Question

The minimum value of ax2 + bx + c is 7/8 at x = 5/4. Find the value of the expression at x = 5, if the value of the expression at x = 1 is 1.

Answer 1

Answer:

29

Step-by-step explanation:

$ax^2+bx+c$

$=a[x^2+\frac{b}{a}x+\frac{c}{a}]$

$=a[x^2+2\frac{b}{2a}x+\frac{c}{a}+(\frac{b}{2a})^2-(\frac{b}{2a})^2]$

$=a[(x+\frac{b}{2a})^2+\frac{c}{a}-(\frac{b}{2a})^2]$

Minimum value occurs when

$x=-\frac{b}{2a}$

$\implies -\frac{b}{2a}=\frac{5}{4}$

$\implies \frac{b}{a}=-\frac{5}{2}$

And the minimum value is $a[(\frac{c}{a}-(\frac{b}{2a})^2]$

$a[(\frac{c}{a}-(-\frac{5}{4})^2]=\frac{7}{8}$

$\implies c-\frac{25a}{16}=\frac{7}{8}$

$\implies 16c-25a=14$  ......................................(1)

Again value of the expression at x = 1 is 1

thus

$a\times1^2+b\times1+c=1$

$\implies a-\frac{5a}{2} +c=1$

$\implies 2a-5a+2c=2$

$\implies -3a+2c=2$  ......................................(2)

Multiplying eq (2) by 8 and subtracting it from eq (1)

-a = -2

or a = 2

Thus, b = -5/2 × 2 = -5

Putting the value of a in eq (2)

-6 + 2c = 2

2c = 8

c = 4

Thus, the expressions is

$2x^2-5x+4$

Value of the expression at x = 5 is

$2\times 5^2-5\times 5+4$

$=50-25+4$

$=25+4$

$=29$

Answer 2

Answer:

Step-by-step explanation: