16. If $\,x\,$ is real, find the signs of each of each of the following expressions:
$\,(i)\,3x^2-2x+1 \,\,\, (ii)\,\,3x-2x^2-2 \\ (iii)\,5x^2-14x+10 \,\,(iv)\, 10x-3x^2-9$
Sol. (i) Let $\,\,y=3x^2-2x+1 \\ \Rightarrow 3x^2-2x+(1-y)=0 \cdots(1)$
Since $\,x\,$ is real, discriminant of (1) : $\,D \geq 0 \\ \Rightarrow (-2)^2-4 \times 3 \times (1-y)\geq 0 \\ \Rightarrow 4[1-3(1-y)]\geq 0 \\ \Rightarrow 1-3(1-y) \geq 0 \\ \Rightarrow -2 +3y \geq 0 \\ \Rightarrow y \geq \frac 23 \cdots(2)$
Hence, from (2), we can conclude that given expression is positive for any value of $\,\,x.$
Sol. (ii) Let $\,y=3x-2x^2-2 \\ \Rightarrow -2x^2+3x-(2+y)=0 \\ \Rightarrow 2x^2-3x+(2+y)=0 \cdots(1)$
Since $\,x\,$ is real, discriminant of (1) : $\,D \geq 0 \\ \Rightarrow (-3)^2-4 \times 2 \times (2+y) \geq 0 \\ \Rightarrow 9-8(2+y) \geq 0 \\ \Rightarrow -7-8y \geq 0 \\ \Rightarrow 8y \leq -7 \\ \Rightarrow y \leq -\frac 78 \cdots(2)$
Hence, from (2), we can conclude that given expression is negative for any value of $\,\,x.$
Sol. (iii) Let $\,y=5x^2-14x+10 \\ \Rightarrow 5x^2-14x+(10-y)=0 \cdots(1)$
Since $\,x\,$ is real, discriminant of (1) : $\,D \geq 0 \\ \Rightarrow (-14)^2-4 \times 5 \times (10-y) \geq 0 \\ \Rightarrow 4[49-5(10-y)] \geq 0 \\ \Rightarrow 49-50+5y \geq 0 \\ \Rightarrow -1+5y \geq 0 \\ \Rightarrow y \geq \frac 15 \cdots(2)$
Hence, from (2), we can conclude that given expression is positive for any value of $\,\,x.$
Sol. (iv) Let $\,\,y=10x-3x^2-9 \\ \Rightarrow -3x^2+10x-(9+y)=0 \\ \Rightarrow 3x^2-10x+(9+y)=0 \cdots(1)$
Since $\,x\,$ is real, discriminant of (1) : $\,D \geq 0 \\ \Rightarrow (-10)^2- 4 \times 3 \times (9+y) \geq 0 \\ \Rightarrow 4[25-3(9+y)] \geq 0 \\ \Rightarrow 25-27-3y \geq 0 \\ \Rightarrow -2-3y \geq 0 \\ \Rightarrow 3y \leq -2 \\ \Rightarrow y \leq -\frac 23 \cdots(2)$
Hence, from (2), we can conclude that given expression is negative for any value of $\,\,x.$
17. If $\,x\,$ be real, find the maximum values of each of the following expressions and corresponding values of $\,x :$
$\,(i)\, 1-2x-x^2\,\,\,\,\,\,(ii)\, 3-20x-25x^2 \\ \,(iii)\,3+2x-x^2$
Sol. (i) Let $\,y=1-2x-x^2=-x^2-2x+1 \cdots(1)\,\,$
We know for a quadratic expression $\,y=ax^2+bx+c \cdots(2),\,\, y\,$ assumes its maximum value if $\,a<0\,$ and $\,y\,$ assumes its minimum value if $\,a<0\,$. For $\,a>0,\,\,$ the value of $\,x\,$ will be $\, -\frac{b}{2a}\,$ for which min. value have been attained and min. value or least value is given by $\,\,\frac{4ac-b^2}{4a} \cdots(3)$
Comparing (1) and (2), we get $\,\,a=-1(<0),\,b=-2,c=1.$
Hence, using the fact given in (3), we get the least value of $\,y=\frac{4 \times (-1) \times 1-(-2)^2}{4 \times (-1)}=2\,\,$ and corresponding value of $\,\,x=-\left(\frac{-2}{2 \times (-1)}\right)=-1.$
Sol. (ii) Let $\,y= 3-20x-25x^2=-25x^2-20x+3 \cdots(1)\,\,$
We know for a quadratic expression $\,y=ax^2+bx+c \cdots(2),\,\, y\,$ assumes its maximum value if $\,a<0\,$ and $\,y\,$ assumes its minimum value if $\,a<0\,$. For $\,a>0,\,\,$ the value of $\,x\,$ will be $\, -\frac{b}{2a}\,$ for which min. value have been attained and min. value or least value is given by $\,\,\frac{4ac-b^2}{4a} \cdots(3)$
Comparing (1) and (2), we get $\,\,a=-25(<0),\,b=-20,c=3.$
Hence, using the fact given in (3), we get the least value of $\,y=\frac{4 \times (-25) \times 3-(-20)^2}{4 \times (-25)}=7\,\,$ and corresponding value of $\,\,x=-\left(\frac{-20}{2 \times (-25)}\right)=-\frac 25.$
Sol. (iii) Let $\,y= 3+2x-x^2=-x^2+2x+3 \cdots(1)\,\,$
We know for a quadratic expression $\,y=ax^2+bx+c \cdots(2),\,\, y\,$ assumes its maximum value if $\,a<0\,$ and $\,y\,$ assumes its minimum value if $\,a<0\,$. For $\,a>0,\,\,$ the value of $\,x\,$ will be $\, -\frac{b}{2a}\,$ for which min. value have been attained and min. value or least value is given by $\,\,\frac{4ac-b^2}{4a} \cdots(3)$
Comparing (1) and (2), we get $\,\,a=-1(<0),\,b=2,c=3.$
Hence, using the fact given in (3), we get the least value of $\,y=\frac{4 \times (-1) \times 3-(2)^2}{4 \times (-1)}=4\,\,$ and corresponding value of $\,\,x=-\left(\frac{2}{2 \times (-1)}\right)=1.$
$\,(i)\, 4x^2-4x+1\,\, (ii)\,3x^2-6x+8 \\ (iii)\,3x^2+6x+7$
Sol. (i) Let $\,y=4x^2-4x+1 \cdots(1)\,\,$
We know for a quadratic expression $\,y=ax^2+bx+c \cdots(2),\,\, y\,$ assumes its maximum value if $\,a<0\,$ and $\,y\,$ assumes its minimum value if $\,a<0\,$. For $\,a>0,\,\,$ the value of $\,x\,$ will be $\, -\frac{b}{2a}\,$ for which min. value have been attained and min. value or least value is given by $\,\,\frac{4ac-b^2}{4a} \cdots(3)$
Comparing (1) and (2), we get $\,\,a=4(>0),\,b=-4,c=1.$
Hence, using the fact given in (3), we get the least value of $\,y=\frac{4 \times 4 \times 1-(-4)^2}{4 \times 4}=0\,\,$ and corresponding value of $\,\,x=-\frac{-4}{2 \times 4}=\frac 12.$
Sol. (ii) Let $\,y=3x^2-6x+8 \cdots(1)\,\,$
We know for a quadratic expression $\,y=ax^2+bx+c \cdots(2),\,\, y\,$ assumes its maximum value if $\,a<0\,$ and $\,y\,$ assumes its minimum value if $\,a<0\,$. For $\,a>0,\,\,$ the value of $\,x\,$ will be $\, -\frac{b}{2a}\,$ for which min. value have been attained and min. value or least value is given by $\,\,\frac{4ac-b^2}{4a} \cdots(3)$
Comparing (1) and (2), we get $\,\,a=3(>0),\,b=-6,c=8.$
Hence, using the fact given in (3), we get the least value of $\,y=\frac{4 \times 3 \times 8-(-6)^2}{4 \times 3}=5\,\,$ and corresponding value of $\,\,x=-\frac{-6}{2 \times 3}=1.$
Sol. (iii) Let $\,y=3x^2+6x+7 \cdots(1)\,\,$
We know for a quadratic expression $\,y=ax^2+bx+c \cdots(2),\,\, y\,$ assumes its maximum value if $\,a<0\,$ and $\,y\,$ assumes its minimum value if $\,a<0\,$. For $\,a>0,\,\,$ the value of $\,x\,$ will be $\, -\frac{b}{2a}\,$ for which min. value have been attained and min. value or least value is given by $\,\,\frac{4ac-b^2}{4a} \cdots(3)$
Comparing (1) and (2), we get $\,\,a=3(>0),\,b=6,c=7.$
Hence, using the fact given in (3), we get the least value of $\,y=\frac{4 \times 3 \times 7-(6)^2}{4 \times 3}=4\,\,$ and corresponding value of $\,\,x=-(\frac{6}{2 \times 3})=-1.$
19. For what values of $\,x\,$ the expression $\,\,x^2-2x+3\,$ is negative ?
Sol. Let $\,\,y=x^2-2x+3\cdots(1)$
We know, for any real values of $\,x,\,$ the discriminant of $\,\,ax^2+bx+c\,\,, D=b^2-4ac \geq 0,\,\,$ but by (1), we notice that the discriminant of $\,y=0\,$ is : $\,=(-2)^2-4 \times 1 \times 3\\=4-12\\=-8<0 \cdots(2)$
Hence, by (2) we can conclude that the expression $\,\,x^2-2x+3\,$ can not be negative for any real values of $\,x.$
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