Question

(2p-7q)^2 +56pq=(2p+7q)^2 verify​

Answer 1

$\bf\underline\bold{ If \: the \: equations \: has \: equal \: roots \: discriminat }$

$\sf{ D = 0 }$

${b}^{2} - 4ac = 0 \\ (7p + 2)^{2} - 4(2p + 1)(7p - 3) = 0 \\ 43^{2} + 428p - 4(149^{2} - 6p - 24p - 12 + 28p = 0 \\ 49^{2} + 4 + 28p - 56p^{2} - 4p + 12 = 0 \\ - 7p^{2} + 24p + 16 = 0 \\ 7p^{2} - 24p - 16 = 0 = > p = 4, \frac{ - 4}{7}$

$\bf{ No \: roots \: A }$

I) P = 4. II) P= -4/7

$\bf{ Equation \: 9x^{2} -30x+25=0 }$

${ -1/7x^{2}+2x-7=0}$

${ 3x-5=0}$

${ x^{2}-14x+49=0}$

${ => x= \frac{ 5}{3}}$

${ => x= 7 }$

$\sf\large\underline\bold \purple{ Mark \: me \: as \: brainlist }$