Question 13 Marks
Solve the following quadratic equation:$3\text{x}^2-2\sqrt6\text{x}+2=0$
Answer$3\text{x}^2-2\sqrt6\text{x}+2=0$$\Rightarrow3\text{x}^2-\sqrt6\text{x}-\sqrt6\text{x}+2=0$
$\Rightarrow\sqrt3\text{x}\big(\sqrt3\text{x}-\sqrt2\big)-\sqrt2\big(\sqrt3\text{x}-\sqrt2\big)=0$
$\Rightarrow\big(\sqrt3\text{x}-\sqrt2\big)\big(\sqrt3\text{x}-\sqrt2\big)=0$
$\Rightarrow\big(\sqrt3\text{x}-\sqrt2\big)^2=0$
$\Rightarrow\sqrt3\text{x}-\sqrt2=0$
$\Rightarrow\text{x}=\frac{\sqrt2}{\sqrt3}$ (repeated root)
View full question & answer→Question 23 Marks
Solve the following quadratic equation:$x^2 + 5x - (a^2 + a - 6) = 0$
Answer$x^2 + 5x - (a^2 + a - 6) = 0$
$\Rightarrow x^2 + 5x - (a^2 + 3a - 2a - 6) = 0$
$\Rightarrow x^2 + 5x - [a(a + 3) - 2(a + 3)] = 0$
$\Rightarrow x^2 + 5x - [(a + 3)(a - 2)] = 0$
$\Rightarrow x^2 + (a + 3)x - (a - 2)x - (a + 3)(a - 2) = 0$
$\Rightarrow x[x + (a + 3)] - (a + 2)[x + (a + 3)] = 0$
$\Rightarrow [x + (a + 3)][x - (a - 2)] = 0$
$\Rightarrow x + (a + 3) = 0 or x - (a - 2) = 0$
$\Rightarrow x = -(a + 3) or x = a - 2$
View full question & answer→Question 33 Marks
Solve the following quadratic equation:$x^2 - (2b - 1)x + (b^2 - b - 20) = 0$
Answer$x^2 - (2b - 1)x + (b^2 - b - 20) = 0$
$\Rightarrow x^2 - (2b - 1)x + (b^2 - 5b + 4b - 20) = 0$
$\Rightarrow x^2 - (2b - 1)x + [b(b - 5) + 4(b - 5)] = 0$
$\Rightarrow x^2- (2b - 1)x + (b - 5)(b + 4) = 0$
$\Rightarrow x^2 - (2b - 1)x - (b + 4)x + (b - 5)(b + 4) = 0$
$\Rightarrow x[x - (b - 5)] - (b + 4)[x - (b - 5)] = 0$
$\Rightarrow [x - (b - 5)][x - (b + 4)] = 0$
$\Rightarrow x - (b - 5) = 0 or x - (b + 4) = 0$
$\Rightarrow x = b - 5 or x = b + 4$
View full question & answer→Question 43 Marks
Solve the following quadratic equation:$2x^2 + ax - a^2 = 0$
Answer$2x^2 + ax - a^2 = 0$
$\Rightarrow 2x^2 + 2ax - ax - a^2 + 2 = 0$
$\Rightarrow 2x(x + a) - a(x + a) = 0$
$\Rightarrow (x + a)(2x - a) = 0$
$\Rightarrow x + a = 0 or 2x - a = 0$
$\Rightarrow x = -a$, $\text{x}=\frac{\text{a}}{2}$
View full question & answer→Question 53 Marks
If -4 is a root of the quadratic equation $x^2+2 x+4 p=0$, find the value of $k$ for which the quadratic equation $x^2+p x(1+3 k)+7(3$ $+2 k)=0$ has equal roots.
AnswerIt is given that -4 is a root of the quadratic equation $x^2 + 2x + 4p = 0.$
$\therefore$ $(-4)^2 + 2 \times (-4) + 4p = 0$
$\Rightarrow 16 - 8 + 4p = 0$
$\Rightarrow 4p + 8 = 0$
$\Rightarrow p = -2$
The equation $x^2 + px(1 + 3k) + 7(3 + 2k) = 0$ has equal roots.
$\therefore$ $D = 0$
$\Rightarrow [p(1 + 3k)]^2 - 4 \times 1 \times 7(3 + 2k) = 0$
$\Rightarrow -[2(1 + 3k)]^2 - 28(3 + 2k) = 0$
$\Rightarrow 4(1 + 6k + 9k^2) - 28(3 + 2k) = 0$
$\Rightarrow 4(1 + 6k + 9k^2 - 21 - 14k) = 0$
$\Rightarrow 9k^2 - 8k - 20 = 0$
$\Rightarrow 9k^2 - 18k + 10k - 20 = 0$
$\Rightarrow 9k(k - 2) + 10(k - 2) = 0$
$\Rightarrow (k - 2)(9k + 10) = 0$
$\Rightarrow k - 2 = 0 or 9k + 10 = 0$
$\Rightarrow k = 2$ or $\text{k}=-\frac{10}{9}$
Hence, the required value of k is 2 or $-\frac{10}{9}.$
View full question & answer→Question 63 Marks
Solve the following quadratic equation:$\text{10x}-\frac{1}{\text{x}}=3$
Answer$\text{10x}-\frac{1}{\text{x}}=3$
$\Rightarrow 10x^2 - 1 = 3x$
$\Rightarrow 10x^2 - 5x + 2x + 1 = 0$
$\Rightarrow 5x(2x - 1) + 1(2x - 1) = 0$
$\Rightarrow (5x + 1)(2x - 1) = 0$
$\Rightarrow (5x + 1)(2x - 1) = 0$
$\Rightarrow 5x + 1 = 0 or 2x - 1 = 0$
$\Rightarrow\text{x}=\frac{-1}{5}$ or $\text{x}=\frac{1}{2}$
Hence, $\frac{-1}{5}$ and $\frac{1}{2}$ are the roots of the given equation.
View full question & answer→Question 73 Marks
Solve the following quadratic equation:$3x^2 - 2x - 1 = 0$
Answer$3x^2 - 2x - 1 = 0$
$\Rightarrow 3x^2 - 3x + 1x - 1 = 0$
$\Rightarrow 3x(x - 1) + 1(x - 1) = 0$
$\Rightarrow (x - 1)(3x + 1) = 0$
$\Rightarrow x - 1 = 0$ or $3x + 1 = 0$
$\Rightarrow x = 1$ or $\text{x}=\frac{-1}{3}$
Hence, 1 and $\frac{-1}{3}$ are the roots of the equation $3x^2 - 2x - 1 = 0$
View full question & answer→Question 83 Marks
Solve the following equations by using the method of completing the square:
$x^2 - 4x + 1 = 0$
Answer$x^2 - 4x + 1 = 0$
$\Rightarrow x^2 - 4x = -1$
$\Rightarrow x^2 - 2 \times x \times 2 + 2^2 = -1 + 2^2$ (Adding $2^2$ on both sides)
$\Rightarrow (x - 2)^2= -1 + 4 = 3$
$\Rightarrow\text{x}-2=\pm\sqrt3$ (Taking square root on both sides)
$\Rightarrow\text{x}-2=\sqrt3$ or $\text{x}-2=-\sqrt3$
$\Rightarrow\text{x}=2+\sqrt3$ or $\text{x}=2-\sqrt3$
Hence, $2+\sqrt3$ and $2-\sqrt3$ are the roots of the given equation.
View full question & answer→Question 93 Marks
Determine the values of p for which the quadratic equation $2x^2 + px + 8 = 0$ has real roots.
AnswerGiven:
$2x^2 + px + 8 = 0$
Here,
a = 2, b = p and c = 8
Discriminant D is given by:
$D = (b^2 - 4ac)$
$D = p^2- 4 \times 2 \times 8$
$D = (p^2 - 64)$
If $\text{D}\ge0,$ the roots of the equation will be real.
$\Rightarrow(\text{p}^2-64)\ge0$
$\Rightarrow(\text{p}+8)(\text{p}-8)\ge0$
$\Rightarrow\text{p}\ge8$ and $\text{p}\le-8$
Thus, the roots of the equation are real for $\text{p}\ge8$ and $\text{p}\le-8.$
View full question & answer→Question 103 Marks
Solve the following equations by using the method of completing the square:
$8x^2 - 14x - 15 = 0$
Answer$8x^2 - 14x - 15 = 0$
$\Rightarrow 16x^2 - 28x - 30 = 0$ (Multiplying both sides by 2)
$\Rightarrow 16x^2 - 28x = 30$
$\Rightarrow(\text{4x})^2-2\times\text{4x}\times\frac{7}{2}+\Big(\frac{7}{2}\Big)^2\\=30+\Big(\frac{7}{2}\Big)^2$ $\Big[$Adding $\Big(\frac{7}2{}\Big)^2$ on both sides$\Big]$
$\Rightarrow\Big(\text{4x}-\frac{7}{2}\Big)^2$
$=30+\frac{49}{4}$
$=\frac{169}{4}=\Big(\frac{13}{2}\Big)^2$
$\Rightarrow\text{4x}-\frac{7}{2}=\pm\frac{13}{2}$ (Taking square root on both sides)
$\Rightarrow\text{4x}-\frac{7}{2}=\frac{13}{2}$ or $\text{4x}-\frac{7}{2}=-\frac{13}{2}$
$\Rightarrow\text{4x}=\frac{13}{2}+\frac{7}{2}=\frac{20}{2}=10$ or $\text{4x}=-\frac{13}{2}+\frac{7}{2}=-\frac{6}{2}=-3$
$\Rightarrow\text{x}=\frac{5}{2}$ or $\text{x}=-\frac{3}4{}$
Hence, $\frac{5}{2}$ and $-\frac{3}{4}$ are the roots of the given equation.
View full question & answer→Question 113 Marks
Find the values of k for which the given quadratic equation has real and distinct roots:
$kx^2 + 6x + 1 = 0$
AnswerThe given equation is $kx^2 + 6x + 1 = 0$
$\therefore$ $D = 6^2 - 4 \times k \times 1$
$D = 36 - 4k$
The given equation has real and distinct roots if D > 0.
$\therefore$ $36 - 4k > 0$
$\Rightarrow 4k < 36$
$\Rightarrow k < 9$
View full question & answer→Question 123 Marks
Solve the following quadratic equation:$\text{2x}^2-\text{x}+\frac{1}{8}=0$
Answer$\text{2x}^2-\text{x}+\frac{1}{8}=0$
$\Rightarrow 16x^2 - 8x + 1 = 0$
$\Rightarrow 16x^2 - 4x - 4x + 1 = 0$
$\Rightarrow 4x(4x - 1) - 1(4x - 1) = 0$
$\Rightarrow (4x - 1)(4x - 1) = 0$
$\Rightarrow (4x - 1)^2 = 0$
$\Rightarrow 4x - 1 = 0$
$\Rightarrow\text{x}=\frac{1}{4}$ (repeated root)
View full question & answer→Question 133 Marks
If the roots of the equation $(c^2 - ab)x^2 - 2(a^2 - bc)x + (b^2 - ac) = 0$ are real and equal, show that either a = 0 or $(a^3 + b^3 + c^3) = 3abc.$
AnswerGiven:
$(c^2 - ab)x^2 - 2(a^2 - bc)x + (b^2 - ac) = 0$
Here,
$a = (c^2 - ab), b = -2(a^2 - bc), c = (b^2 - ac)$
It is given that the roots of the equation are real and equal; therefore, we have:
$D = 0$
$\Rightarrow (b^2 - 4ac) = 0$
$\Rightarrow {-2(a^2 - bc)}^2 - 4 \times (c^2 - ab) \times (b^2 - ac) = 0$
$\Rightarrow 4(a^4 - 2a^2bc + b^2c^2) - 4(b^2c^2 - ac^3 - ab^3 + a^2bc) = 0$
$\Rightarrow a^4 - 2a^2bc + b^2c^2 - b^2c^2 + ac^3 + ab^3 - a^2bc = 0$
$\Rightarrow a^4 - 3a^2bc + ac^3 + ab^3 = 0$
$\Rightarrow a(a^3 - 3abc + c^3 + b^3) = 0$
Now,
$a = 0 or a^3 - 3abc + c^3 + b^3 = 0$
$a = 0 or a^3 + b^3 + c^3 = 3abc$
View full question & answer→Question 143 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$36x^2 - 12ax + (a^2 - b^2) = 0$
AnswerThe given equation is $36x^2 - 12ax + (a^2 - b^2) = 0$. Comparing it with $Ax^2 + Bx + C = 0$, we get $A = 36, B = -12a$ and $C = a^2 - b^2$
$\therefore$ Discriminant, $D = B^2 - 4AC$
$= (-12a)^2 - 4 \times 36 \times (a^2 - b^2) = 144a^2 - 144a^2 + 144b^2 = 144b^2 > 0$ So, the given equation has real roots.Now, $\sqrt{\text{D}}=\sqrt{144\text{b}^2}=12\text{b}$
$\therefore\alpha=\frac{-\text{B}+\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-(-12\text{a}+12\text{b})}{2\times36}$
$=\frac{12(\text{a}+\text{b})}{72}$
$=\frac{\text{a}+\text{b}}{6}$
$\beta=\frac{-\text{B}-\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-(-12\text{a})-12\text{b}}{2\times36}$
$=\frac{12(\text{a}-\text{b})}{72}$
$=\frac{\text{a}-\text{b}}{6}$
Hence, $\frac{\text{a}+\text{b}}{6}$ and $\frac{\text{a}-\text{b}}{6}$ are the roots of the given equation.
View full question & answer→Question 153 Marks
The following are the roots of $3x^2 + 2x - 1 = 0?$
$-\frac{1}{2}$
AnswerThe given equation is $3x^2 + 2x - 1 = 0$
On substituting $\text{x}=-\frac{1}{2}$ in the equation, we get
$\text{LHS}=3\times\Big(\frac{-1}{2}\Big)^2+2\times\Big(\frac{-1}{2}\Big)-1=0$
$=\frac{3}{4}-1+1\neq0$
$\therefore\ \text{LHS}\neq\text{RHS}$
$\therefore\ \text{x}=\frac{-1}{2}$ is not a solution of $3x^2 + 2x - 1 = 0$
View full question & answer→Question 163 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$2x^2 + x - 4 = 0$
AnswerGiven, $2x^2 + x - 4 = 0$ On comparing it
with $ax^2 + bx + c = 0$,
we get a = 2, b = 1 and c = -4 Discriminant D is given by:
$D = (b^2 - 4ac) = (1)^2 - 4 \times 2 \times (-4) = 1 + 32 = 33 > 0$ So, the
given equation has real roots.
Now, $\sqrt{\text{D}}=\sqrt{33}$
$\therefore\ \alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-1+\sqrt{33}}{2\times2}$
$=\frac{-1+\sqrt{33}}{4}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-1-\sqrt{33}}{2\times2}$
$=\frac{-1-\sqrt{33}}{4}$
Hence $\frac{-1+\sqrt{33}}{4}$ and $\frac{-1-\sqrt{33}}{4}$ are the roots of the given equation.
View full question & answer→Question 173 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:$4\sqrt3\text{x}^2+5\text{x}-2\sqrt3=0$
AnswerThe given equation is:$4\sqrt3{\text{x}}^2+5\text{x}-2\sqrt3=0$
Comparing it with $ax^2+ bx + c = 0$, we get
$\text{a}=4\sqrt3,\ \text{b}=5$ and $\text{c}=-2\sqrt3$
$\therefore$ Discriminant, $\text{D}=\text{b}^2-4\text{ac}$
$=5^2-4\times4\sqrt3\times\big(-2\sqrt3\big)$
$=25+96$
$=121>0$
So, the given equation has real roots.
Now, $\sqrt{\text{D}}=\sqrt{121}=11$
$\therefore\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-5+11}{2\times4\sqrt3}$
$=\frac{6}{8\sqrt3}$
$=\frac{\sqrt3}{4}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-5-11}{2\times4\sqrt3}$
$=\frac{-16}{8\sqrt3}$
$=-\frac{2\sqrt3}{3}$
Hence, $\frac{\sqrt3}{4}$ and $-\frac{2\sqrt3}{3}$ are the roots of the given equation.
View full question & answer→Question 183 Marks
Solve the following equations by using the method of completing the square:
$5x^2 - 6x - 2 = 0$
Answer$5x^2 - 6x - 2 = 0 $
$\Rightarrow 25x^2 - 30x - 10 = 0$ (Multiplying both sides by 5)
$\Rightarrow 25x^2 - 30x = 10$
$\Rightarrow (5x)^2 - 2 \times 5x \times 3 + 3^2 = 10 + 3^2$ [Adding $3^2$ on both sides]
$\Rightarrow (5x - 3)^2 = 10 + 9 = 19$ $\Rightarrow\text{5x}-3=\pm19$ (Taking square root on both sides)$\Rightarrow\text{5x}-3=\sqrt{19}$ or $\text{5x}- 3=-\sqrt{19}$
$\Rightarrow\text{5x}=3+\sqrt{19}$ or $\text{5x}=3-\sqrt{19}$
$\Rightarrow\text{x}=\frac{3+\sqrt{19}}{5}$ or $\text{x}=\frac{3-\sqrt{19}}{5}$
Hence, $\frac{3+\sqrt{19}}{5}$ and $\frac{3-\sqrt{19}}{5}$ are the roots of the given equation.
View full question & answer→Question 193 Marks
Find the non-zero value of k for which the roots of the quadratic equation $9x^2 - 3kx + k = 0$ are real and equal.
AnswerThe given equation is $9x^2 - 3kx + k = 0.$
This is of the form $ax^2 + bx + c = 0$, where $a = 9, b = -3k$ and $c = k.$
$\therefore$ $D = b^2 - 4ac$
$D = (-3k)^2 - 4 \times 9 \times k$
$D = 9k^2 - 36k$
The given equation will have real and equal roots if D = 0.
$\therefore$ $9k^2 - 36k = 0$
$\Rightarrow 9k(k - 4) = 0$
$\Rightarrow k = 0 or k - 4 = 0$
$\Rightarrow k = 0 or k = 4$
But, k ≠ 0 (Given)
Hence, the required value of k is 4.
View full question & answer→Question 203 Marks
Solve the following quadratic equation:$\sqrt2\text{x}^2+7\text{x}+5\sqrt2=0$
Answer$\sqrt2\text{x}^2+7\text{x}+5\sqrt2=0$$\Rightarrow\sqrt2\text{x}^2+5\text{x}+2\text{x}+50\sqrt2=0$
$\Rightarrow\text{x}\big(\sqrt2\text{x}+5\big)-\sqrt2\big(\sqrt2\text{x}+5\big)=0$
$\Rightarrow\big(\sqrt2\text{x}+5\big)\big(\text{x}+\sqrt2\big)=0$
$\Rightarrow\sqrt2\text{x}+5=0$ or $\text{x}+\sqrt2=0$
$\Rightarrow\text{x}=\frac{-5}{\sqrt2}$ or $\text{x}=-\sqrt2$
View full question & answer→Question 213 Marks
Solve the following quadratic equation:$\sqrt3\text{x}^2+\text{10x}+7\sqrt3=0$
Answer$\sqrt3\text{x}^2+\text{10x}+7\sqrt3=0$$\Rightarrow\sqrt3\text{x}^2+12\text{x}-2\text{x}-8\sqrt3=0$
$\Rightarrow\sqrt3\text{x}\big(\text{x}+4\sqrt3\big)-2\big(\text{x}+4\sqrt{3}\big)=0$
$\Rightarrow\big(\sqrt{3}\text{x}-2\big)\big(\text{x}+4\sqrt3\big)=0$
$\Rightarrow\sqrt3\text{x}-2=0$ or $\text{x}+4\sqrt3=0$
$\Rightarrow\text{x}=\frac{2}{\sqrt3}$ or $\text{x}=-4\sqrt3$
View full question & answer→Question 223 Marks
Solve the following quadratic equation:$4\sqrt6\text{x}^2-\text{13x}-2\sqrt6=0$
Answer$4\sqrt6\text{x}^2-\text{13x}-2\sqrt6=0$$\Rightarrow4\sqrt6\text{x}^2-16\text{x}+3\text{x}-2\sqrt6=0$
$\Rightarrow4\sqrt2\text{x}\big(\sqrt3\text{x}-2\sqrt2\big)+\sqrt3\big(\sqrt3\text{x}-2\sqrt2\big)=0$
$\Rightarrow\big(\sqrt3\text{x}-2\sqrt2\big)\big(4\sqrt2\text{x}+\sqrt3\big)=0$
$\Rightarrow\Big(\frac{2\sqrt2}{\sqrt3}\times\frac{\sqrt3}{\sqrt3}\Big)=\frac{2\sqrt6}{3}$ or $\Big(\frac{-\sqrt3}{4\sqrt2}\times\frac{\sqrt2}{\sqrt2}\Big)=\frac{-\sqrt6}{8}$
Hence, $\frac{2\sqrt6}{3}$ and $\frac{-\sqrt6}{8}$ are the roots of the given equation.
View full question & answer→Question 233 Marks
If -5 is a root of the quadratic equation $2x^2 + px - 15 = 0$ and the quadratic equation $p(x^2 + x) + k = 0$ has equal roots, find the value of k.
AnswerIt is given that -5 is a root of the quadratic equation $2x^2 + px - 15 = 0.$
$\therefore$ $2(-5)^2 + p \times (-5) - 15 = 0$
$\Rightarrow -5p + 35 = 0$
$\Rightarrow p = 7$
The roots of the equation $px^2 + px + k = 0$ are equal.
$\therefore$ $D = 0$
$\Rightarrow p^2 - 4pk = 0$
$\Rightarrow (7)^2 - 4 \times 7 \times k = 0$
$\Rightarrow 49 - 28k = 0$
$\Rightarrow\text{k}=\frac{49}{28}=\frac{7}{4}$
Thus, the value of k is $\frac{7}{4}.$
View full question & answer→Question 243 Marks
Solve the following quadratic equation:$\text{x}^2-3\sqrt5\text{x}+10=0$
Answer$\text{x}^2-3\sqrt5\text{x}+10=0$$\Rightarrow\text{x}^2-\sqrt5\text{x}-2\sqrt{5}\text{x}+10=0$
$\Rightarrow\text{x}\big(\text{x}-\sqrt5\big)-2\sqrt5\big(\text{x}-\sqrt5\big)=0$
$\Rightarrow\big(\text{x}-\sqrt5\big)\big(\text{x}-2\sqrt5\big)=0$
$\Rightarrow\text{x}-\sqrt5=0$ or $\text{x}-2\sqrt5=0$
$\Rightarrow\text{x}=\sqrt{5}$ or $\text{x}=2\sqrt5$
View full question & answer→Question 253 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$\sqrt3\text{x}^2+10\text{x}-8\sqrt3=0$
AnswerGiven: $\sqrt3\text{x}^2+10\text{x}-8\sqrt3=0$
On comparing it with $ax^2 + bx + c = 0$, we get:
$\text{x}=\sqrt3,\ \text{b}=10$ and $\text{c}=-8\sqrt3$
Discriminant D is given by:
$D = (b^2 - 4ac)$
$=(10)^2-4\times\sqrt3\times\big(8\sqrt3\big)$
$=100+96$
$=196>0$
Hence, the roots of the equation are real.
Roots $\alpha$ and $\beta$ are given by:
$\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{\text{2a}}$
$=\frac{-10+\sqrt{196}}{2\sqrt3}$
$=\frac{-10+14}{2\sqrt3}$
$=\frac{4}{2\sqrt3}$
$=\frac{2}{\sqrt3}$
$=\frac{2}{\sqrt3}\times\frac{\sqrt3}{\sqrt3}$
$=\frac{2\sqrt3}{3}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{\text{2a}}$
$=\frac{-10-\sqrt{196}}{2\sqrt3}$
$=\frac{-10-14}{2\sqrt3}$
$=\frac{-24}{2\sqrt3}$
$=\frac{-12}{\sqrt3}$
$=\frac{-12}{\sqrt3}\times\frac{\sqrt3}{\sqrt3}$
$=\frac{-12\sqrt3}{3}$
$=-4\sqrt3$
Thus, the roots of the equation are $\frac{2\sqrt3}{3}$ and $-4\sqrt3.$
View full question & answer→Question 263 Marks
Solve the following quadratic equation:$5x^2 + 13x + 8 = 0$
Answer$5x^2 + 13x + 8 = 0$
$\Rightarrow 5x^2 + 5x + 8x + 8 = 0$
$\Rightarrow 5x(x + 1) + 8(x + 1) = 0$
$\Rightarrow (x + 1)(5x + 8) = 0$
$\Rightarrow x + 1 = 0 or 5x + 8 = 0$
$\Rightarrow x = -1$ or $\text{x}=\frac{-8}{5}$
View full question & answer→Question 273 Marks
If the equation $(1 + m^2)x^2 + 2mcx + (c^2 - a^2) = 0$ has equal roots, prove that $c^2 = a^2(1 + m^2).$
AnswerGiven:
$(1 + m^2)x^2 + 2mcx + (c^2- a^2) = 0$
Here,
$a = (1 + m^2), b = 2mc$ and $c = (c^2 - a^2)$
It is given that the roots of the equation are equal; therefore, we have:
$D = 0$
$\Rightarrow (b^2 - 4ac) = 0$
$\Rightarrow (2mc)^2 - 4 \times (1 + m^2) \times (c^2 - a^2) = 0$
$\Rightarrow 4m^2c^2 - 4(c^2 - a^2 + m^2c^2- m^2a^2) = 0$
$\Rightarrow 4m^2c^2 - 4c^2 + 4a^2 - 4m^2c^2 + 4m^2a^2 = 0$
$\Rightarrow -4c^2 + 4a^2 + 4m^2a^2 = 0$
$\Rightarrow a^2 + m^2a^2 = c^2$
$\Rightarrow a^2(1 + m^2) = c^2$
$\Rightarrow c^2 = a^2(1 + m^2)$
Hence proved.
View full question & answer→Question 283 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$x^2 - 6x + 4 = 0$
AnswerGiven, $x^2 - 6x + 4 = 0$ On comparing it with $ax^2 + bx + c = 0$, we get
Discriminant D is given by: $a = 1, b = -6$ and $c = 4 D = (b^2 - 4ac) = (-6)^2 - 4 \times 1 \times 4 = 36 - 16 = 20 > 0$
Hence, the roots of the equation are real. Roots $\alpha$ and $\beta$ are
given by, $\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-6)+\sqrt{20}}{2\times1}$
$=\frac{6+2\sqrt5}{2}$
$=\frac{2\big(3+\sqrt5\big)}{2}$
$=\big(3+\sqrt5\big)$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-6)-\sqrt{20}}{2\times1}$
$=\frac{6-2\sqrt5}{2}$
$=\frac{2\big(3-\sqrt5\big)}{2}$
$=\big(3-\sqrt5\big)$
Thus, the roots of the equation are $\big(3+2\sqrt5\big)$ and $\big(3-2\sqrt5\big).$
View full question & answer→Question 293 Marks
Solve the following equations by using the method of completing the square:
$3x^2 - x - 2 = 0$
Answer$3x^2 - x - 2 = 0$
$\Rightarrow 9x^2 - 3x - 6 = 0$ (Multiplying both sides by 3)
$\Rightarrow 9x^2 - 3x = 6$
$\Rightarrow(\text{3x})^2-2\times\text{3x}\times\frac{1}{2}+\Big(\frac{1}{2}\Big)^2\\=6+\Big(\frac{1}{2}\Big)^2$ $\Big[$Adding $\Big(\frac{1}2{}\Big)^2$ on both sides$\Big]$
$\Rightarrow\Big(\text{3x}-\frac{1}{2}\Big)^2$
$=6+\frac{1}{4}$
$=\frac{25}{4}=\Big(\frac{5}{2}\Big)^2$
$\Rightarrow\text{3x}-\frac{1}{2}=\pm\frac{5}{2}$ (Taking square root on both sides)
$\Rightarrow\text{3x}-\frac{1}{2}=\frac{5}{2}$ or $\text{3x}-\frac{1}{2}=-\frac{5}{2}$
$\Rightarrow\text{3x}=\frac{5}{2}+\frac{1}{2}=\frac{6}{2}=3$ or $\text{3x}=-\frac{5}{2}+\frac{1}{2}=-\frac{4}{2}=-2$
⇒ x = 1 or $\text{x}=-\frac{2}3{}$
Hence 1 and $-\frac{2}{3}$ are the roots of the given equation.
View full question & answer→Question 303 Marks
For what values of k are the roots of the quadratic equation $3x^2 + 2kx +27 = 0$ real and equal?
AnswerGiven:
$3x^2 + 2kx + 27 = 0$
Here,
$a = 3, b = 2k$ and $c = 27$
It is given that the roots of the equation are real and equal; therefore, we have:
$D = 0$
$\Rightarrow (2k)^2 - 4 \times 3 \times 27 = 0$
$\Rightarrow 4k^2 - 324 = 0$
$\Rightarrow 4k^2 = 324$
$\Rightarrow k^2 = 81$
$\Rightarrow\text{k}=\pm9$
$\therefore$ k = 9 or k = -9
View full question & answer→Question 313 Marks
Find the value of k for which x = 1 is a root of the equation $x^2 + kx + 3 = 0$ Also, find the other root.
AnswerSince x = 1 is a solution of $x^2 + kx + 3 = 0$, it must be satisfy the equation.
$(1)^2 + k(1) + 3 = 0$
$\Rightarrow k = -4$
Hence the required value of $k = -4$
Since $\text{x}=\frac{3}{4}$ is a root of $ax^2 + bx - 6 = 0$, we have $\text{a}\times\Big(\frac{3}{4}\Big)^2+\text{b}\times\Big(\frac{3}{4}\Big)-6=0$
$\Rightarrow\frac{\text{9a}}{16}+\frac{\text{3b}}{4}-6=0$
$\Rightarrow 9a + 12b = 96$
$\Rightarrow 3a + 4b = 32 ...(1)$
Again x = -2 being a root of $ax^2 + bx - 6 = 0$, we have
$a \times (-2) + b(-2) - 6 = 0$
$\Rightarrow 4a - 2b = 6$
$\Rightarrow 2a - b = 3 ...(2)$
View full question & answer→Question 323 Marks
Solve the following equations by using the method of completing the square:
$4x^2 + 4bx - (a^2 - b^2) = 0$
Answer$4x^2 + 4bx - (a^2 - b^2) = 0$
$\Rightarrow 4x^2 + 4bx = a^2 - b^2$
$\Rightarrow (2x)^2 + 2 \times 2x \times b + b^2 = a^2- b^2 + b^2$ [Adding $b^2$ on both sides]
$\Rightarrow (2x + b)^2 = a^2$
$\Rightarrow\text{2x}+\text{b}=\pm\text{a}$ (Taking square root on both sides)
$\Rightarrow 2x + b = a or 2x + b = -a$
$\Rightarrow 2x = a - b or 2x = -a - b$
$\Rightarrow\text{x}=\frac{\text{a}-\text{b}}{2}$ or $\text{x}=-\frac{\text{a}+\text{b}}{2}$
Hence, $\frac{\text{a}-\text{b}}{2}$ and $-\frac{\text{a}+\text{b}}{2}$ are the roots of the given equation.
View full question & answer→Question 333 Marks
Solve the following equations by using the method of completing the square:
$x^2 - 6x + 3 = 0$
Answer$x^2 - 6x + 3 = 0$
$\Rightarrow x^2 - 6x = -3$
$\Rightarrow x^2 - 2 \times x \times 3 + 3^2 = -3 + 3^2$ (Adding $3^2$ on both sides)
$\Rightarrow (x - 3)^2= -3 + 9 = 6$
$\Rightarrow\text{x}-3=\pm\sqrt6$ (Taking square root on both sides)
$\Rightarrow\text{x}-3=\sqrt6$ or $\text{x}-3=-\sqrt6$
$\Rightarrow\text{x}=3+\sqrt6$ or $\text{x}=3-\sqrt6$
Hence, $3+\sqrt6$ and $3-\sqrt6$ are the roots of the given equation.
View full question & answer→Question 343 Marks
Solve the following quadratic equation:$4x^2 + 4bx - (a^2 - b^2) = 0$
Answer$4x^2 + 4bx - (a^2 - b^2) = 0\Rightarrow 4x^2 + 4bx + (b^2 - a^2) = 0$
$\Rightarrow 4x^2 + 2(b + a)x + 2(b - a)x + (b^2 - a^2) = 0$
$\Rightarrow 2x[2x + (b + a)] + (b - a)[2x + (b + a)] = 0$
$\Rightarrow [2x + (b + a)][2x + (b - a)] = 0$
$\Rightarrow 2x + (b - a) = 0 or 2x + 9b - a = 0$
$\Rightarrow\text{x}=\frac{-(\text{b}+\text{a})}{2}$ or $\text{x}=\frac{-(\text{b}-\text{a})}{2}$
$\Rightarrow\text{x}=\frac{-(\text{a}+\text{b})}{2}$ or $\text{x}=\frac{(\text{a}-\text{b})}{2}$
View full question & answer→Question 353 Marks
The following are quadratic equations in x?
$\text{x}^2-\frac{1}{\text{x}^2}=5$
Answer$\text{x}^2-\frac{1}{\text{x}^2}=5$
$\Rightarrow x^4- 1 = 5x^2$
$\Rightarrow x^4 - 5x^2 - 1 = 0$
And $(x^4 - 5x^2 - 1)$ Being a polynomial of degree 4
$\therefore\ \text{x}^2-\frac{1}{\text{x}^2}=5$ is not a quadratic equation.
View full question & answer→Question 363 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:$3\text{x}^2-2\sqrt6\text{x}+2=0$
AnswerThe given equation is:$3\text{x}^2-2\sqrt6\text{x}+2=0$
Comparing it with $ax^2+ bx + c = 0$, we get
$\text{a}=3,\ \text{b}=-2\sqrt6$ and $\text{c}=2$
$\therefore$ Discriminant, $\text{D}=\text{b}^2-4\text{ac}$
$=\big(-2\sqrt6\big)^2-4\times3\times2$
$=24-24=0$
So, the given equation has real roots.
Now, $\sqrt{\text{D}}=0$
$\therefore\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-\big(2\sqrt6\big)+0}{2\times3}$
$=\frac{2\sqrt6}{6}$
$=\frac{\sqrt6}{3}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-\big(2\sqrt6\big)-0}{2\times3}$
$=\frac{2\sqrt6}{6}$
$=\frac{\sqrt6}{3}$
Hence, $\frac{\sqrt6}{3}$ is the repeated roots of the given equation.
View full question & answer→Question 373 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:$\frac{1}{\text{x}}-\frac{1}{\text{x}-2}=3,\ \text{x}\neq0,\ 2$
AnswerThe given equation is:$\frac{1}{\text{x}}-\frac{1}{\text{x}-2}=3,\ \text{x}\neq0,\ 2$
$\Rightarrow\frac{\text{x}-2-\text{x}}{\text{x}(\text{x}-2)}=3$
$\Rightarrow\frac{-2}{\text{x}^2-2\text{x}}=3$
$\Rightarrow-2=3\text{x}^2-6\text{x}$
$\Rightarrow3\text{x}^2-6\text{x}+2=0$
This equation is of the form $ax^2 + bx + c = 0$, where
$a = 3, b = -6$ and $c = 2.$
$\therefore$ Discriminant, $D = b^2 - 4ac$
$= (-6)^2 - 4 \times 3 \times 2$
$= 36 - 24$
$= 12 > 0$
so, the given equation has real roots.
Now,$\sqrt{\text{D}}=\sqrt{12}=2\sqrt3$
$\therefore\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-6)+2\sqrt3}{2\times3}$
$=\frac{6+2\sqrt3}{6}$
$=\frac{3+\sqrt3}{3}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-6)-2\sqrt3}{2\times3}$
$=\frac{6-2\sqrt3}{6}$
$=\frac{3-\sqrt3}{3}$
Hence, $\frac{3+\sqrt3}{3}$ and $\frac{3-\sqrt3}{3}$ are the roots of the given equation.
View full question & answer→Question 383 Marks
Solve the following quadratic equation$:x^2 - 2ax - (4b^2 - a^2) = 0$
Answer$x^2 - 2ax - (4b^2 - a^2) = 0$
$\Rightarrow x^2 - 2ax + (a^2 - 4b^2) = 0$
$\Rightarrow x^2 - 2ax + (a - 2b)(a + 2b) = 0$
$\Rightarrow x^2 - (a - 2b)x - (a + 2b)x + (a - 2b)(a + 2b) = 0$
$\Rightarrow x[x + (a - 2b)] - (a + 2b)[x - (a - 2b)] = 0$
$\Rightarrow [x - (a - 2b)][x - (a + 2b)] = 0$
$\Rightarrow x - (a - 2b) = 0 or x - (a + 2b) = 0$
$\Rightarrow x = a - 2b or x = a + 2b$
View full question & answer→Question 393 Marks
Solve the following quadratic equation:$4x^2 - 9x = 100$
Answer$4x^2 - 9x = 100$
$\Rightarrow 4x^2 - 9x - 100 = 0$
$\Rightarrow 4x^2 - 25x + 16x - 100 = 0$
$\Rightarrow x(4x - 25) + 4(4x - 25) = 0$
$\Rightarrow (4x - 25)(x + 4) = 0$
$\Rightarrow 4x - 25 = 0 or x + 4 = 0$
$\Rightarrow\text{x}=\frac{25}{4}$ or x = -4
Hence, $\frac{25}{4}$ and -4 are the roots of the equation $4x^2 - 9x = 100$
View full question & answer→Question 403 Marks
Find the values k for which of roots of $9x^2 + 8kx + 16 = 0$ are real and equal
AnswerGiven:
$9x^2 + 8kx + 16 = 0$
Here,
a = 9, b = 8k and c = 16
It is given that the roots of the equation are real and equal; therefore, we have:
$D = 0$
$\Rightarrow (b^2 - 4ac) = 0$
$\Rightarrow (8k)^2 - 4 \times 9 \times 16 = 0$
$\Rightarrow 64k^2 - 576 = 0$
$\Rightarrow 64k^2 = 576$
$\Rightarrow k^2 = 9$
$\Rightarrow k = ±3$
$\therefore$ k = 3 or k = -3
View full question & answer→Question 413 Marks
Solve the following quadratic equation:$9x^2 + 6x + 1 = 0$
Answer$9x^2 + 6x + 1 = 0$
$\Rightarrow 9x^2 + 3x + 3x + 1 = 0$
$\Rightarrow 3x(3x + 1) + 1(3x + 1) = 0$
$\Rightarrow (3x + 1)(3x + 1) = 0$
$\Rightarrow (3x + 1)^2 = 0$
$\Rightarrow 3x + 1 = 0$
$\Rightarrow\text{x}=\frac{-1}{3}$ (repeated root)
View full question & answer→Question 423 Marks
Solve the following quadratic equation:$x^2 + 6x - (a^2 + 2a - 8) = 0$
Answer$x^2 + 6x - (a^2 + 2a - 8) = 0$
$\Rightarrow x^2 + 6x - (a^2 + 4a - 2a - 8) = 0$
$\Rightarrow x^2 + 6x - [a(a + 4) - 2(a + 4)] = 0$
$\Rightarrow x^2+ 6x - (a + 4)(a - 2) = 0$
$\Rightarrow x^2 + (a + 4)x - (a - 2)x - (a + 4)(a - 2) = 0$
$\Rightarrow x[x + (a + 4)] - (a - 2)[x + (a + 4)] = 0$
$\Rightarrow [x + (a + 4)][x - (a - 2)] = 0$
$\Rightarrow x + (a + 4) = 0 or x - (a - 2) = 0$
$\Rightarrow x = -(a + 4) or x = (a - 2)$
View full question & answer→Question 433 Marks
The following are quadratic equations in x?
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=2\Big(\text{x}+\frac{1}{\text{x}}\Big)+3$
Answer$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=2\Big(\text{x}+\frac{1}{\text{x}}\Big)+3$
$\Rightarrow\text{x}^2+2+\frac{1}{\text{x}^2}=\text{2x}+\frac{2}{\text{x}}+3$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}-\text{2x}-\frac{2}{\text{x}}-1=0$
$\Rightarrow\text{x}^4+1-\text{2x}^3-\text{2x}-\text{x}^2=0$
$\Rightarrow\text{x}^4-\text{2x}^3-\text{x}^2-\text{2x}+1=0$
Clearly, $x^4 - 2x^3 - x^2- 2x + 1$ is a polynomial of degree 4
This is not of the form $ax^2 + bx + c = 0$
Hence, the given equation is not a quadratic equation.
View full question & answer→Question 443 Marks
Solve the following quadratic equation:$\text{x}^2-\big(1+\sqrt2\big)\text{x}+\sqrt2=0$
Answer$\text{x}^2-\big(1+\sqrt2\big)\text{x}+\sqrt2=0$
$\Rightarrow\text{x}^2-1.\text{x}-\sqrt2\text{x}+\sqrt2=0$
$\Rightarrow\text{x}(\text{x}-1)-\sqrt2(\text{x}-1)=0$
$\Rightarrow(\text{x}-1)\big(\text{x}-\sqrt2\big)=0$
$\Rightarrow(\text{x}-1)=0$ or $\text{x}-\sqrt2=0$
$\Rightarrow\text{x}=1$ or $\text{x}=\sqrt2$
Hence, 1 and $\sqrt2$ are the roots of the given equation.
View full question & answer→Question 453 Marks
Solve the following quadratic equation:$\text{x}^2+2\sqrt{2}\text{x}-6=0$
Answer$\text{x}^2+2\sqrt{2}\text{x}-6=0$$\Rightarrow\text{x}^2+3\sqrt{2}\text{x}-\sqrt{2}\text{x}-6=0$
$\Rightarrow\text{x}\big(\text{x}+3\sqrt{2}\big)-\sqrt{2}\big(\text{x}+3\sqrt{2}\big)$
$\Rightarrow\big(\text{x}+3\sqrt{2}\big)\big(\text{x}-\sqrt2\big)=0$
$\Rightarrow\text{x}=-3\sqrt2$ or $\text{x}=\sqrt2$
View full question & answer→Question 463 Marks
Solve the following quadratic equation:$\sqrt3\text{x}^2-2\sqrt2\text{x}-2\sqrt3=0$
Answer$\sqrt3\text{x}^2-2\sqrt2\text{x}-2\sqrt3=0$$\Rightarrow\sqrt{3}\text{x}^2-3\sqrt2\text{x}-\sqrt{2}\text{x}-2\sqrt{3}=0$
$\Rightarrow\sqrt3\text{x}\big(\text{x}-\sqrt6\big)-\sqrt2\big(\text{x}-\sqrt6\big)=0$
$\Rightarrow\big(\text{x}-\sqrt6\big)\big(\sqrt3\text{x}+\sqrt2\big)=0$
$\Rightarrow\text{x}-\sqrt6=0$ or $\sqrt{3}\text{x}+\sqrt2=0$
$\Rightarrow\text{x}=\sqrt{6}$ or $\text{x}=\frac{-\sqrt{2}}{\sqrt{3}}$
View full question & answer→Question 473 Marks
Find the values of k for which the quadratic equation $(3k + 1)x^2 + 2(k + 1)x + 1 = 0$ has real and equal roots.
AnswerThe given equation is $(3k + 1)x^2 + 2(k + 1)x + 1 = 0.$
This is of the form $ax^2 + bx + c = 0$, where a = 3k +1, b = 2(k + 1) and c = 1.
$\therefore$ $D = b^2 - 4ac$
$D = [2(k + 1)]^2 - 4 \times (3k + 1) \times 1$
$D = (k^2 + 2k + 1) - 4(3k + 1)$
$D = 4k^2 + 8k + 4 - 12k - 4$
$D = 4k^2- 4k$
The given equation will have real and equal roots if D = 0.
$\therefore$ $4k^2 - 4k = 0$
$\Rightarrow 4k(k - 1) = 0$
$\Rightarrow k = 0 or k - 1 = 0$
Hence, 0 and 1 are the required values of k.
View full question & answer→Question 483 Marks
Solve the following quadratic equation:$\frac{2}{\text{x}^2}-\frac{5}{\text{x}}+2=0$
Answer$\frac{2}{\text{x}^2}-\frac{5}{\text{x}}+2=0$ Multiplying by $x^22 - 5x + 2x^2 = 0 or 2x^2 - 5x + 2 = 0$
$\Rightarrow 2x^2 - 4x - x + 2 = 0$
$\Rightarrow 2x(x - 2) - 1(x - 2) = 0$
$\Rightarrow (x - 2)(2x - 1) = 0$
$\therefore$ x - 2 = 0 or 2x - 1 = 0
⇒ x = 2, $\text{x}=\frac{1}{2}$
Hence, 2 and $\frac{1}{2}$ are the roots of the given equation.
View full question & answer→Question 493 Marks
If 3 is a root of the quadratic equation $x^2-x+k=0$, find the value of $p$ so that the roots of the equation $x^2+k(2 x+k+2)+p=0$ are equal.
AnswerIt is given that 3 is a root of the quadratic equation $x^2 - x + k = 0.$
$\therefore$ $(3)^2 - 3 + k = 0$
$\Rightarrow k + 6 = 0$
$\Rightarrow k = -6$
The roots of the equation $x^2 + 2kx + (k^2 + 2k + p) = 0$ are equal.
$\therefore$$ D = 0$
$\Rightarrow (2k)^2 - 4 \times 1 \times (k^2 + 2k + p) = 0$
$\Rightarrow 4k2 - 4k2 - 8k - 4p = 0$
$\Rightarrow -8k - 4p = 0$
$\Rightarrow\text{p}=\frac{\text{8k}}{-4}=-\text{2k}$
⇒ p = -2 × (-6) = 12
Hence, the value of p is 12.
View full question & answer→Question 503 Marks
Solve the following equations by using the method of completing the square:
$\frac{2}{\text{x}^2}-\frac{5}{\text{x}}+2=0$
Answer$\frac{2}{\text{x}^2}-\frac{5}{\text{x}}+2=0$
$\Rightarrow\frac{2-\text{5x}+\text{2x}^2}{\text{x}^2}=0$
$\Rightarrow 2x^2 - 5x + 2 = 0$
$\Rightarrow 4x^2 - 10x + 4 = 0$ (Multiplying both sides by 2)
$\Rightarrow 4x^2 - 10x = -4$
$\Rightarrow(\text{2x})^2-2\times\text{2x}\times\frac{5}{2}+\Big(\frac{5}{2}\Big)^2\\=-4+\Big(\frac{5}{2}\Big)^2$ $\Big[$Adding $\Big(\frac{5}2{}\Big)^2$ on both sides$\Big]$
$\Rightarrow\Big(\text{2x}-\frac{5}{2}\Big)^2$
$=-4+\frac{25}{4}$
$=\frac{9}{4}=\Big(\frac{3}{2}\Big)^2$
$\Rightarrow\text{2x}-\frac{5}{2}=\pm\frac{3}{2}$ (Taking square root on both sides)
$\Rightarrow\text{2x}-\frac{5}{2}=\frac{3}{2}$ or $\text{2x}-\frac{5}{2}=-\frac{3}{2}$
$\Rightarrow\text{2x}=\frac{3}{2}+\frac{5}{2}=\frac{8}{2}=4$ or $\text{2x}=-\frac{3}{2}+\frac{5}{2}=\frac{2}{2}=1$
$\Rightarrow x = 2$ or $\text{x}=\frac{1}{2}$
Hence, 2 and $\frac{1}{2}$ are the roots of the given equation.
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