ then the other root is $4 - 2\sqrt{3}$Sum of roots $= 8 = -\dfrac{m}{a}$ or $m = -8a$product of roots $= 4 = \dfrac{n}{a}$ or $n = 4a$($\frac{b}{m} + \frac{(c-2b)}{n}$)= $\frac{6a}{8a} + \frac{a+12a}{4a}$ = $\frac34 + \frac{13}{4} = \frac{16}{4} = 4$

Given a, b, c are rational numbers.Hence a, b, c are three numbers that can be written in the form of p/q.Hence if one both the root is $2 + \sqrt3$ and considering the other root to be x.The sum of the roots and the product of the two roots must be rational numbers.For this to happen the other root must be the conjugate of&nbsp;$2 + \sqrt3$so the sum and the product of the roots are rational numbers which are represented by:&nbsp;-$\frac{b}a$,&nbsp;$\frac{c}a$Since the quadratic equation has negative b, it will cancel the negative sign of the sum of the roots. Hence, the sum of the roots and the products of the roots will be represented by&nbsp;-$\frac{b}a$,&nbsp;$\frac{c}a$Hence the sum of the roots is $2 + \sqrt3 + 2 - \sqrt3 = 4$ The product of the roots is&nbsp;$(2 + \sqrt3) + (2 - \sqrt3) = 1$ b/a = 4, c/a = 1.b = 4*a, c = a.Since b = $c^3$4*a = $a^3$$a^2$ = 4a = 2 or -2.|a| = 2

CAT Quadratic Equations

Question 1.

Question 2.

Question 3.

Question 4.

Question 5.

Question 6.

Question 7.

Question 8.

Question 9.

Question 10.

Question 11.

Question 12.

Question 13.

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Free CAT Quant Questions

CAT Quadratic Equations

Question 1.

<div><p>Let $x, y,$ and $z$ be real numbers satisfying $4(x^{2}+y^{2}+z^{2})=a$, $4(x-y-z)=3+a$ The a equals</p></div>

Question 2.

<div><p>lf the equations $x^{2}+mx+9=0, x^{2}+nx+17=0$ and $x^{2}+(m+n)x+35=0$ have a common negative root, then the value of $(2m+3n)$ is</p></div>

Question 3.

<div><p>The roots $\alpha, \beta$ of the equation $3x^2 + \lambda x - 1 = 0$, satisfy $\cfrac{1}{\alpha^2} + \cfrac{1}{\beta^2} = 15$.The value of $(\alpha^3 + \beta^3)^2$, is</p></div>

Question 4.

<div><p>If x and y are real numbers such that $4x^2 + 4y^2 - 4xy - 6y + 3 = 0$, then the value of $(4x + 5y)$ is</p></div>

Question 5.

<div><p>Let &alpha; and &beta; be the two distinct roots of the equation $2 x^2-6 x+k=0$, such that (&alpha;+&beta;) and &alpha;&beta; are the distinct roots of the equation $x^2+p x+p=0$. Then, the value of 8(k&minus;p) is</p></div>

Question 6.

<div><p>The equation $x^3+(2 r+1) x^2+(4 r-1) x+2=0$ has -2 as one of the roots. If the other two roots are real, then the minimum possible non-negative integer value of r is</p></div>

Question 7.

<div><p>The sum of all possible values of x satisfying the equation $2^{4 x^2}-2^{2 x^2+x+16}+2^{2 x+30}=0$, is</p></div>

Question 8.

<div><p>Let k be the largest integer such that the equation $(x-1)^2+2 k x+11=0$ has no real roots. If y is a positive real number, then the least possible value of $\frac{k}{4 y}+9 y$ is</p></div>

Question 9.

<div><p>A quadratic equation $x^2+b x+c=0$ has two real roots. If the difference between the reciprocals of the roots is $\frac{1}{3}$, and the sum of the reciprocals of the squares of the roots is $\frac{5}{9}$, then the largest possible value of (b + c) is</p></div>

Question 10.

<div><p>Let a,b,c be non-zero real numbers such that $b^2 \lt 4 a c$ and $f(x)=a x^2+b x+c$. If the set S consists of al integers m such that f(m) &lt; 0, then the set S must necessarily be</p></div>

Question 11.

<div><p>Let r and c be real numbers. If r and &minus;r are roots of $5 x^3+c x^2-10 x+9=0$, then c equals</p></div>

Question 12.

<div><p>Suppose k is any integer such that the equation $2 x^2+k x+5=0$ has no real roots and the equation $x^2+(k-5) x+1=0$ has two distinct real roots for x . Then, the number of possible values of k is</p></div>

Question 13.

<div><p>The minimum possible value of $\frac{x^2-6 x+10}{3-x}$, for x&lt;3, is</p></div>

Question 14.

<div><p>If $(3+2 \sqrt{2})$ is a root of the equation $a x^2+b x+c=0$, and $(4+2 \sqrt{3})$ is a root of the equation $a y^2+m y+n=0$, where a, b, c, m and n are integers, then the value of $\left(\frac{b}{m}+\frac{c-2 b}{n}\right)$ is</p></div>

Question 15.

<div><p>Suppose one of the roots of the equation a $x^2$ - b x + c = 0 is 2 + &radic;3, where a, b and c are rational numbers and a &ne; 0. If b = $c^3$ then |a| equals</p></div>

Question 16.

<div><p>The number of non-negative integer values of $k$ for which the quadratic equation $x^2 - 5x + k = 0$ has only integer roots, is .</p></div>

Free CAT Quant Questions